Approximately 60.38% of 90Sr still exists in Oct. 2001.
Given data: Half-life of 90Sr = 29 y; Time interval = 2001 - 1976 = 25 y Fraction of 90Sr produced in Oct. 1976 that still existed in Oct. 2001 can be calculated as follows:
Number of half-lives = Total time passed / Half-life
Number of half-lives = 25 years / 29 years
Number of half-lives ≈ 0.8621
Since we want to find the fraction that still exists, we can use the formula:
Fraction remaining = (1/2)^(Number of half-lives)
Fraction remaining = (1/2)^(0.8621)
Fraction remaining ≈ 0.6038
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A supertanker can hold 3.00 ✕ 105 m3 of liquid (nearly 300,000 tons of crude oil). (a) How long (in s) would it take to fill the tanker if you could divert a small river flowing at 2600 ft3/s into it? s (b) How long (in s) for the same river at a flood stage flow of 100,000 ft3/s? s
(a)The time required to fill the supertanker when the speed of the river is 2600 [tex]ft^3/s[/tex]. is [tex]3.62 \times 10^{4}[/tex]seconds to fill the using a small river flowing at
(b) The time required to fill the supertanker when the speed of the river is 100,000 [tex]ft^3/s[/tex]. is [tex]1.08 \times 10^5[/tex] seconds.
To determine the time it takes to fill the supertanker, we can use the concept of flow rate, which is the volume of liquid passing through a given point per unit of time. The flow rate can be calculated by dividing the volume by the time.
(a) For the small river flowing at 2600 [tex]ft^3/s[/tex]., we need to convert the volume of the tanker to the same units. 1 [tex]m^{3}[/tex] is approximately equal to 35.3147 [tex]ft^3[/tex]. Therefore, the volume of the tanker is [tex]3.00 \times 10^5 \times 35.3147[/tex] = [tex]1.06 \times 10^7 \ ft^3[/tex]. Dividing the volume by the flow rate, we get the time:
Time = Volume / Flow rate = [tex]\frac{1.06 \times 10^7 }{2600 }[/tex] ≈ [tex]3.62 \times 10^4[/tex]seconds.
(b) For the flood stage flow of 100,000 [tex]ft^3/s[/tex], we can use the same approach. The time to fill the supertanker would be:
Time = Volume / Flow rate = [tex](1.06 \times 10^7) / (100,000 )[/tex] ≈[tex]1.08 \times 10^5[/tex] seconds.
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Electrical Principles [15] 2.1 An electric desk furnace is required to heat 0,54 kg of copper from 23,3°C to a melting point of 1085°C and then convert all the solid copper into the liquid state (melted state). The whole process takes 2 minutes and 37 seconds. The supply voltage is 220V and the efficiency is 67,5%. Assume the specific heat capacity of copper to be 389 J/kg.K and the latent heat of fusion of copper to be 206 kJ/kg. The cost of Energy is 236c/kWh. 2.1.1 Calculate the energy consumed to raise the temperature and melt of all of the copper.
The energy consumed to raise the temperature and melt all of the copper is 337196.182 J or 0.0937 kWh, and the total cost of energy consumed is 0.0221 R.
The electrical energy consumed to raise the temperature and melt all of the copper is calculated as follows:
Initial temperature of copper, T[tex]_{1}[/tex]= 23.3°C
Final temperature of copper, T[tex]_{2}[/tex] = 1085°C
Specific heat capacity of copper, c = 389 J/kg.K
Latent heat of fusion of copper, L[tex]_{f}[/tex] = 206 kJ/kg
Mass of copper, m = 0.54 kg
Time taken, t = 2 minutes 37 seconds = 157 seconds
Efficiency, η = 67.5% = 0.675
Supply voltage, V = 220 V
Cost of energy, CE = 236 c/kWh = 0.236 R/kWh
The energy required to raise the temperature of the copper from T[tex]_{1}[/tex] to T[tex]_{2}[/tex] is given by:
Q[tex]_{1}[/tex] = mc(T[tex]_{2}[/tex] - T[tex]_{1}[/tex])= 0.54 × 389 × (1085 - 23.3) = 0.54 × 389 × 1061.7= 225956.182 J
The energy required to melt the copper is given by:
Q[tex]_{2}[/tex] = mL[tex]_{f}[/tex]= 0.54 × 206 × 1000Q[tex]_{2}[/tex] = 111240 J
The total energy consumed is the sum of Q[tex]_{1}[/tex] and Q[tex]_{2}[/tex], that is:
Q[tex]_{tot}[/tex] = Q[tex]_{1}[/tex] + Q[tex]_{2}[/tex] = 225956.182 + 111240= 337196.182 J
The energy consumed is then converted from Joules to kWh:
Energy (kWh) = Q[tex]_{tot}[/tex] ÷ 3.6 × 10⁶
Energy (kWh) = 337196.182 ÷ 3.6 × 10⁶
Energy (kWh) = 0.0937 kWh
The total cost of energy consumed is calculated by multiplying the energy consumed (in kWh) by the cost of energy (in R/kWh):
Cost = Energy × CE = 0.0937 × 0.236
Cost = 0.0221 R
Therefore, the energy consumed to raise the temperature and melt all of the copper is 337196.182 J or 0.0937 kWh, and the total cost of energy consumed is 0.0221 R.
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electron maving in the negative *-birection is undeflected. K/im (b) What In For the value of E found in part (a), what would the kinetc energy of a proton have to be (in Mev) for is to move undefiected in the negative x-direction? MeV
Therefore, the kinetic energy of a proton that moves undeflected in the negative x-direction is 2.5 MeV.
In the case of an electron moving in the negative x-direction, which remains undeflected, the magnitude of the magnetic force, FB is balanced by the magnitude of the electrostatic force, FE. Therefore,FB= FEwhere,FB = qvB, andFE = qE Where,q = 1.60 × 10-19 C (charge on an electron).The kinetic energy of a proton that would move undeflected in the negative x-direction is found from the expression for the kinetic energy of a particle;KE = (1/2)mv2where,m is the mass of the proton,v is its velocity.To find the value of kinetic energy, the following expression may be used;KE = qE d /2where,d is the distance travelled by the proton. The electric field strength, E is equal to the ratio of the potential difference V across the two points in space to the distance between them, d. Thus,E = V/dWe know that,V = E × d (potential difference), where the value of potential difference is obtained by substituting the values of E and d.V = E × d = 5 × 10^3 V = 5 kVA proton will be able to move undeflected if it has a kinetic energy of KE = qE d/2 = 4.0 × 10^-13 J. This value can be converted to MeV by dividing it by the electron charge and multiplying by 10^6.MeV = KE/q = (4.0 × 10^-13 J) / (1.60 × 10^-19 J/eV) × 10^6 eV/MeV = 2.5 MeV. Therefore, the kinetic energy of a proton that moves undeflected in the negative x-direction is 2.5 MeV.
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The temperature is -9 °C, the air pressure is 95 kPa, and the vapour pressure is 0.4 kPa. Calculate the following
a)What is the dew-point temperature?
b)What is the relative humidity? c)What is the absolute humidity?
d)What is the mixing ratio?
e)What is the saturation mixing ratio?
f)Use your answers to d) and e) to recalculate the relative humidity.
The relative humidity is 63.46%. We can recalculate the relative humidity (RH) using the mixing ratio. The relative humidity is given by RH = [(w/ws) × 100%].
a) The dew-point temperature (Td) is the temperature at which the water vapor in the air becomes saturated and begins to condense into liquid water. It is the temperature at which the air becomes fully saturated with water vapor. The dew-point temperature can be calculated using the Clausius-Clapeyron equation:
Td = [B / (A - log e)]
Where A and B are constants for water, and e is the vapor pressure in kilopascals. Given the temperature of the air (T = -9 °C) and water vapor pressure (e = 0.4 kPa), we can calculate the dew-point temperature as Td = -13.87 °C.
b) The relative humidity (RH) is the ratio of the amount of water vapor in the air to the amount of water vapor the air can hold when it is saturated at a particular temperature. It is expressed as a percentage. The relative humidity can be calculated using the actual vapor pressure (e) and the saturation vapor pressure (es) at the given temperature. Given the actual vapor pressure (e = 0.4 kPa), we can calculate the saturation vapor pressure (es) using the equation es = [610.78 * exp((-9 * 17.27)/(237.3 - 9))] = 1.91 kPa. The relative humidity is RH = [(0.4/1.91) × 100%] = 20.94%.
c) The absolute humidity (Ah) is the mass of water vapor per unit volume of air. It represents the total amount of water vapor present in the air and is expressed in grams of water vapor per cubic meter of air. The absolute humidity can be calculated using the actual vapor pressure (e) and the temperature (T). Given the actual vapor pressure (e = 0.4 kPa) and temperature (T = -9 °C), we can calculate the absolute humidity as Ah = [(0.4 * 1000)/(0.287 * (264.15))] = 4.37 g/m³.
d) The mixing ratio (w) is the ratio of the mass of water vapor to the mass of dry air in a sample of air. It is a measure of the moisture content in the air. The mixing ratio can be calculated using the actual vapor pressure (e), the total pressure (p), and a constant ratio. Given the actual vapor pressure (e = 0.4 kPa) and total pressure (p = 95 kPa), we can calculate the mixing ratio as w = [(0.622 * 0.4)/(95 - 0.4)] = 0.0033 kg/kg.
e) The saturation mixing ratio (ws) is the maximum amount of water vapor that the air can hold at a given temperature. It is the ratio of the mass of water vapor in the air to the mass of dry air when the air is saturated. The saturation mixing ratio can be calculated using the saturation vapor pressure (es), the temperature (T), and a constant ratio. Given the saturation vapor pressure (es = 1.91 kPa) and temperature (T = -9 °C), we can calculate the saturation mixing ratio as ws = [(0.622 * es)/(95 - es)] = 0.0052 kg/kg.
f) Using the values from parts d) and e), we can recalculate the relative humidity (RH) using the mixing ratio. The relative humidity is given by RH = [(w/ws) × 100%]. Using the mixing ratio (w = 0.0033 kg/kg) and the saturation mixing ratio (ws = 0.0052 kg/kg), we can calculate the relative humidity as RH = [(0.0033/0.0052) × 100%] = 63.46%.
Therefore, the relative humidity is 63.46%.
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An object is located 72 cm from a thin diverging lens along the axis. If a virtual image forms at a distance of 18 cm from the lens, what is the focal length of the lens? in cm Is the image inverted or upright?
The focal length of the lens is -24 cm (negative sign indicates a diverging lens). Regarding the orientation of the image, for a diverging lens, the image formed is always virtual and upright.
To determine the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the distance of the virtual image from the lens (positive for virtual images),
u is the distance of the object from the lens (positive for objects on the same side as the incident light).
Given that the object is located 72 cm from the lens (u = -72 cm) and the virtual image forms at a distance of 18 cm from the lens (v = -18 cm), we can substitute these values into the lens formula:
1/f = 1/-18 - 1/-72
Simplifying this expression:
1/f = -1/18 + 1/72
= (-4 + 1) / 72
= -3/72
= -1/24
Now, taking the reciprocal of both sides of the equation:
f = -24 cm
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A 171 g ball is tied to a string. It is pulled to an angle of 6.8° and released to swing as a pendulum. A student with a stopwatch finds that 13 oscillations take 19 s.
The period of the pendulum is approximately 1.46 seconds per oscillation, the frequency is approximately 0.685 oscillations per second, and the angular frequency is approximately 4.307 radians per second.
To analyze the given situation, we can apply the principles of simple harmonic motion and use the provided information to determine relevant quantities.
First, let's calculate the period of the pendulum, which is the time it takes for one complete oscillation.
We can divide the total time of 19 seconds by the number of oscillations, which is 13:
Period (T) = Total time / Number of oscillations
T = 19 s / 13 = 1.46 s/oscillation
Next, let's calculate the frequency (f) of the pendulum, which is the reciprocal of the period:
Frequency (f) = 1 / T
f = 1 / 1.46 s/oscillation ≈ 0.685 oscillations per second
Now, let's calculate the angular frequency (ω) of the pendulum using the formula:
Angular frequency (ω) = 2πf
ω ≈ 2π * 0.685 ≈ 4.307 rad/s
The relationship between the angular frequency (ω) and the period (T) of a pendulum is given by:
ω = 2π / T
Solving for T:
T = 2π / ω
T ≈ 2π / 4.307 ≈ 1.46 s/oscillation
Since we already found T to be approximately 1.46 seconds per oscillation, this confirms our calculations.
In summary, the period of the pendulum is approximately 1.46 seconds per oscillation, the frequency is approximately 0.685 oscillations per second, and the angular frequency is approximately 4.307 radians per second.
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A 1.0nF air-filled parallel plate capacitor is charged up by a 100V battery. While still connected to the battery, a dielectric with κ=3 is filled between the plates. What is the final energy stored in the capacitor?
Answer Choices:
A. 15 μJ
B. 1.6 μJ
C. It is not possible to answer the question without knowing the charge on each plate
D. 5 μJ
A 1.0nF air-filled parallel plate capacitor is charged up by a 100V battery. While still connected to the battery, a dielectric with κ=3 is filled between the plates. Therefore, the correct option is A. 15 μJ.
The capacitance of an air-filled parallel plate capacitor is given by the formula C = εA/d,
where ε is the permittivity of air, A is the area of the plates, and d is the distance between them.
The permittivity of air is 8.85 x 10^-12 F/m.So,C = εA/d = 8.85 x 10^-12 * A/d = 1.0 x 10^-9nF = 1 x 10^-12 F So, A/d = 1.13 x 10^-3 m^-1 = capacitance per meter.
Since the capacitor is charged to 100V, the energy stored in it is given by the formulaE = 1/2 * CV^2 = 1/2 * 1 x 10^-12 * (100)^2 = 5 x 10^-9 J.
When a dielectric material with a dielectric constant (κ) is introduced between the plates, the capacitance of the capacitor increases by a factor of κ, which means the capacitance of the capacitor becomes κC, and the final energy stored in the capacitor is E' = 1/2 * κCV^2 = 1/2 * 3 * 1 x 10^-12 * (100)^2 = 1.5 x 10^-8 J = 15 μJ.
Therefore, the correct option is A. 15 μJ.
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The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.4-cm-diameter lens with f = 10 cm to cast an image of the sun on a piece of paper.
What is the intensity of sunlight in the projected image? Assume that all of the light captured by the lens is focused into the image
The intensity of the incoming sunlight is 1050 W/m²
The intensity of sunlight in the projected image is calculated to bee 7.271x10¹⁹ W/m².
In this scenario, a student utilizes a lens with a diameter of 5.4 cm and a focal length of 10 cm to project an image of the sun onto a piece of paper. The sun, positioned 150,000,000 km away from Earth, has a diameter of 1,400,000 km. The image formed is obtained using the lens formula. It is given by the relation as:
[tex]$$\frac {1}{f}=\frac {1}{u}+\frac {1}{v}$$[/tex]
Using the relation, the image distance (v) can be calculated as;
[tex]$$\frac {1}{10}=\frac {1}{150000000}-\frac {1}{u}$$[/tex]
The value of u comes out to be 1.4999 x 10⁹ m.
Using the formula for magnification, the magnification can be calculated as,
[tex]$$\frac {v}{u}=\frac {h'}{h}$$[/tex]
Where h' is the height of the imageh is the height of the object.
Height of the object, h = 1.4 x 10⁹ mHeight of the image, h' = 0.054 m
Therefore, the magnification is,
[tex]$$M=\frac {0.054}{1.4 x 10^9}=-3.8571x10^{-8}$$[/tex]
The negative sign in the magnification value indicates that the image is formed upside down or inverted. The intensity of sunlight in the projected image is given by the relation;
[tex]$$I'=\frac {I}{M^2}$$[/tex]
Where, I is the intensity of the incoming sunlight
The value of I is given to be 1050 W/m²
Therefore, substituting the given values in the above relation, we get,
[tex]$$I'=\frac {1050}{(-3.8571x10^{-8})^2}$$[/tex]
The value of I' comes out to be 7.271x10¹⁹ W/m².
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Match each of the following lapse rates with the appropriate
condition for which they should be used. Each answer is used only
once.
a. Normal/environmental lapse rate (NLR/ELR; 3.5°F/1,000 ft)
b. Dr
The normal or average lapse rate is called the environmental lapse rate (ELR). It's the standard rate at which the temperature decreases with height in the troposphere. The average ELR is -6.5°C per kilometer or -3.5°F per 1,000 feet
The question refers to matching the correct lapse rate with the condition for which they are applied. Here are the lapse rates and the appropriate conditions:
Normal/Environmental Lapse Rate (NLR/ELR; 3.5°F/1,000 ft) - It is used to calculate the average temperature decrease in the troposphere, which is -6.5°C or -3.5°F per 1000 feet of altitude.
Dry Adiabatic Lapse Rate (DALR; 5.5°F/1,000 ft) - This is the rate at which unsaturated air masses decrease their temperature with an increase in altitude. It is applicable in dry air conditions.
Wet Adiabatic Lapse Rate (WALR; 3.3°F/1,000 ft) - It is used to calculate the rate at which saturated air cools as it rises. This rate varies depending on the amount of moisture in the air.Therefore, the main answer is to match the given lapse rates with the appropriate condition for which they should be used. The lapse rates include the Normal/Environmental Lapse Rate (NLR/ELR), Dry Adiabatic Lapse Rate (DALR), and Wet Adiabatic Lapse Rate (WALR).
The change of temperature with height is called the lapse rate. Lapse rates come in various forms, and each has its application. A lapse rate is a measure of how temperature changes with height in the Earth's atmosphere. When the temperature decreases with height, it is referred to as the environmental lapse rate (ELR). The ELR is calculated by dividing the decrease in temperature by the increase in height. In contrast, the dry adiabatic lapse rate (DALR) is the rate at which the temperature of a parcel of unsaturated air decreases as it ascends. When a parcel of unsaturated air rises, it expands adiabatically (without exchanging heat with the surrounding air). The expanding parcel of air cools at the DALR rate. The DALR for unsaturated air is 5.5°F per 1,000 feet.
Wet adiabatic lapse rate (WALR) is the rate at which the temperature of a saturated parcel of air decreases as it rises. This rate varies depending on the amount of moisture in the air. As an air mass rises and cools, the moisture in it will eventually condense to form clouds. The heat released during this process offsets some of the cooling, causing the temperature to decrease at a lower rate, which is the WALR. The WALR is around 3.3°F per 1,000 feet.
Finally, the normal or average lapse rate is called the environmental lapse rate (ELR). It's the standard rate at which the temperature decreases with height in the troposphere. The average ELR is -6.5°C per kilometer or -3.5°F per 1,000 feet. It is used to calculate the average temperature decrease in the troposphere.
There are three different types of lapse rates, and each one is used to calculate temperature changes with height in the atmosphere under different conditions. The ELR, DALR, and WALR are calculated to determine the rate at which air temperature changes with altitude.
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A block of mass 4.0 kg and a block of mass 6.0 kg are linked by a spring balance of negligible mass. The blocks are placed on a frictionless horizontal surface. A force of 18.0 N is applied to the 6.0 kg block as shown. What is the reading on the spring balance?
The reading on the spring balance is 0.4 N.
When a force of 18.0 N is applied to the 6.0 kg block and there is no friction between the blocks and the horizontal surface. A spring balance is connected between two blocks. We are required to find the reading on the spring balance.
For that, we can use the formula of force that acts between the blocks connected by a spring balance. The formula is given as below:
F = kx where F is the force that acts between two blocks, k is the spring constant, and x is the displacement of the spring.The force that acts on the blocks is equal to the force applied on the heavier block. i.e., 18.0 N
The mass of the two blocks is M = 4.0 + 6.0 = 10.0 kg
The acceleration of the two blocks is given as follows:
For the heavier block 6.0 kg:
F = m₁a where m₁ is mass of the block
F = 18 N, m₁ = 6.0 kg
So, a = 18.0/6.0 = 3.0 m/s²
For the lighter block 4.0 kg:F = m₂a where m₂ is mass of the block m₂ = 4.0 kg
So, a = 3.0 m/s²
Using the force formula F = kxk = F/x = 18.0/0.4 = 45.0 N/m
The force on the spring is given as:F = kx
So, x = F/k = 18.0/45.0 = 0.4 m
Therefore, the reading on the spring balance is 0.4 m or 0.4 N (because 1 N/m = 1 N/m)
Answer: The reading on the spring balance is 0.4 N.
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you are riding a Ferris Wheel with a diameter of 19.3 m. You count the time it takes to go all the way around to be 38 s. How fast (in m/s) are you moving?
Round your answer to two (2) decimal places.
The speed (in m/s) of the Ferris wheel is 1.59.
The circumference of the Ferris wheel is given by the formula 2πr where r is the radius of the Ferris wheel.Calculation of the radius isR = d/2R = 19.3/2R = 9.65 m
The circumference can be given byC = 2πrC = 2 * 3.14 * 9.65C = 60.47 mNow the time taken to move around the Ferris wheel is given as 38 s.Now the speed of the Ferris wheel can be given asSpeed = distance/timeSpeed = 60.47/38Speed = 1.59 m/s.
Therefore, the speed (in m/s) of the Ferris wheel is 1.59.
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Suppose a beam of 5 eV protons strikes a potential energy barrier of height 6 eV and thickness 0.25 nm , at a rate equivalent to a current of 1000A (which is extremely high current!). a. How long would you have to wait, on average, for one proton to be transmitted? Give answer in seconds. b. How long would you have to wait if a beam of electrons with the same energy and current would strike potential barrier of the same height and length? Give answer in seconds.
The calculated times for one proton and one electron to be transmitted through the barrier are 1.23 × 10⁻¹⁶ seconds and 3.61 × 10⁻⁸ seconds, respectively.
a) In order to determine the time taken by one proton to transmit the barrier, we will use the tunneling formula as shown below:
[tex]$$t \approx e^{\frac{2 d}{\hbar}\sqrt{2 m \cdot (V-E)}}$$\\[/tex]
Where, d is the thickness of the barrierh is Planck's constantm is the mass of proton
E is the energy of the proton
V is the height of the potential barrier
Thickness of the barrier, d = 0.25 nm
Height of the potential barrier, V = 6 eV
Mass of a proton, m = 1.67 x 10⁻²⁷ kg
Energy of the proton, E = 5 eV = 5 x 1.6 x 10⁻¹⁹ J
Plugging in the data, we get:
[tex]$$t \approx e^{\frac{2 (0.25 x 10^{-9})}{\hbar}\sqrt{2 \cdot 1.67\times10^{-27} \cdot (6 - 5)\cdot 1.6\times10^{-19}}}$$[/tex]
The value of Planck's constant is 6.626 x 10⁻³⁴ Js
Plugging in the data, we get:
[tex]$$t \approx e^{\frac{2 (0.25 x 10^{-9})}{6.626 \times 10^{-34}}\sqrt{2 \cdot 1.67\times10^{-27} \cdot 1.6\times10^{-19}}}$$[/tex]
t ≈ 1.23 × 10⁻¹⁶ seconds
Therefore, we have to wait for 1.23 × 10⁻¹⁶ seconds for one proton to be transmitted through the barrier.
b) Electrons are a lot lighter than protons, so we can assume the mass of the electron to be 9.11 x 10^-31 kg. Hence, we can use the same formula as above to determine the time taken by one electron to transmit the barrier by using the following values:
Thickness of the barrier, d = 0.25 nmHeight of the potential barrier, V = 6 eV
Energy of the electron, E = 5 eV = 5 x 1.6 x 10⁻¹⁹ J
Mass of an electron, m = 9.11 x 10⁻³¹ kg
Plugging in the data, we get:
[tex]$$t \approx e^{\frac{2 (0.25 x 10^{-9})}{\hbar}\sqrt{2 \cdot 9.11\times10^{-31} \cdot (6 - 5)\cdot 1.6\times10^{-19}}}$$[/tex]
t ≈ 3.61 × 10⁻⁸ seconds
Therefore, we have to wait for 3.61 × 10⁻⁸ seconds for one electron to be transmitted through the barrier.
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you jumped from the same height into water as opposed to onto concrete, the impulse required to reduce your speed to zero velocity, assuming you have same posture at impact in cach case, would be the same True / False
False. The impulse required to reduce your speed to zero velocity would be different when jumping into water compared to jumping onto concrete.
The impulse required to reduce your speed to zero velocity would not be the same when jumping into water compared to jumping onto concrete. The impulse is equal to the change in momentum, which is determined by the mass and velocity of the object.
When jumping into water, the water exerts a greater resistive force compared to the concrete, which results in a longer deceleration time and a smaller impulse. The water acts as a cushion, spreading out the force over a longer duration.
On the other hand, when jumping onto concrete, the deceleration time is shorter, resulting in a larger impulse and potentially higher impact forces. The concrete does not provide the same cushioning effect as water.
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An object is located a distance of d0=19 cm in front of a concave mirror whose focal length is f=10.5 cm. A 50% Part (a) Write an expression for the image distance, d1. di= __________
Part (b) Numerally, what is this distance in cm?
Part (a) The expression for the image distance, d1 is di = 23.4 cm
Part (b) )Numerically, the distance of the image, d1 is 23.4 cm.
d0 = 19 cm
focal length is f = 10.5 cm.
The formula used for the distance of the image for a concave mirror is given as follows:
1/f = 1/do + 1/di
Where,
f = focal length
do = object distance from the mirror, and
di = image distance from the mirror
Part (a)
we substitute the given values in the above formula.
1/10.5 = 1/19 + 1/di
Multiplying both sides by 10.5 × 19 × di, we get:
19 × di = 10.5 × di + 10.5 × 19
Subtracting 10.5 from both sides, we get:
19 × di - 10.5 × di = 10.5 × 19
Combining like terms, we get:
di(19 - 10.5) = 10.5 × 19
Dividing both sides by (19 - 10.5), we get:
di = 10.5 × 19/(19 - 10.5)
di = 10.5 × 19/8.5
di = 23.4 cm
Therefore, the expression for the image distance, d1 is di = 23.4 cm
Part (b)
Numerically, the distance of the image, d1 is 23.4 cm.
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In each of the following situations a bar magnet is either moved toward or away from a coil of wire attached to a galvanometer. The polarity of the magnet and the direction of the motion are indicated. Do the following on each diagram: Indicate whether the magnetic flux (Φ) through the coil is increasing or decreasing. Indicate the direction of the induced magnetic field in the coil. (left or right?) Indicate the direction of the induced current in the coil. (up or down?) A) B) C)
Here are the explanations for the given diagrams: Diagrams (a) and (b) show the same scenario, where a north pole of a magnet is brought near to a coil or taken away from it. The change in the magnetic field causes a change in flux in the coil, which induces an emf. When the magnet is moved near the coil, the flux increases, and the induced magnetic field opposes the magnet's motion.
When the magnet is moved away, the flux decreases and the induced magnetic field is in the same direction as the magnet's motion, as shown in the following diagram: [tex]\downarrow[/tex] means the induced magnetic field is in the downward direction.
(a) For the first diagram, the magnetic flux is increasing, the induced magnetic field is to the left, and the induced current is downwards.
(b) For the second diagram, the magnetic flux is decreasing, the induced magnetic field is to the right, and the induced current is upwards.
(c) represents a different scenario, where a magnet is held stationary near a coil, but the coil is moved towards or away from the magnet. When the coil is moved towards the magnet, the magnetic flux increases, and the induced magnetic field opposes the motion of the coil. When the coil is moved away, the flux decreases and the induced magnetic field supports the motion of the coil, as shown in the following diagram: (c) For the third diagram, the magnetic flux is increasing, the induced magnetic field is to the left, and the induced current is downwards.
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A rocket, constructed on Earth by Lockheed engineers with a design length of 200.m, is launched into space and now moves past the Earth at a speed of 0.970c. What is the length of the rocket as measured by Bocing engineers observing the rocket from Earth?
The length of the rocket as measured by Bocing engineers observing the rocket from Earth is 47.98 m.
Given information:Length of the rocket on Earth = 200 m
A rocket is a vehicle or apparatus that moves forward on its own power by ejecting high-speed exhaust gases produced by the burning of propellants. The third rule of motion, which asserts that there is an equal and opposite response to every action, is the foundation upon which rockets are operated. Rockets move forward by experiencing a thrust in the opposite direction as they eject gases at high speeds through a nozzle.
Space exploration, satellite deployment, scientific research, military applications, and transportation are just a few of the uses for rockets. To accomplish their intended goals, they rely on exact engineering, cutting-edge propulsion systems, and sophisticated guidance mechanisms.
Speed of the rocket as measured by an observer on Earth = 0.970 cThe length of the rocket as measured by Bocing engineers observing the rocket from Earth is asked.
So, we have to determine the length of the rocket as measured by Bocing engineers observing the rocket from Earth.Solution:Given,Length of the rocket on Earth = 200 m
Speed of the rocket as measured by an observer on Earth = 0.970 cLet,Length of the rocket as measured by Bocing engineers observing the rocket from Earth = L'
Now, Length contraction formula is given by,[tex]L' = L√(1 - v²/c²)[/tex]
Where,v = 0.970c (speed of the rocket as measured by an observer on Earth )c = speed of lightL =[tex]200 mL' = L√(1 - v²/c²)L' = 200 m √(1 - (0.970c)²/c²)L' = 200 m √(1 - 0.970²)L' = 200 m √(1 - 0.94249)L' = 200 m √0.05751L' = 200 m × 0.2399L' = 47.98 m[/tex]
[tex]200 mL' = L√(1 - v²/c²)L' = 200 m √(1 - (0.970c)²/c²)L' = 200 m √(1 - 0.970²)L' = 200 m √(1 - 0.94249)L' = 200 m √0.05751L' = 200 m × 0.2399L' = 47.98 m[/tex]
Therefore, the length of the rocket as measured by Bocing engineers observing the rocket from Earth is 47.98 m.
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An initially uncharged capacitor with a capacitance of C=2.50μF is connected in series with a resistor with a resistance of R=5.5kΩ. If this series combination of circuit elements is attached to an ideal battery with an emf of E=12.0 V by means of a switch S that is closed at time t=0, then answer the following questions. (a) What is the time constant of this circuit? (b) How long will it take for the capacitor to reach 75% of its final charge? (c) What is the final charge on the capacitor?
Therefore, the final charge on the capacitor is:Q = CE = (2.50 x 10^-6 F) x (12.0 V) = 3.00 x 10^-5 C.
(a) Time Constant:Initially, the capacitor is uncharged. At t=0, the switch is closed, and a current begins to flow in the circuit. The current is equal to E/R, and the charge on the capacitor builds up according to the equation Q = CE(1 - e^(-t/RC)).Since the initial charge on the capacitor is zero, the final charge on the capacitor is equal to CE. Therefore, the time constant of the circuit is RC = 2.5 x 10^-6 F x 5.5 x 10^3 Ω = 0.01375 s(b) Time to reach 75% of final charge:The equation for charge on a capacitor is Q = CE(1 - e^(-t/RC)). To find the time at which the capacitor has reached 75% of its final charge, we can set Q/CE equal to 0.75, and solve for t.0.75 = 1 - e^(-t/RC) => e^(-t/RC) = 0.25 => -t/RC = ln(0.25) => t = RC ln(4)The value of RC is 0.01375 s, so t = 0.01375 ln(4) = 0.0189 s(c) Final charge on the capacitor: We know that the final charge on the capacitor is CE, where C = 2.50μF and E = 12.0 V. Therefore, the final charge on the capacitor is:Q = CE = (2.50 x 10^-6 F) x (12.0 V) = 3.00 x 10^-5 C.
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A hollow aluminum cylinder 17.0 cm deep has an internal capacity of 2.000 L at 21.0°C. It is completely filled with turpentine at 21.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 79.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10−6°C−1, and the average volume expansion coefficient for turpentine is 9.0 ✕ 10−4°C−1.)
(a) How much turpentine overflows? ----------- cm3
(b) What is the volume of turpentine remaining in the cylinder at 79.0°C? (Give your answer to at least four significant figures.)
---------- L
(c) If the combination with this amount of turpentine is then cooled back to 21.0°C, how far below the cylinder's rim does the turpentine's surface recede?
---------------- cm
The amount of turpentine that overflows can be calculated using the volume expansion coefficients of turpentine and the change in temperature.
(a) To calculate the amount of turpentine that overflows, we need to find the change in volume of the aluminum cylinder and the change in volume of the turpentine. The change in volume of the aluminum cylinder can be calculated using the linear expansion coefficient and the change in temperature: ΔV_aluminum = V_aluminum * α_aluminum * ΔT. Substituting the given values, ΔV_aluminum = (2.000 L) * (24 * 10^-6 °C^-1) * (79.0°C - 21.0°C).
The change in volume of the turpentine can be calculated using the volume expansion coefficient and the change in temperature: ΔV_turpentine = V_turpentine * β_turpentine * ΔT. Substituting the given values, ΔV_turpentine = (2.000 L) * (9.0 * 10^-4 °C^-1) * (79.0°C - 21.0°C).
The amount of turpentine that overflows is the difference between the change in volume of the turpentine and the change in volume of the aluminum cylinder: Overflow = ΔV_turpentine - ΔV_aluminum.
(b) The volume of turpentine remaining in the cylinder at 79.0°C is the initial volume of turpentine minus the amount that overflows: V_remaining = V_initial - Overflow.
(c) When cooled back to 21.0°C, the volume of the turpentine remains the same, but the volume of the aluminum cylinder shrinks. The volume change of the aluminum cylinder can be calculated using the linear expansion coefficient and the change in temperature: ΔV_aluminum = V_aluminum * α_aluminum * ΔT. Substituting the given values, ΔV_aluminum = (2.000 L) * (24 * 10^-6 °C^-1) * (21.0°C - 79.0°C).
The turpentine's surface recedes below the cylinder's rim by the difference between the change in volume of the aluminum cylinder and the change in volume of the turpentine: Recession = ΔV_aluminum - ΔV_turpentine.
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A commuter airplane starts from an airport and takes the following route: The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it flies for 150. km 20.0° west of north, to city B. Finally, the plane flies 190. km due west, to city C. Find the location of city C relative to the location of the starting point. (Hint: Draw a diagram on an xy-plane. Draw the start of a new vector from the end of the previous one, also known as, tip - to - toe method.)
The location of city C relative to the location of the starting point is (-30 km, 255 km).
The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it flies for 150. km 20.0° west of north, to city B. Finally, the plane flies 190. km due west, to city C.To find the location of city C relative to the location of the starting point, we need to draw a diagram on an xy-plane using the tip-to-toe method. This is the route of plane.
Let us assume the starting point as O and point C as (x, y) and find the values of x and y using the given data.From the starting point O, the plane flies to city A, located 175 km away in a direction 30.0° north of east.Now, from the starting point, O draw a vector 175 km in the direction 30.0° north of east. Let the end of this vector be P.From the end of the vector OP, draw a vector 150 km in the direction 20.0° west of north. Let the end of this vector be Q.From the end of the vector OQ, draw a vector 190 km in the due west direction. Let the end of this vector be R.Then, join OR as shown in the figure below. From the figure, we can see that the coordinates of R are (-30 km, 255 km).
Therefore, the location of city C relative to the location of the starting point is (-30 km, 255 km).
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I am driving to CSU at 23 m/s. I'm 100 m from the intersection when I see the light turn red. My reaction time is 0.73 s. Assuming my car has a constant acceleration for its brakes, what is the total time needed to bring my car to rest right at the edge of the intersection. Answer in seconds.
The total distance is 100 m - 16.79 m = 83.21 m. The total time needed to bring your car to rest at the edge of the intersection, we can break down the problem into two parts: the reaction time and the braking time. Since you are driving at a constant speed of 23 m/s, in 0.73 seconds your car would have traveled a distance of:
Distance = Speed × Time
Distance = 23 m/s × 0.73 s
Distance = 16.79 m
Now, let's calculate the remaining distance you need to cover to reach the edge of the intersection, considering that your car is coming to a stop. The total distance is 100 m - 16.79 m = 83.21 m.
Since your car is braking with a constant acceleration, we can use the following kinematic equation to find the braking time (t):
Distance = (Initial Velocity × t) + (0.5 × Acceleration ×[tex]t^2)[/tex]
In this case, the initial velocity is 23 m/s, the distance is 83.21 m, and the acceleration is negative (since it opposes the motion):
83.21 m = (23 m/s × t) + (0.5 × (-acceleration) × [tex]t^2)[/tex]
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A helicopter is flying North-West at 80 m/s relative to the ground and the wind velocity is 15 m/s from the East. The helicopter's main rotor lies in a horizontal plane, has a radius of 6 m, and is rotating at 20 rad/s in a clockwise sense looking down on it. a) Calculate the helicopter's air speed and apparent heading through the air (i.e. both relative to the air). b) Calculate the maximum and minimum velocities of the blade-tips relative to the air. Hint: In both parts, draw sketches to visualise what's happening. In the second part, only consider the helicopter's motion through the air and the blade-tips' motion relative to the helicopter (i.e. the air becomes your main reference frame, not the ground).
The helicopter's air speed is 59.4 m/s and apparent heading through the air is 45°
The maximum velocity of the blade tip relative to the air is 179.4 m/s and the minimum velocity of the blade tip relative to the air is 40.6 m/s.
Speed of helicopter relative to ground (VHG) = 80 m/s
Wind velocity = 15 m/sR
otor radius = 6 m
Rotor speed = 20 rad/s
a) The airspeed of the helicopter can be obtained by calculating the resultant of the helicopter velocity vector and wind velocity vector. Let us take North as y-axis and West as x-axis.The vector components of VHG along the x-axis and y-axis respectively will be as follows:
Vx = VHG * cos 45°Vy = VHG * sin 45°
The vector components of wind velocity along the x-axis and y-axis respectively will be as follows:
V'x = 15 m/sVy' = 0
The resultant vector of the helicopter velocity and the wind velocity will be as follows:
V = Vx + V'yV = 80(cos 45°) + 15V = 59.4 m/s
The apparent heading of the helicopter through the air can be calculated as follows:tan θ = Vy / Vxθ = tan⁻¹(Vy / Vx)θ = tan⁻¹(1)θ = 45°
b) The maximum velocity occurs when the blade is perpendicular to the direction of motion and the minimum velocity occurs when the blade is parallel to the direction of motion.
Let v1 and v2 be the maximum and minimum velocities of the blade-tips relative to the air.
Velocity of the tip of a rotor blade relative to the air is given by the formula,v = (ωr) ± V
where,v = velocity of the blade tip
ω = angular velocity of the rotor
r = radius of the rotor
V = airspeed of the helicopter
Taking velocity in the upward direction as positive, we get:
v1 = (ωr) + Vv2 = (ωr) - V
Let us substitute the given values in the above two formulas.
v1 = (20 * 6) + 59.4
v1 = 179.4 m/s
v2 = (20 * 6) - 59.4
v2 = 40.6 m/s
Hence, the maximum velocity of the blade tip relative to the air is 179.4 m/s and the minimum velocity of the blade tip relative to the air is 40.6 m/s.
Thus :
(a) The helicopter's air speed is 59.4 m/s and apparent heading through the air is 45°
(b) The maximum velocity of the blade tip relative to the air is 179.4 m/s and the minimum velocity of the blade tip relative to the air is 40.6 m/s.
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Required information A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 37.7- μF capacitor is charged to 5.40kV. Paddles are used to make an electric connection to the patient's chest. A pulse of current lasting 1.00 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 240 Q. How much energy is dissipated in the patient during the 1.00 ms?
The amount of energy dissipated in the patient during the 1.00 ms is approximately [tex]0.545 \mu J (or 5.45 * 10^-^7 J).[/tex].
To calculate the energy dissipated in the patient, we can use the formula:
[tex]Energy = (1/2) * C * (V^2)[/tex],
where C represents the capacitance and V represents the voltage. In this case, the capacitance is 37.7 μF (or [tex]37.7 * 10^-^6 F[/tex]) and the voltage is 5.40 kV (or [tex]5.40 * 10^3 V[/tex]). Plugging in these values into the formula, we get:
Energy = ([tex]1/2) * (37.7 * 10^-^6) * (5.40 * 10^3)^2[/tex].
Simplifying the expression, we find:
Energy = [tex]0.5 * 37.7 * 10^-^6 * (5.40 * 10^3)^2[/tex].
After calculating the values inside the parentheses, we have:
Energy [tex]= 0.5 * 37.7 *10^-^6 * 29.16 * 10^6[/tex].
Multiplying these values together, we obtain:
Energy ≈ [tex]0.5 * 1.09 * 10^-^6 J[/tex].
Therefore, the amount of energy dissipated in the patient during the 1.00 ms is approximately [tex]0.545 \mu J (or 5.45 * 10^-^7 J).[/tex]
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A wave has a frequency of 5.0x10-1Hz and a speed of 3.3x10-1m/s. What is the wavelength of this wave?
The wavelength of a wave with a frequency of [tex]5.0*10^-^1Hz[/tex] and a speed of [tex]3.3*10^-^1m/s[/tex] is 0.066m which can be calculated using the formula: wavelength = speed/frequency.
To find the wavelength of a wave, we can use the formula: wavelength = speed/frequency. In this case, the frequency is given as [tex]5.0*10^-^1Hz[/tex] and the speed is given as [tex]3.3*10^-^1m/s[/tex]. We can plug these values into the formula to calculate the wavelength.
wavelength = speed/frequency
wavelength = [tex]3.3*10^-^1m/s[/tex] / [tex]5.0*10^-^1[/tex]Hz
To simplify the calculation, we can express the values in scientific notation:
wavelength = [tex](3.3 / 5.0) * 10^-^1^-^(^-^1^)[/tex]m
Simplifying the fraction gives us:
wavelength = [tex]0.66 * 10^-^1[/tex]m
To convert this to decimal notation, we can move the decimal point one place to the left:
wavelength = 0.066m
Therefore, the wavelength of the wave is 0.066m.
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A radio transmitter broadcasts at a frequency of 96,600 Hz. What is the wavelength of the wave in meters? What is the wavelength (in nanometers) of the peak of the blackbody radiation curve for something at 1,600 kelvins?
The wavelength of a radio wave with a frequency of 96,600 Hz is approximately 3.10 meters. The peak wavelength of blackbody radiation for an object at 1,600 kelvins is around 1,810 nanometers.
To calculate the wavelength of a radio wave, we can use the formula: wavelength = speed of light / frequency. The speed of light is approximately 299,792,458 meters per second. Therefore, for a radio wave with a frequency of 96,600 Hz, the calculation would be: wavelength = 299,792,458 m/s / 96,600 Hz ≈ 3.10 meters.
Blackbody radiation refers to the electromagnetic radiation emitted by an object due to its temperature. The peak wavelength of this radiation can be determined using Wien's displacement law, which states that the peak wavelength is inversely proportional to the temperature of the object. The formula for calculating the peak wavelength is: peak wavelength = constant / temperature. The constant in this equation is approximately 2.898 × 10^6 nanometers * kelvins.
Plugging in the temperature of 1,600 kelvins, the calculation would be: peak wavelength = 2.898 × 10^6 nm*K / 1,600 K ≈ 1,810 nanometers. Thus, for an object at 1,600 kelvins, the peak wavelength of its blackbody radiation curve would be around 1,810 nanometers.
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Definition of Lenz's law According to Lenz's law, a) the induced current in a circuit must flow in such a direction to oppose the magnetic flux. b) the induced current in a circuit must flow in such a direction to oppose the change in magnetic flux. c) the induced current in a circuit must flow in such a direction to enhance the change in magnetic flux. d) the induced current in a circuit must flow in such a direction to enhance the magnetic flux. e) There is no such law, the prof made it up specifically to fool gullible students that did not study.
According to Lenz's law, the correct option is (b) the induced current in a circuit must flow in such a direction to oppose the change in magnetic flux.
Lenz's law is a fundamental principle in electromagnetism named after the Russian physicist Heinrich Lenz. It states that when there is a change in magnetic flux through a circuit, an induced electromotive force (EMF) is produced, which in turn creates an induced current.
The direction of this induced current is such that it opposes the change in magnetic flux that produced it. This means that the induced current creates a magnetic field that acts to counteract the change in the original magnetic field.
Option (a) is incorrect because the induced current opposes the magnetic flux, not the magnetic field itself. Option (c) is incorrect because the induced current opposes the change in magnetic flux, rather than enhancing it.
Option (d) is also incorrect because the induced current opposes the change in magnetic flux, not enhances it. Finally, option (e) is a false statement. Lenz's law is a well-established principle in electromagnetism that has been experimentally confirmed and is widely accepted in the scientific community.
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The cable of a high-voltage power line is 21 m above the ground and carries a current of 1.66×10 3
A. (a) What maqnetic field does this current produce at the ground? T x
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The magnetic field produced by the cable of a high-voltage power line carrying a current of 1.66×10^3 A and is 21 m above the ground is 5.88×10^-5 T. This value can also be written as 0.0588 mT.
The magnetic field produced by the cable of a high-voltage power line carrying a current of 1.66×10^3 A and is 21 m above the ground is 5.88×10^-5 T. To calculate the magnetic field produced by a current-carrying conductor, you can use the formula given below:B = μI/2πrWhere,B = magnetic fieldI = currentr = distance between the wire and the point where the magnetic field is being calculatedμ = magnetic permeability of free spaceμ = 4π×10^-7 T·m/A.
Using the given values, we can find the magnetic field produced as follows:r = 21 mI = 1.66×10^3 Aμ = 4π×10^-7 T·m/AB = μI/2πrB = 4π×10^-7 × 1.66×10^3/(2π × 21)B = 5.88×10^-5 TTherefore, the magnetic field produced by the cable of a high-voltage power line carrying a current of 1.66×10^3 A and is 21 m above the ground is 5.88×10^-5 T. This value can also be written as 0.0588 mT.
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You have a series RLC circuit connected in series to an
oscillating voltage source ( Vrms0.120=e) which is driving it.
FCmHLRμ50.4,0.960,0.144 ==W=. The circuit is being driven initially at resonance.
(a) (2 pts) What is the impedance of the circuit?
(b) (3 pts) What is the power dissipated by the resistor?
(c) (6 pts) The inductor is removed and replaced by one of lower value such that the
impedance doubles with no other changes. What is the new inductance?
(d) (4 pts) What is the power dissipated by the resistor now?
(e) (3 pts) What is the phase angle?
An oscillating voltage source is coupled in series with a series RLC circuit. 50.41 is the estimated impedance of the circuit, 2.857 x 10-5 W is the power wasted by the resistor, and 0.0157 radians is the approximate phase angle.
(a) The impedance of the series RLC circuit can be calculated using the formula:
Z = √(R^2 + (Xl - Xc)^2)
Where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. In this case, the values are given as:
R = 50.4 Ω (resistance)
Xl = 0.960 Ω (inductive reactance)
Xc = 0.144 Ω (capacitive reactance)
Plugging these values into the impedance formula, we have:
Z = √(50.4^2 + (0.960 - 0.144)^2)
Z = √(2540.16 + 0.739296)
Z ≈ √2540.899296
Z ≈ 50.41 Ω
So, the impedance of the circuit is approximately 50.41 Ω.
(b) The power dissipated by the resistor can be calculated using the formula:
P = (Vrms^2) / R
Where Vrms is the rms voltage of the source. In this case, the rms voltage is given as 0.120 V, and the resistance is 50.4 Ω.
P = (0.120^2) / 50.4
P = 0.00144 / 50.4
P ≈ 2.857 x 10^-5 W
So, the power dissipated by the resistor is approximately 2.857 x 10^-5 W
(c) When the impedance of the circuit doubles by replacing the inductor, we can find the new inductance by using the impedance formula and considering the new impedance as twice the original value:
Z_new = 2Z = 2 * 50.41 Ω = 100.82 Ω
To calculate the new inductance, we can rearrange the inductive reactance formula:
Xl_new = Z_new - Xc = 100.82 - 0.144 = 100.676 Ω
Using the inductive reactance formula:
Xl_new = 2πfL_new
Solving for L_new:
L_new = Xl_new / (2πf) = 100.676 / (2π * 50) ≈ 0.321 H
So, the new inductance is approximately 0.321 H.
(d) The power dissipated by the resistor remains the same even after changing the inductance because the resistance value and the voltage across the resistor have not changed. Therefore, the power dissipated by the resistor remains approximately 2.857 x 10^-5 W.
(e) The phase angle of the circuit can be determined using the formula:
θ = arctan((Xl - Xc) / R)
Plugging in the values:
θ = arctan((0.960 - 0.144) / 50.4)
θ = arctan(0.816 / 50.4)
θ ≈ 0.0157 radians
So, the phase angle of the circuit is approximately 0.0157 radians.
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The sun makes up 99.8% of all of the mass in the solar system at 1.989×10 30
kg. This means that for many of the objects that orbit well outside the outer planets they can be treated as a satellite orbiting a single mass (the sun). a) If the radius of the sun is 700 million meters calculate the gravitational field near the 'surface'? b) If a fictional comet has an orbital period of 100 years calculate the semi-major axis length for its orbit? c) Occasionally the sun emits a "coronal mass ejection". If CME's have an average speed of 550 m/s how far away would this material make it from the center of the sun before the suns gravity brings it o rest?
a) The gravitational field strength near the "surface" of the Sun is approximately 274.7 N/kg b) The semi-major axis length for the fictional comet's orbit is approximately 7.78 × 10^11 meters. c) The material from the coronal mass ejection (CME) would travel approximately 4.14 × 10^8 meters from the center of the Sun before coming to rest due to the Sun's gravity.
a) Gravitational field near the "surface" of the Sun:
Using the formula:
[tex]\[ g = \frac{{G \cdot M}}{{r^2}} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M \)[/tex] is the mass of the Sun, and [tex]\( r \)[/tex] is the radius of the Sun. Substituting the given values, we have:
[tex]\[ g = \frac{{(6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (1.989 \times 10^{30} \, \text{kg})}}{{(700 \, \text{million meters})^2}} \approx 274.7 \, \text{N/kg} \][/tex]
Therefore, the gravitational field near the "surface" of the Sun is approximately 274.7 N/kg.
b) Semi-major axis length for the fictional comet's orbit:
Using Kepler's third law equation:
[tex]\[ a = \left( \frac{{T^2 \cdot GM}}{{4\pi^2}} \right)^{1/3} \][/tex]
where [tex]\( T \)[/tex]is the orbital period of the comet,[tex]\( G \)[/tex] is the gravitational constant, and [tex]\( M \)[/tex] is the mass of the Sun. Substituting the given values, we get:
[tex]\[ a = \left( \frac{{(100 \, \text{years})^2 \cdot (6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (1.989 \times 10^{30} \, \text{kg})}}{{4\pi^2}} \right)^{1/3} \approx 7.78 \times 10^{11} \, \text{m} \][/tex]
Therefore, the semi-major axis length for the fictional comet's orbit is approximately [tex]\( 7.78 \times 10^{11} \) meters.[/tex]
c) Distance traveled by material from a coronal mass ejection (CME):
Using the equation:
[tex]\[ r = \frac{{GM}}{{2v^2}} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant,[tex]\( M \) i[/tex]s the mass of the Sun, and [tex]\( v \)[/tex] is the average speed of the CME. Substituting the given values, we have:
[tex]\[ r = \frac{{(6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \cdot (1.989 \times 10^{30} \, \text{kg})}}{{2 \cdot (550 \, \text{m/s})^2}} \approx 4.14 \times 10^{8} \, \text{m} \][/tex]
Therefore, the material from the coronal mass ejection (CME) would travel approximately [tex]\( 4.14 \times 10^8 \)[/tex]meters from the center of the Sun before coming to rest due to the Sun's gravity.
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A car accelerates from a speed of 12 m/s at 3.0 m/s/s for 150m. After that, it continues along at the same velocity for 310 more meters. How long does it take for the car to go the whole distance?
A car accelerates from a speed of 12 m/s at 3.0 m/s/s for 150m. it takes the car approximately 33.15 seconds to cover the entire distance.
To find the total time it takes for the car to cover the entire distance, we need to consider the two stages of its motion: the acceleration phase and the constant velocity phase.
First, let's calculate the time taken during the acceleration phase:
Given initial velocity (vi) = 12 m/s, acceleration (a) = 3.0 m/s², and distance (d) = 150 m.
We can use the equation of motion: d = vit + (1/2)at²
Rearranging the equation, we get:
t = (sqrt(2ad - vi²)) / a
Plugging in the values, we find:
t = (sqrt(2 * 3.0 * 150 - 12²)) / 3.0 = 7.32 s
Next, we calculate the time taken during the constant velocity phase:
Given distance (d) = 310 m and velocity (v) = 12 m/s.
We can use the equation: t = d / v
Plugging in the values, we get:
t = 310 / 12 = 25.83 s
Finally, we add the times from both phases to find the total time:
Total time = t_acceleration + t_constant_velocity = 7.32 s + 25.83 s = 33.15 s
Therefore, it takes the car approximately 33.15 seconds to cover the entire distance.
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Light from a HeNe LASER (λ=633nm) passes through a narrow slit and is seen on a screen 2.0 m behind the slit. The first minimum in the diffraction pattern is 1.2 cm from the central maximum. How wide is the slit?
The width of the slit is approximately 52.75 μm.
The width of the slit can be calculated using the formula for the diffraction pattern. In this case, the first minimum is observed 1.2 cm from the central maximum when light from a HeNe laser with a wavelength of 633 nm passes through the slit and is projected onto a screen 2.0 m away.
The position of the first minimum in a diffraction pattern can be determined using the equation:
θ = λ / (2 * a),
where θ is the angular position of the first minimum, λ is the wavelength of light, and a is the width of the slit.
To find the width of the slit, we need to convert the angular position of the first minimum into radians. Since the screen is located 2.0 m away from the slit, we can use the small angle approximation:
θ = y / D,
where y is the distance from the central maximum to the first minimum (1.2 cm = 0.012 m) and D is the distance from the slit to the screen (2.0 m).
Rearranging the equation and substituting the values, we have:
a = λ * D / (2 * y) = (633 nm * 2.0 m) / (2 * 0.012 m) = 52.75 μm.
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