The given toaster is rated at 660 W when it is connected to a 220 V source. We can find the current that the toaster as follows,
P = VI or I=P/V, where P is the power, V is the voltage, I is the current
So, I=660/220
I=3A
Therefore, the current that the toaster carries C. 3.0 A.
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Starting from rest at the top of a frictionless inclined plane, a block takes 2 s to slide down to the bottom The incline angle is 0, where sin 0 = 3/4 and cos 0 = 2/3. What is the length of this inclined plane? 7.5 m 10 m 15 m 30 m 20 m
Starting from rest at the top of a frictionless inclined plane, a block takes 2 s to slide down to the bottom The incline angle is 0, where sin 0 = 3/4 and cos 0 = 2/3. Thus, the length of the inclined plane is 20 m
The given incline angle is θ = 0 where sin θ = 3/4 and cos θ = 2/3 and the block slides down without any friction.
We are to find out the length of the inclined plane.
Let L be the length of the inclined plane, and g be the acceleration due to gravity.
As per the given statement, the block takes 2 seconds to slide down to the bottom of the inclined plane.
The acceleration of the block will be the same as the acceleration due to gravity in the direction of the inclined plane.
Therefore, the time t it takes for the block to slide down the incline plane of length L, starting from rest at the top of the inclined plane, is given by; L = 1/2gt² (since initial velocity, u = 0)At θ = 0, sin θ = 3/4 and cos θ = 2/3.
Therefore, the length of the inclined plane is; L = 1/2 × 9.8 m/s² × (2 s)² = 19.6 m
Thus, the length of the inclined plane is 20 m (approximated to one significant figure).Hence, the correct option is (e) 20 m.
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Two large parallel conducting plates are separated by d = 10 cm, causing a uniform electric field between them. The voltage difference between the two plates is 500 V. An electron is released at rest from the edge of the negative plate inside. a) What is the magnitude of the electric field between the two plates? b) Find the work done by the electric field on the electron as it moves from the negative plate to the positive plate. Express your answer in both electron volts (eV) and Joules c) What is the change in potential energy of the electron as it moves from the negative plate to the positive plate? d) What is the kinetic energy of the electron when it reaches the positive plate?
The magnitude is 5000 V/m. The work done by the electric field on the electron is -5 x 10^2 eV or -8 x 10^-17 J. The change in potential energy is -8 x 10^-17 J.The kinetic energy of the electron when it reaches the positive plate will be 8 x 10^-17 J.
a) The magnitude of the electric field between the two plates can be determined using the formula:
E = V / d
where E is the electric field, V is the voltage difference, and d is the distance between the plates.
Given that V = 500 V and d = 10 cm = 0.1 m, we can calculate the electric field:
E = 500 V / 0.1 m = 5000 V/m
b) The work done by the electric field on the electron as it moves from the negative plate to the positive plate can be calculated using the formula:
Work = q * V
where Work is the work done, q is the charge of the electron, and V is the voltage difference.
The charge of an electron is approximately -1.6 x 10^-19 C (coulombs). The voltage difference is given as V = 500 V.
Work = (-1.6 x 10^-19 C) * (500 V) = -8 x 10^-17 J
To express the answer in electron volts (eV), we can convert from joules to electron volts using the conversion factor:
1 eV = 1.6 x 10^-19 J
Work = (-8 x 10^-17 J) / (1.6 x 10^-19 J/eV) = -5 x 10^2 eV
c) The change in potential energy of the electron as it moves from the negative plate to the positive plate is equal to the work done by the electric field. From part (b), we found that the work done is -8 x 10^-17 J.
d) The change in potential energy of the electron is equal to the change in kinetic energy. Therefore, when the electron reaches the positive plate, its kinetic energy will be equal to the magnitude of the change in potential energy.
Since the change in potential energy is -8 x 10^-17 J, the kinetic energy of the electron when it reaches the positive plate will be 8 x 10^-17 J.
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What is the magnetic field at the center of a single (N=1 turn) circular loop of wire or radius 10 cm carrying a current of 2.5 A ? 2.41×10 −4
T 5.0×10 −6
T 1.57×10 −7
T 3.14×10 −5
T
The magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A is 3.14 × 10-5 T.
Magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A can be calculated using the formula;
B=μ0I/2R
where B is the magnetic field, I is the current flowing, R is the radius of the loop and μ0 is the permeability of free space.The given values are;I = 2.5 AR = 10 cm = 0.1 mμ0 = 4π × 10-7 T m/A.
Substitute the values into the formula; B = μ0I/2R = (4π × 10-7 T m/A) × (2.5 A)/2(0.1 m)= 3.14 × 10-5 T
Therefore, the magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A is 3.14 × 10-5 T.
Answer: 3.14×10^−5T.
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Find the range in wavelengths (in vacuum) for visible light in the frequency range between 7.9 × 10¹⁴ Hz (violet light) Express the answers in nanometers. (Express your answer in whole number)
The range in wavelengths (in vacuum) for visible light in the frequency range between 7.9 × 10¹⁴ Hz (violet light) is 380 nm (approx).
The formula is given as:
frequency = (speed of light) / (wavelength)
Where:
frequency = 7.9 x 10¹⁴ Hz
speed of light = 3 x 10⁸ m/s (in vacuum)
Solving for wavelength:
wavelength = (speed of light) / (frequency)
Therefore, wavelength = (3 x 10⁸) / (7.9 x 10¹⁴) = 3.80 x 10⁻⁷ m or 380 nm (approx)
Hence, the range in wavelengths (in vacuum) for visible light in the frequency range between 7.9 × 10¹⁴ Hz (violet light) is 380 nm (approx).
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Why Does Elasticity Matter?
Often, a lot of what is covered in courses has little application in the so-called "real world". In this discussion board, you need to post an entry to the discussion board stating why elasticity actually does matter in the everyday lives of businesses and consumers, using an example of a good or service as part of your explanation.
Part I
Using an example of a good or service, you will state why elasticity is applicable in the everyday lives of businesses and consumers. Please be clear in your explanation
Elasticity is of significant importance in the everyday lives of businesses and consumers as it helps them understand and respond to changes in prices and demand for goods or services. By considering elasticity, businesses can make informed decisions regarding pricing strategies, production levels, and resource allocation. Consumers, on the other hand, can assess the impact of price changes on their purchasing decisions and adjust their consumption patterns accordingly.
Elasticity, specifically price elasticity of demand, measures the responsiveness of consumer demand to changes in price. It indicates the percentage change in quantity demanded resulting from a one percent change in price. Understanding price elasticity allows businesses to determine how sensitive consumers are to changes in price and adjust their pricing strategies accordingly.
For example, let's consider the market for gasoline. Gasoline is a highly price-sensitive good, meaning that changes in its price have a significant impact on consumer demand. If the price of gasoline increases, consumers may reduce their consumption and seek alternatives such as carpooling or using public transportation. In this scenario, businesses need to consider the price elasticity of gasoline to predict and respond to changes in consumer behavior. They might lower prices to stimulate demand or introduce more fuel-efficient options to cater to price-conscious consumers.
In conclusion, elasticity matters because it provides valuable insights into the dynamics of supply and demand, enabling businesses and consumers to make informed decisions in response to price changes. By understanding elasticity, businesses can adapt their strategies to maintain competitiveness, while consumers can optimize their purchasing choices based on price sensitivity.
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Convert 47 deg into radian.
47 degrees is equal to 0.8203 radians.
To convert degrees to radians, we can use the following conversion formula:
radians = (degrees * π) / 180
Where:
degrees is the measurement in degrees
π (pi) is a mathematical constant approximately equal to 3.14159
To convert 47 degrees into radians, we will use the following formula;
Radian = (Degree × π) / 180 Where π = 3.14 radians
47 degrees is given, so we can substitute it into the formula:
Radian = (Degree × π) / 180
Radian = (47 × 3.14) / 180
Radian = 0.8203 radians
Therefore, 47 degrees is equal to 0.8203 radians.
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lamp and a 30 Q lamp are connected in series with a 10 V battery. Calculate the following: the power dissipated by the 20 02 lamp ] A 20 lamp and a 30 02 lamp are connected in series with a 10 V battery. Calculate the following: the power dissipated by the 30 Q lamp
The power dissipated by the 20 ohm lamp is 0.5556 W and the power dissipated by the 30 ohm lamp is 0.8333 W.
Two lamps having resistances of 20 ohm and 30 ohm are connected in series with a 10V battery. The current in the circuit is given by:I = V/R (series circuit)Resistance of the circuit, R = R₁ + R₂I = 10/(20 + 30)I = 0.1667ANow, using Ohm's Law:Power dissipated by the 20 ohm lamp:P = I²R = (0.1667)² × 20P = 0.5556WattsPower dissipated by the 30 ohm lamp:P = I²R = (0.1667)² × 30P = 0.8333WattsTherefore, the power dissipated by the 20 ohm lamp is 0.5556 W and the power dissipated by the 30 ohm lamp is 0.8333 W.
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An air-track glider of mass 0.150 kg is attached to the end of a horizontal air track by a spring with force constant 45.0 N/m (Figure 1). Initially the spring is unstretched and the glider is moying at 1.25 m/s to the right. Find the maximum distance d that the glider moves to the right if the air track is turned on, so that there is no friction. Express your answer with the appropriate units. All attempts used; correct answer displayed Part B Find the maximum distance d that the glider moves to the right if the air is turned off, so that there is kinetic friction with coefficient 0.320. Express your answer with the appropriate units.
Part A. The maximum distance (d) that the glider moves to the right when the air track is turned on is approximately 0.082 m.
Part B. The maximum distance (d) that the glider moves to the right when there is kinetic friction with a coefficient of 0.320 is approximately 0.069 m.
Part A:
To find the maximum distance (d) that the glider moves to the right when the air track is turned on, we can use the conservation of mechanical energy. The initial mechanical energy of the system is equal to the maximum potential energy stored in the spring.
The formula for potential energy stored in a spring is given by:
[tex]\[ PE_{\text{spring}} = \frac{1}{2} k x^2 \][/tex]
where PE is the potential energy, k is the force constant of the spring, and x is the displacement from the equilibrium position.
Initially, the glider is moving to the right, so the displacement (x) is negative. The initial kinetic energy (KE) is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where m is the mass of the glider and v is its velocity.
Since mechanical energy is conserved, the initial mechanical energy ([tex]\rm ME_{initial[/tex]) is equal to the maximum potential energy ([tex]PE_{max[/tex]). Therefore:
[tex]\[ ME_{\text{initial}} = PE_{\text{max}} = KE + PE_{\text{spring}} \][/tex]
Substituting the given values:
[tex]\[ \frac{1}{2} m v^2 + \frac{1}{2} k x^2 = \frac{1}{2} (0.150 \, \text{kg})(1.25 \, \text{m/s})^2 + \frac{1}{2} (45.0 \, \text{N/m})(x)^2 \][/tex]
Simplifying the equation, we can solve for x:
[tex]\[ 0.150 \, \text{kg} \times (1.25 \, \text{m/s})^2 + 45.0 \, \text{N/m} \times (x)^2 = 0.5 \, \text{kg} \times v^2 \]\[ 0.234375 + 45x^2 = 0.9375 \]\[ 45x^2 = 0.703125 \]\[ x^2 = \frac{0.703125}{45} \]\[ x = \sqrt{\frac{0.703125}{45}} \][/tex]
Calculating x, we find:
[tex]\[ x \approx 0.082 \, \text{m} \][/tex]
Therefore, the maximum distance (d) that the glider moves to the right when the air track is turned on is approximately 0.082 m.
Part B:
To find the maximum distance (d) that the glider moves to the right when there is kinetic friction, we need to consider the work done by friction.
The work done by friction can be calculated using the formula:
[tex]\[ W_{\text{friction}} = \mu_k N d \][/tex]
where [tex]\( \mu_k \)[/tex] is the coefficient of kinetic friction, N is the normal force (equal to the weight of the glider), and d is the distance traveled.
The work done by friction is equal to the change in mechanical energy:
[tex]\[ W_{\text{friction}} = \Delta ME \][/tex]
Therefore:
[tex]\[ \mu_k N d = \Delta ME \][/tex]
Substituting the given values:
[tex]\[ 0.320 \times (0.150 \, \text{kg} \times 9.8 \, \text{m/s}^2) \times d = \frac{1}{2} (0.150 \, \text{kg}) (1.25 \, \text{m/s})^2 + \frac{1}{2} (45.0 \, \text{N/m}) (d)^2 \][/tex]
Simplifying the equation, we can solve for d:
[tex]\[ 0.320 \times 0.150 \times 9.8 \times d = \frac{1}{2} \times 0.150 \times 1.25^2 + \frac{1}{2} \times 45.0 \times d^2 \]\[ 0.4704d = 0.1171875 + 22.5d^2 \]\[ 22.5d^2 - 0.4704d + 0.1171875 = 0 \][/tex]
Using the quadratic formula, we find:
[tex]\[ d \approx 0.069 \, \text{m} \][/tex]
Therefore, the maximum distance (d) that the glider moves to the right when there is kinetic friction with a coefficient of 0.320 is approximately 0.069 m.
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What is the value of the electric field in front of a charged flat plate whose surface charge density σ is 1.2×10 ∧
−12c/m ∧
2. If the plate has a length of 15 cm and a width of 20 cm. A) calculate the total charge on its surface B) if a proton has a charge of 1.6×10 ∧
−19 coulombs, determine the number of protons sitting on its surface. …2×10 −12
c/m 2
The value of the electric field in front of the charged flat plate with a surface charge density is 8 × 10^4 N/C.
There are approximately 2.25 × 10^5 protons sitting on the surface of the plate.
The total charge on the surface of the plate can be calculated by multiplying the surface charge density by the area of the plate. In this case, the plate has a length of 15 cm and a width of 20 cm.
A) The total charge on the surface of the plate is given by Q = σ × A, where Q is the total charge and A is the area of the plate. Substituting the given values, we have Q = (1.2 × 10^(-12) C/m^2) × (0.15 m) × (0.20 m) = 3.6 × 10^(-14) C.
B) To determine the number of protons sitting on the surface of the plate, we need to divide the total charge by the charge of a single proton. The charge of a proton is q = 1.6 × 10^(-19) C.
Number of protons = Q / q = (3.6 × 10^(-14) C) / (1.6 × 10^(-19) C) ≈ 2.25 × 10^5 protons.
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A parallel-plate capacitor with capacitance Co stores charge of magnitude Qoon plates of area A separated by distance do. The potential difference across the plates is AV- HINT AV new> AVO Co If the capacitor is attached to a battery and the charge is doubled to 200, what are the ratios new and (a) Cnew = Co (b) AV new AVO Cnew and Co AV now? AVO A second capacitor is identical to the first capacitor except the plate area is doubled to 2A. If given a charge of Qo, what are the ratios. (c) Cnew Co AV new (d) Cnew and AVO Co A third capacitor is identical to the first capacitor, except the distance between the plates is doubled to 2do. If the third capacitor is then given a charge of Qo, what are the ratios (e) Cnew = Co (f) = = AV new = AVO AV new? AVO
A parallel-plate capacitor with capacitance Co stores charge of magnitude Qoon plates of area A separated by distance do. (a)Cnew / Co = 200 / Qo(b)AV new / AVo = 200 / Qo(c)Cnew / Co = 2.(d)AV new / AVo = Qo / Qo = 1. (e)Cnew / Co = do / (2do) = 1/2. (f)AV new / AVo = Qnew / Qo = Qo / Qo = 1
(a) The ratio of the new capacitance (Cnew) to the original capacitance (Co) is equal to the ratio of the new charge (Qnew) to the original charge (Qo):
Cnew / Co = Qnew / Qo
Since the charge is doubled to 200, the ratio becomes:
Cnew / Co = 200 / Qo
(b) The ratio of the new potential difference (AV new) to the original potential difference (AVo) is equal to the ratio of the new charge (Qnew) to the original charge (Qo):
AV new / AVo = Qnew / Qo
Since the charge is doubled to 200, the ratio becomes:
AV new / AVo = 200 / Qo
(c) The ratio of the new capacitance (Cnew) to the original capacitance (Co) is equal to the ratio of the new plate area (2A) to the original plate area (A):
Cnew / Co = (2A) / A
Cnew / Co = 2
(d) The ratio of the new capacitance (Cnew) to the original capacitance (Co) is equal to the ratio of the new plate area (2A) to the original plate area (A), and the ratio of the new potential difference (AV new) to the original potential difference (AVo):
Cnew / Co = (2A) / A = 2
AV new / AVo = Qnew / Qo
Since the charge is given as Qo, the ratio becomes:
AV new / AVo = Qo / Qo = 1
(e) The ratio of the new capacitance (Cnew) to the original capacitance (Co) is equal to the ratio of the new distance between the plates (2do) to the original distance between the plates (do):
Cnew / Co = do / (2do) = 1/2
(f) The ratio of the new potential difference (AV new) to the original potential difference (AVo) is equal to the ratio of the new charge (Qnew) to the original charge (Qo):
AV new / AVo = Qnew / Qo = Qo / Qo = 1
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A light plane must reach a speed of 35 m/s for take off. How long a runway is needed if the (constant) acceleration is 3 m/s27
The required runway length for a light plane to take off if the constant acceleration is 3 m/s² is 408.33 m.
How to solve the problem?
Here's a step-by-step solution to the problem:
Step 1: Write down the given variables
The plane needs to reach a speed of 35 m/s, and the constant acceleration is 3 m/s².
Step 2: Choose an appropriate kinematic equation to solve the problem
The equation v² = u² + 2as is appropriate for this problem since it relates the final velocity (v), initial velocity (u), acceleration (a), and distance traveled (s).
Step 3: Substitute the known variables and solve for the unknowns
The initial velocity is zero since the plane is starting from rest.
v = 35 m/s
u = 0 m/s
a = 3 m/s²
s = ?
v² = u² + 2as
s = (v² - u²) / 2a
Plug in the values:
v² = 35² = 1225
u² = 0² = 0
a = 3
s = (1225 - 0) / (2 x 3) = 408.33 m
Therefore, the required runway length for a light plane to take off if the constant acceleration is 3 m/s² is 408.33 m.
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A 1C charge is originally a distance of 1m from a 0.2C charge, but is moved to a distance of 0.1 m. What is the change in electric potential energy? OJ -9.0x10^9 J 1.6x10^10 J 9.0x10^9 J
Therefore, the change in electric potential energy is $1.62 \times 10^{10} J$, which is approximately $1.6 \times 10^{10} J$.Hence, the correct option is $1.6 \times 10^{10} J$.
Electric potential energy is calculated using the formula :$E_{p}=k \frac{q_{1} q_{2}}{r}$where,$k$ is Coulomb's constant, $9 \times 10^9 Nm^2/C^2$$q_1$ is the magnitude of charge 1$q_2$ is the magnitude of charge 2$r$ is the distance between the chargesFrom the above formula,$E_{p} \propto \frac{1}{r}$ which implies that when the distance between the two charges decreases, the electric potential energy will increase.
The change in electric potential energy, $\Delta E_{p}$ can be calculated using the formula,$\Delta E_{p} = E_{p final} - E_{p initial}$Given,$q_{1} = 1C$$q_{2} = 0.2C$$r_{initial} = 1m$$r_{final} = 0.1m$Let's find the initial electric potential energy:$E_{p initial} = k \frac{q_{1} q_{2}}{r_{initial}}$$E_{p initial} = 9 \times 10^9 \frac{(1)(0.2)}{1}$$E_{p initial} = 1.8 \times 10^9 J$Now,
let's find the final electric potential energy:$E_{p final} = k \frac{q_{1} q_{2}}{r_{final}}$$E_{p final} = 9 \times 10^9 \frac{(1)(0.2)}{0.1}$$E_{p final} = 1.8 \times 10^{10} J$The change in electric potential energy is $\Delta E_{p} = E_{p final} - E_{p initial}$$\Delta E_{p} = (1.8 \times 10^{10}) - (1.8 \times 10^9)$$\Delta E_{p} = 1.62 \times 10^{10} J$
Therefore, the change in electric potential energy is $1.62 \times 10^{10} J$, which is approximately $1.6 \times 10^{10} J$.Hence, the correct option is $1.6 \times 10^{10} J$.
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Use the density of strontium (d = 2. 60 g/cm3) to determine the volume in cubic centimeters of a sample that has a mass of 47. 2 pounds
To determine the volume of a sample of strontium with a given mass, we can use the formula:
Volume = Mass / Density
Given:
Density of strontium (d) = 2.60 g/cm^3
Mass of the sample = 47.2 pounds
Before we proceed, let's convert the mass from pounds to grams, as the density is given in grams per cubic centimeter (g/cm^3).
1 pound is approximately equal to 453.592 grams.
Mass of the sample in grams = 47.2 pounds * 453.592 grams/pound
Now, we can calculate the volume using the formula:
Volume = Mass / Density
Volume = (47.2 * 453.592) / 2.60
By performing the calculations, we can determine the volume of the strontium sample in cubic centimeters.
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A Find the Resistance of 100 meters of # 18 AWG Copper wire at 20° C ? B Find the Area you need to calculate the Resistance ? C Find the Resistance of 600 meters of solid Copper wire with a diameter of 5 mm ? P Find the Area you need to calculate the Resistance ? If the Resistance of some Copper wire is 80 ohms at 20° C, what is it's Resistance at 100° C ?
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C is 0.2098 Ω
b. To calculate the resistance of a wire, the cross-sectional area of the wire is required.
c. The area required to calculate the resistance is 2.155 × [tex]10^{-10}[/tex] m². The resistance of copper wire at 100°C is 92.2 Ω.
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C can be determined using the formula;
R = ρL/A
A = πr²ρ
where;
R = resistance
ρ = resistivity
L = length of the wire
A = area of cross-section
r = radius of the wire
Substituting the given values;
Length of wire L = 100 meters
Area of cross-section A = ?
Diameter of wire d = 0.0403 inches or 1.02462 mm
Cross-sectional area A = πd²/4 = π(1.02462 mm)²/4 = 0.8231 mm²
Resistivity ρ = 1.724 x [tex]10^{-8}[/tex] Ω-m (at 20°C for copper)
Thus;
R = ρL/A = 1.724 x [tex]10^{-8}[/tex] Ω-m x 100 meters / 0.8231 mm²R = 0.2098 Ω
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C is 0.2098 Ω
b. To calculate the resistance of a wire, the cross-sectional area of the wire is required.
c. To find the resistance of 600 meters of solid Copper wire with a diameter of 5 mm, we need to know the cross-sectional area of the wire. The formula for the cross-sectional area is;
A = πr²A = π(5/2)²A = 19.63 mm²
The resistivity of copper is 1.724 × [tex]10^{-8}[/tex] Ωm. Using the formula;
R = ρL/A
where;
L = 600 mA = 19.63 mm²
ρ = 1.724 × [tex]10^{-8}[/tex] Ωm
R = 0.16 ΩP.
To find the area required to calculate the resistance, the cross-sectional area of the wire is required. If the resistance of copper wire is 80 ohms at 20°C, we can use the above formula for resistivity.
ρ = RA/L
where;
R = 80 Ω
A = ?
L = 1 m
ρ = 1.724 × [tex]10^{-8}[/tex] Ωm
A = ρL/R = 1.724 × [tex]10^{-8}[/tex] × 1/80A = 2.155 × [tex]10^{-10}[/tex] m²
The resistance of copper wire at 100°C can be determined using the formula;
Rt = R0 [1 + α(T[tex]_{t}[/tex] - T[tex]_{0}[/tex])]
where;
R0 = resistance at 20°C = 80 Ω
T0 = temperature at 20°C = 293 K (20 + 273)
Tt = temperature at 100°C = 373 K (100 + 273)
α = temperature coefficient of copper = 0.00393/°C
Rt = 80 [1 + 0.00393(373 - 293)]R[tex]_{t}[/tex] = 92.2 Ω
Answer:
Therefore area required to calculate the resistance is 2.155 × [tex]10^{-10}[/tex] m². The resistance of copper wire at 100°C is 92.2 Ω.
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The only force acting on a 3.3 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a velocity of 2.4 m/s in the positive x direction, and some time later has a velocity of 5.6 m/s in the positive y direction. How much work is done on the canister by the 3.0 N force during this time? Number ___________ Units _____________
The work done on the canister by the 3.0 N force during this time is 0 J (joules).
To calculate the work done on the canister by the 3.0 N force during this time, we need to find the displacement of the canister and the angle between the force and the displacement.
The mass of the canister (m) is 3.3 kg.
The magnitude of the force (F) is 3.0 N.
The initial velocity (v₁) is 2.4 m/s.
The final velocity (v₂) is 5.6 m/s.
The work done (W) by the force can be calculated using the formula:
W = F * d * cosθ
To find the displacement (d), we need to calculate the change in position of the canister. Since the canister moves from the positive x direction to the positive y direction, we can consider the displacement as the vector sum of the initial and final velocities:
d = √((Δx)² + (Δy)²)
Δx represents the difference or change in the x-coordinate (horizontal direction) of the canister's position, while Δy represents the difference or change in the y-coordinate (vertical direction) of the canister's position.
Δx = 0 (since the canister does not move in the x direction)
Δy = v₂ - v₁ = 5.6 m/s - 2.4 m/s = 3.2 m/s
By substituting the given values into the formula mentioned above, we can determine the work done on the canister by the 3.0 N force during this time.
d = √((0)² + (3.2)²) = √10.24 = 3.2 m
Now, we need to find the angle θ between the force and the displacement. Since the force is acting in the xy plane and the displacement is in the positive y direction, the angle θ is 90 degrees.
Cosine of 90 degrees is 0, so cosθ = 0.
Substituting the values into the work formula, we get:
W = 3.0 N * 3.2 m * cos90° = 0 J
Therefore, the work done on the canister by the 3.0 N force during this time is 0 J (joules).
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hen two rainbows form, there is a dark region in-between them. What is the reason for this dark region? light is being reflected away from you the rainbow needs a certain temperature to have color you do not have the biology in your eyes to see those wavelenghts it is due to the critical angle a rainbow is not real
The dark region between the two rainbows is due to the critical angle at which light rays are reflected away from the observer's eye, and this angle depends on the size of the rain droplets.
When two rainbows form, there is a dark region in-between them because of the critical angle. This critical angle is the minimum angle of incidence beyond which total internal reflection of a light ray occurs from the water droplets in the atmosphere. Because of this angle, the light that reflects from the rain droplets moves away from the observer's eye, so a dark region is formed between the two rainbows.
The light that enters the drop slows down and bends, and the angle of bending is dependent on the color of the light. Red light is bent the least, while violet is bent the most, causing the separation of the colors in a rainbow. The angle of incidence can vary based on the size of the rain droplets, which is why two rainbows can form with different angles of incidence producing the different colors.
Thus, the dark region between the two rainbows is due to the critical angle at which light rays are reflected away from the observer's eye, and this angle depends on the size of the rain droplets.
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A series RLC circuit has an impedance of 1209 and a resistance of 642. What average power is delivered to this circuit when Vrms = 90 volts? O 36W O 100 W O 192 W 0 360 W O 12 W
A series RLC circuit has an impedance of 1209 and a resistance of 642. The average power delivered to the circuit is 12 W (Option E)
Given;
Impedance, Z = 1209 Ω
Resistance, R = 642 Ω
Voltage, Vrms = 90 volts
We are to calculate the average power delivered to the circuit.
P = Vrms2 / R *cos(Φ) ---(1)
Where Φ = angle of phase difference between the current and voltage
Since it is not given whether the circuit is capacitive or inductive or purely resistive, we will have to calculate the value of Φ to determine the nature of the circuit.
Cos(Φ) = R/Z = 642/1209 = 0.531<0.08
Thus, the circuit is inductive (since cos(Φ) is positive and < 1)
We can determine the value of angle Φ using the following equation;
Cos(Φ) = R/ZΦ = cos-1(R/Z)Φ = cos-1(642/1209)Φ = 0.08 rad
Average power delivered to the circuit;
P = Vrms2 / R *cos(Φ)
Substituting the values of Vrms, R and cos(Φ)P = (90)2 / 642 *0.531P = 12.6 W ≈ 12 W
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Using the balance of forces and derive the formula for hydrostatic equilibrium
a. Diagram and label each force, b. State the equation for each force c. Combine the forces to derive the hydrostatic relationship d. Compute the strength of the vertical pressure gradient force knowing that the pressure 850mb and the temperature is 0°C.
The hydrostatic equilibrium formula is derived by considering the balance of forces acting on a column of air. These forces include the pressure force, gravity force, and vertical pressure gradient force. The vertical pressure gradient force can be calculated using the hydrostatic equation.
In a specific example, when the pressure is 850 mb and the temperature is 0°C, the strength of the vertical pressure gradient force is found to be 7.1 N/m².
Using the balance of forces and derive the formula for hydrostatic equilibrium.
A) Diagram and label each force
A diagram of the forces acting on a column of air is shown below:
b. State the equation for each force
1. Pressure force
The pressure force is the force that the air exerts on a given area, represented by the symbol "P." This force acts at right angles to the surface and in the direction of the force. The formula for pressure force is:
Fp = P * A
where:
Fp is the pressure force in Newtons (N)
P is the pressure in Pascals (Pa)
A is the area in square meters (m²)
2. Gravity force
The force of gravity on an object is given by its weight. The force of gravity acts in a downward direction on the object. The formula for the gravitational force is:
Fg = mg
where:
Fg is the gravitational force in Newtons (N)
m is the mass in kilograms (kg)
g is the acceleration due to gravity, 9.8m/s²
3. Vertical pressure gradient force
The vertical pressure gradient force is the difference in pressure between two points, divided by the distance between them. This force is directed from high pressure to low pressure. The formula for the vertical pressure gradient force is:
Fv = -1/ρ * ΔP/Δz
where:
Fv is the vertical pressure gradient force in Newtons (N)
ρ is the density of air in kg/m³
ΔP is the pressure difference between two points in Pascals (Pa)
Δz is the distance between the two points in meters (m)
C) Combine the forces to derive the hydrostatic relationship
The balance of the forces in the vertical direction is:
ΣF = Fp + Fg + Fv = 0
The hydrostatic relationship is given by:
Fv = Fg + Fp - ΣF
v = -1/ρ * ΔP/Δz = mg + P * A
where:
m is the mass of the column of air
g is the acceleration due to gravity
P is the pressure in Pascals (Pa)
A is the area in square meters (m²)
ρ is the density of air in kg/m³
D) Compute the strength of the vertical pressure gradient force knowing that the pressure 850mb and the temperature is 0°C.
The hydrostatic equation can be used to calculate the vertical pressure gradient force when the pressure and temperature of a column of air are known.
Using the ideal gas law, the density of air at 850 mb and 0°C can be calculated as:
ρ = P/RT
where:
R is the gas constant
T is the temperature in Kelvin
For air at 0°C, R = 287 J/kg.K and T = 273 K, so:
ρ = P/RT = 850 * 100 Pa / (287 J/kg.K * 273 K) = 1.199 kg/m³
Using the hydrostatic equation:
Fv = -1/ρ * ΔP/Δz = -1/1.199 kg/m³ * (0 - 850 * 100 Pa) / 1000 m
= 7.1 N/m²
Therefore, the strength of the vertical pressure gradient force is 7.1 N/m².
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A 1581.64 kg tank runs out of brakes when it achieves a speed of 34.83 mi/hr. What linear momentum will you be experiencing?
Remember to perform the necessary conversions before solving.
Express your answer WITHOUT DECIMALS.
A 41 kg metal ball with a radius of 6.8 m is rolling at 19 m/s on a level surface when it reaches a 25 degree incline. How high does the ball go?
The ball rises to a height of 18.5 meters when it reaches a 25-degree incline.
When the 41 kg metal ball reaches a 25 degree incline, the height it goes to can be calculated. Here's how you can calculate the height of the ball:
First, we will calculate the potential energy of the ball by utilizing the formula: potential energy = mass * gravity * height
PE = mgh
Where m = 41 kg, g = 9.81 m/s² (the acceleration due to gravity), and h is the height in meters.
Since the ball is rolling at 19 m/s on a level surface, its kinetic energy will be:
kinetic energy = 0.5 * mass * velocity²
KE = 0.5 * m * v²
KE = 0.5 * 41 * 19²
KE = 7383.5 J
Now, we will equate the potential energy to the kinetic energy since the energy is conserved:
PE = KE => mgh = 7383.5Jh = 7383.5 / (41 * 9.81)h = 18.5 m
Therefore, the ball rises to a height of 18.5 meters when it reaches a 25-degree incline.
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Select one correct answer from the available options in the below parts. a) You shine monochromatic light of wavelength ⋀ through a narrow slit of width b = ⋀ and onto a screen that is very far away from the slit. What do you observe on the screen? A. Two bright fringes and three dark fringes B. one bright band C. A series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes D. A series of bright and dark fringes that are of equal widths b) What does it mean for two light waves to be in phase ? A. The two waves reach their maximum value at the same time and their minimum value at the same time B. The two waves have the same amplitude C. The two waves propagate in the same direction D. The two waves have the same wavelength and frequency
a) The correct answer is C. A series of bright and dark fringes with the central bright fringe being wider and brighter than the other bright fringes.
b) The correct answer is A. The two waves reach their maximum value at the same time and their minimum value at the same time.
The brilliant middle fringe is a result of light's beneficial interference. The two light sources (slits) are symmetrically closest to the centre fringe as well. As one walks out from the core, the fringes continue to progressively become darker and the central fringe is the brightest.
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:If we can't build a telescope on Earth to image the Apollo footprints, let's solve the problem by putting a telescope in orbit around the Moon instead. By being in the vacuum of space, our lunar satellite will avoid all the problems of astronomical seeing and will actually be able to achieve its theoretical diffraction limit. By being so much closer to the Moon, the footprints themselves will be much, much larger in angular size, allowing us to resolve them with a much, much smaller telescope mirror. So, let's imagine you place a telescope in an orbit that is d=50.0km above the surface of the Moon, such that as it passes directly overhead of the Apollo landing sites, it can record images from that distance. [This is the actual distance that the Lunar Reconnaissance Orbiter satellite orbits above the Moon's surface.] Following the work in Part II, calculate the angular size of the footprints from this new, much closer distance. The length units must match, so use the fact that 1.00 km=1.00×103 m to convert the orbital radius/viewing distance, d=50.0 km, from kilometers to meters: d=( km)×[ /. ]=
The angular size of the footprints from the new, much closer distance of 50.0 km above the surface of the Moon is 4 × 10¹⁰.
Given data:
Orbital radius/viewing distance, d = 50.0 km = 50.0 × 10³ m
To convert the orbital radius/viewing distance from kilometers to meters, we use the conversion factor:
1 km = 1 × 10³ m
Thus, d = 50.0 × 10³ m
The formula for calculating the angular size of footprints is given below:
θ = d / D
Where,
θ = Angular size of footprints.
d = Distance of telescope from the footprints.
D = Length of the footprints.
The Lunar Reconnaissance Orbiter satellite orbits 50 km above the surface of the Moon. So, the distance of the telescope from the footprints is d = 50.0 × 10³ m.
From Part II, the length of the footprints is D = 1.25 × 10⁻³ m.
Using the above formula, we can calculate the angular size of footprints as:
θ = d / D
θ = (50.0 × 10³) / (1.25 × 10⁻³)
θ = (50.0 × 10³) × (10³ / 1.25)
θ = (50.0 × 10³) × (8 × 10²)
θ = 4 × 10¹⁰
Therefore, the angular size of footprints from this new, much closer distance is 4 × 10¹⁰.
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Two volleyballs each carry a charge of 1.0 x 10-7 C. The magnitude of the electric force between them is 3.0 x 10-3 N. Calculate the distance between these two charged objects. Write your answer using two significant figures. m Show Calculator
The distance between the two charged objects is approximately 547 meters, rounded to two significant figures.
To calculate the distance between the two charged objects, we can use Coulomb's law, which states that the magnitude of the electric force between two charged objects is given by the equation:
F = k * (|q1| * |q2|) / [tex]r^2[/tex]
where F is the electric force, k is the electrostatic constant (9.0 x [tex]10^9[/tex] N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
In this case, we have:
F = 3.0 x [tex]10^{-3}[/tex] N
|q1| = |q2| = 1.0 x [tex]10^{-7}[/tex] C
Plugging these values into the equation, we can solve for r:
3.0 x [tex]10^{-3}[/tex] N = (9.0 x [tex]10^9[/tex] N m^2/C^2) * (1.0 x [tex]10^{-7}[/tex] C) * (1.0 x [tex]10^{-7}[/tex] C) / r^2
Simplifying the equation:
3.0 x [tex]10^{-3}[/tex] N = 9.0 x 10^2 N m^2 / r^2
Cross-multiplying and rearranging:
r^2 = (9.0 x 10^2 N m^2) / (3.0 x [tex]10^{-3}[/tex] N)
[tex]r^2 = 3.0 * 10^5 m^2[/tex]
Taking the square root of both sides:
r = [tex]\sqrt{3.0 * 10^5 m^2}[/tex]
r ≈ 547 m
Therefore, the distance between the two charged objects is approximately 547 meters, rounded to two significant figures.
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Two batteries V1 = 18 V, V2 = 15 V are connected to resisters R1 = 109, R2 = 209, and R3 = 30 Q Use Kirchhoff's Rules to find the current through Ry in the following circuit R w R. R Select one: a. 0.63 A O b. 0.55 A Oc. 0.08 A O d. None of these
Answer:
The correct option is (c) 0.08 A.
To find the current through Ry in the following circuit, we will apply Kirchhoff's Rules.
Kirchhoff's Rules are the basic rules used to analyze a circuit.
There are two rules:
Kirchhoff’s First Law (KCL) and Kirchhoff’s Second Law (KVL).
Kirchhoff’s First Law (KCL) states that the total current entering a junction is equal to the total current leaving the junction.
Kirchhoff’s Second Law (KVL) states that the total voltage around a closed circuit is zero.
For Junction A, the current entering the junction is equal to the current leaving the junction:
For junction B, the current entering the junction is equal to the current leaving the junction:
From the above two equations, we get:
This is equation 1.
We apply Kirchhoff's Second Law to the outer loop as shown below:
This is equation 2
Putting the values of equations 1 and 2, we get:
The current through Ry is:
Ry = R2 || R3
=> Ry = 209*30/(209+30)
=> Ry = 25.14Ω
Iy = 0.0795 A ≈ 0.08
Therefore, the correct option is (c) 0.08 A.
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Physics
The Gravity Force Fgrav between two objects with masses M1 and
M2 is 100 N. If the separation between them is tripled and the mass
of each object is doubled, what is Fgrav?
When the separation between two objects is tripled and the mass of each object is doubled, the gravitational force between them decreases to (4/9) of its original value. In this case, the force decreases from 100 N to approximately 44.44 N.
The gravitational force between two objects is given by the equation:
Fgrav = G * (M₁ * M₂) / r²,
where G is the gravitational constant, M₁ and M₂ are the masses of the objects, and r is the separation between them.
In this scenario, we have Fgrav = 100 N. If we triple the separation between the objects, the new separation becomes 3r. Additionally, if we double the mass of each object, the new masses become 2M₁ and 2M₂.
Substituting these values into the gravitational force equation, we get:
Fgrav' = G * ((2M₁) * (2M₂)) / (3r)²
= (4 * G * (M₁ * M₂)) / (9 * r²)
= (4/9) * Fgrav.
Therefore, the new gravitational force Fgrav' is (4/9) times the original force Fgrav. Substituting the given value Fgrav = 100 N, we find:
Fgrav' = (4/9) * 100 N
= 44.44 N (rounded to two decimal places).
Hence, the new gravitational force is approximately 44.44 N.
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What is the reasons that called the capacitor is an ideal parallel plate capacitor?
The reasons for calling a capacitor an ideal parallel plate capacitor are: (1) It assumes infinite plate area, resulting in uniform electric field between the plates; (2) It assumes no dielectric or conducting material between the plates, minimizing losses and fringing effects.
An ideal parallel plate capacitor is a theoretical concept used to simplify the analysis of real-world capacitors. It is called "ideal" because it assumes certain conditions that may not be fully achievable in practice. The key reasons for labeling it as an ideal parallel plate capacitor are as follows.
Firstly, it assumes infinite plate area. This assumption implies that the plates are infinitely large, ensuring a uniform electric field between them. In reality, the plates of a capacitor have finite dimensions, leading to non-uniform electric fields near the edges, known as fringing effects. However, by assuming infinite plate area, these edge effects are disregarded, simplifying the analysis.
Secondly, the ideal parallel plate capacitor assumes no dielectric or conducting material between the plates. This assumption eliminates losses due to dielectric absorption or leakage currents, which can occur in real capacitors. In practice, capacitors employ dielectric materials between the plates to enhance capacitance, but these materials may introduce non-ideal characteristics.
While an ideal parallel plate capacitor serves as a useful theoretical model, real-world capacitors deviate from these assumptions. Factors like finite plate area, dielectric properties, and parasitic effects influence the behavior of practical capacitors. Nonetheless, the ideal parallel plate capacitor provides a valuable starting point for understanding the fundamental principles of capacitance and energy storage.
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The critical angle in air for a particular type of material is 42.0 ∘
. What is the speed of light in this material in 10 8
m/s ? Use three significant digits please.
The speed of light in this material is approximately 2.00 × 10^8 m/s (to three significant digits).
To determine the speed of light in a particular material, we can use Snell's law, which relates the refractive indices of the two media:
n1*sin(theta1) = n2*sin(theta2)
Where:
n1 is the refractive index of the initial medium (air, in this case)
theta1 is the angle of incidence (measured from the normal)
n2 is the refractive index of the second medium (the material)
theta2 is the angle of refraction (measured from the normal)
Given that the critical angle in air for the material is 42.0 degrees, we can find the refractive index (n2) using the equation:
n2 = 1 / sin(critical angle)
Substituting the value, we get:
n2 = 1 / sin(42.0 degrees) ≈ 1.499
Now, the speed of light in a medium is related to the refractive index by the equation:
v = c / n
where:
v is the speed of light in the material
c is the speed of light in vacuum (approximately 3.00 × 10^8 m/s)
Substituting the values, we have:
v = (3.00 × 10^8 m/s) / 1.499 ≈ 2.00 × 10^8 m/s
Therefore, the speed of light in this material is approximately 2.00 × 10^8 m/s (to three significant digits).
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An green hoop with mass mh=2.6 kg and radius Rh=0.14 m hangs from a string that goes over a blue solid disk pulley with mass md=1.9 kg and radius Rd=0.1 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms=4.1 kg and radius R5 =0.21 m. The system is released from rest. 1) What is magnitude of the linear acceleration of the hoop? m/s2 2) What is magnitude of the linear acceleration of the sphere? m/s2 3) What is the magnitude of the angular acceleration of the disk pulley? rad/s2 4) What is the magnitude of the angular acceleration of the sphere? rad/s2 5) What is the tension in the string between the sphere and disk pulley? N 6) What is the tension in the string between the hoop and disk pulley? N 7) The green hoop falls a distance d=1.57 m. (After being released from rest.) How much time does the hoop take to fall 1.57 m ? 5 8) What is the magnitude of the velocity of the green hoop after it has dropped 1.57 m ? m/s 9) What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.57 m )? rad/s
1)Magnitude of the linear acceleration of the hoop= 9.8 m/s²2)the magnitude of the linear acceleration of the sphere is 0. 3)The magnitude of the angular acceleration of the disk pulley α = 0.4 m/s². 4)The magnitude of the angular acceleration of the sphere= 0.23 m/s². 5)The tension in the string between the sphere and disk pulleyT1 = 40.38 N. 6)The tension in the string between the hoop and disk pulleyT = 50.68 N.7)The hoop takes time to fall 1.57 m= 0.56 s. 8)the magnitude of the velocity of the green hoop v² = 6.2 m/s. 9)The magnitude of the final angular speed of the orange sphere is 29.5 rad/s.
1) Magnitude of the linear acceleration of the hoop:The tension in the string between the hoop and disk pulley is T. Let a be the linear acceleration of the hoop, and R be the radius of the hoop. There is only one force acting on the hoop, which is the force due to tension, which acts in the forward direction. Hence,mh * a = TThus, a = T / mh. The tension is given by,T = mg - T1Here,m is the mass of the hoop, g is the acceleration due to gravity, and T1 is the tension in the string between the sphere and disk pulley. Hence,a = (mg - T1) / mhGiven that,mh = 2.6 kgm = 9.8 m/s²g = 9.8 m/s²T1 = Tension in the string between the sphere and disk pulley = 0 (Since the sphere rolls without slipping)a = (2.6 × 9.8 - 0) / 2.6 = 9.8 m/s²
2) Magnitude of the linear acceleration of the sphere:Since the sphere rolls without slipping, the acceleration of the sphere is the same as the linear acceleration of its center of mass. Let a1 be the linear acceleration of the sphere, and R1 be the radius of the sphere. Let T1 be the tension in the string between the sphere and disk pulley. Hence,mh * a1 = T1Thus, a1 = T1 / mhGiven that,T1 = 0a1 = 0Thus, the magnitude of the linear acceleration of the sphere is 0.
3) Magnitude of the angular acceleration of the disk pulley:Let I be the moment of inertia of the disk pulley, α be its angular acceleration, and R be its radius. The disk pulley is rolling without slipping. Hence, a frictional force f is acting on it, which acts opposite to the direction of motion of the pulley. Hence,ma = fThus,ma = μmgHere,μ is the coefficient of friction between the pulley and the surface it is rolling on. Thus,α = a / R = μg / RThus,α = 0.4 m/s².
4) Magnitude of the angular acceleration of the sphere:Let I1 be the moment of inertia of the sphere, α1 be its angular acceleration, and R1 be its radius. Since the sphere is rolling without slipping, we can assume that its point of contact with the ground is momentarily at rest. Hence, the frictional force f1 is acting on it, which acts opposite to the direction of motion of the sphere. Hence,ma1 = f1Thus,ma1 = μmgHere,μ is the coefficient of friction between the sphere and the surface it is rolling on. Thus,α1 = a1 / R1 = μg / R1Thus,α1 = 0.23 m/s².
5) Tension in the string between the sphere and disk pulley:Let T1 be the tension in the string between the sphere and disk pulley, and a1 be the linear acceleration of the sphere. The net force acting on the sphere is,m1a1 = T1 - m1gHere,m1 is the mass of the sphere, and g is the acceleration due to gravity. Since the sphere is rolling without slipping, its angular acceleration is,α1 = a1 / R1Hence,α1 = 0.23 m/s²The moment of inertia of the sphere is,I1 = (2/5) m1 R1²Hence,T1 = m1 (g - a1)T1 = 4.1 (9.8 - 0)T1 = 40.38 N.
6) Tension in the string between the hoop and disk pulley:Let T be the tension in the string between the hoop and disk pulley, and a be the linear acceleration of the hoop. The net force acting on the hoop is,mh a = T - mh gHere,mh is the mass of the hoop, and g is the acceleration due to gravity. Hence,T = mh (g + a)T = 2.6 (9.8 + 9.8)T = 50.68 N.
7) Time taken by the hoop to fall a distance of 1.57 m:Let h be the distance fallen by the hoop, and t be the time taken to fall this distance. Hence,1/2 mgh = mh g h t/2 = sqrt (2h/g)t = sqrt (2 × 1.57 / 9.8)t = 0.56 s.
8) Magnitude of the velocity of the hoop after it has dropped 1.57 m:Let v be the velocity of the hoop after it has dropped 1.57 m. The final velocity of the hoop is given by,v² - u² = 2ghHere,u is the initial velocity of the hoop, which is 0. Hence,v² = 2ghv² = 2 × 9.8 × 1.57v = 6.2 m/s.
9) Magnitude of the final angular speed of the sphere:Let ω be the final angular speed of the sphere, v1 be its final linear velocity, and R1 be its radius. Since the sphere rolls without slipping,ω = v1 / R1Hence,ω = v / R1Here,v is the linear velocity of the hoop just before it hits the sphere. Hence,v = 6.2 m/sAlso,R1 = 0.21 mω = v / R1ω = 29.5 rad/sThus, the magnitude of the final angular speed of the orange sphere is 29.5 rad/s.
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You pull downward with a force of 31 N on a rope that passes over a disk-shaped pulley of mass of 1.5 kg and a radius of 0.075 m. The other end of the rope is attached to a 0.77 kg mass.
(1) Find the tension in the rope on both sides of the pulley. T1,T2 = (?) N
You pull downward with a force of 31 N on a rope that passes over a disk-shaped pulley of mass of 1.5 kg and a radius of 0.075 m . Therefore, the tension in the rope on both sides of the pulley is:T1 = 25.155 N and T2 = 15.345 N
When a 31N force is applied to a rope that passes over a disk-shaped pulley of mass of 1.5 kg and a radius of 0.075 m, the tension in the rope on both sides of the pulley is as follows:
T1 = (m1g + T2)/(1)T2 = (m2g - T1)/(2)
Where,m1=1.5 kgm2=0.77 kg T1 = tension in the rope on the side with the mass m1, T2 = tension in the rope on the side with the mass m2g = acceleration due to gravity = 9.81 m/s²
T1:T1 = (m1g + T2)/(1)T1 = (1.5 kg × 9.81 m/s² + T2)/(1)
Substitute the given value for T2:31 N = (1.5 kg × 9.81 m/s² + T2)/(1)T2 = (31 N - 1.5 kg × 9.81 m/s²)T2 = 15.345 N
Therefore, T1 = (1.5 kg × 9.81 m/s² + 15.345 N)/(1)T1 = 25.155 N
Therefore, the tension in the rope on both sides of the pulley is:T1 = 25.155 N and T2 = 15.345 N
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What are two adaptations that telescope must make to account for
different types of light?
Answer: Reflecting telescopes focus light with a series of mirrors, while refracting telescopes use lenses.
Explanation: