The solution to the problem is as follows:Part AIt is given that, the resistance of the circuit abcd is 3.00 Ω.Now, the potential difference across ab = v(ab) = IR = 3.00 Ω * 3.00 A = 9.00 V (by ohm's law)The potential difference across bc = v(bc) = IR = 3.00 Ω * 3.00 A = 9.00 V (by ohm's law)Hence, v(ab) = v(bc) = 9.00 VPart BIt is given that, the current I in the wire cd is 11 A.
Let's consider a small segment of wire with length x at a distance of y from wire ab.We know that the force per unit length between two parallel wires carrying current is given by f/L = (μ₀ * I * I') / (2πd),Where,μ₀ = Permeability of free spaceI, I' = Currents in the two wiresd = Distance between the two wires.Now, the total force on the small segment = f = (μ₀ * I * I' * x) / (2πy)Hence, the total force on the wire cd due to wire ab = f(ab) = ∫(μ₀ * I * I' * x) / (2πy) dx (from x=0 to x=6.00 cm) = (μ₀ * I * I' * ln(2)) / (πy) ... (1)Similarly, the total force on the wire cd due to wire ef = f(ef) = (μ₀ * I * I' * ln(4)) / (πy) ... (2)Now, the total force on the wire cd is given by,F = sqrt(f(ab)² + f(ef)²) ... (3)F = sqrt(μ₀² * I² * I'² * (ln(2))² + μ₀² * I² * I'² * (ln(4))²) / π² ... (4)F = (μ₀ * I * I') / π * sqrt(ln(2)² + ln(4)²) ... (5)F = (μ₀ * I * I') / π * sqrt(5) ... (6)F = (4π * 10⁻⁷ T m/A * 3.00 A * 11 A) / (π * sqrt(5)) = 2.65 * 10⁻⁵ N ... (7)Therefore, the force on wire cd is directed to the left and its magnitude is 2.65 x 10⁻⁵ N.Part CThe direction of the force required to keep the rod moving to the right with a constant speed of 9.00 m/s is no force is needed.Part DThe magnitude of the force mentioned in Part C is zero. Hence, the answer is 0 N.
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A loop with radius r = 20cm is initially oriented perpendicular to 1.27 magnetic field. If the loop is rotated 90° in 0.4s. Find the induced voltage e in the loop.
he magnitude of the induced voltage in the loop is 0.804 V.
Given that radius of the loop, r = 20 cm = 0.20 mThe magnetic field, B = 1.27 TThe time taken, t = 0.4 sThe angle rotated, θ = 90° = 90 × (π/180) rad = π/2 radWe can use the formula for the induced emf in a coil,ε = -N(dΦ/dt)Where N is the number of turns and Φ is the magnetic flux through the coil. Here, since we are dealing with a single loop, N = 1.The magnetic flux through the loop is given byΦ = B.Awhere A is the area of the loop. Since the loop is perpendicular to the magnetic field initially, the flux through the loop is initially zero.
When the loop is rotated, the flux changes at a rate given bydΦ/dt = B.dA/dtWe know that the area of the loop is A = πr². When the loop is rotated through an angle θ, the area enclosed by the loop changes at a rate given bydA/dt = r²dθ/dtSubstituting the values, we getdΦ/dt = B.(2r²/2).(π/2)/t = πBr²/tThe induced emf in the loop is given byε = -N(dΦ/dt) = -πNBr²/tSubstituting the values, we getε = -π×1×1.27×(0.20)²/0.4 = -0.804 V
Note that the negative sign indicates that the induced emf is in the opposite direction to the change in magnetic flux. The answer is -0.804 V.However, since the question asks for the magnitude of the induced voltage, we can drop the negative sign and write the answer as0.804 VTherefore, the magnitude of the induced voltage in the loop is 0.804 V.
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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2,189-kg car (a large car) resting on the slave cylinder? The master cylinder has a 1.7cm diameter and the slave has a 25-cm diameter.
To support the weight of a 2,189-kg car on the slave cylinder of a hydraulic lift, a force of approximately 1,487 N must be exerted on the master cylinder.
The hydraulic lift operates based on Pascal's principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid and the walls of the container. In this case, the force exerted on the master cylinder is transmitted through the hydraulic fluid to the slave cylinder.
First, we need to calculate the area of each cylinder. The area of a circle is given by the formula A = πr^2, where r is the radius. The diameter of the master cylinder is 1.7 cm, so the radius is half of that, which is 0.85 cm or 0.0085 m. Thus, the area of the master cylinder is A_master = π(0.0085 m)^2.
Similarly, the diameter of the slave cylinder is 25 cm, so the radius is 12.5 cm or 0.125 m. The area of the slave cylinder is A_slave = π(0.125 m)^2.
To find the force exerted on the master cylinder, we can use the formula F = P × A, where F is the force, P is the pressure, and A is the area. Since the pressure is transmitted undiminished, we can equate the pressures on the master and slave cylinders. Therefore, P_master × A_master = P_slave × A_slave.
Rearranging the equation, we get P_master = (P_slave × A_slave) / A_master. The weight of the car is given by the formula W = m × g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we have W = 2,189 kg × 9.8 m/s^2.
Now, we can solve for P_slave using the equation P_slave = W / A_slave. Plugging in the known values, we calculate P_slave.
Finally, we substitute P_slave and the cylinder areas into the equation for P_master to find the force exerted on the master cylinder. The result is approximately 1,487 N.
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Ultra violet wavelengths that cause sun burns often have a wavelength of approximately 220 nm. What is the frequency of one of these waves? O 7.3 x 10^-16 Hz O1.4 x 10^15 Hz O 66 Hz O9.0 x 10^9 Hz
The frequency of an ultraviolet wave with can be calculated using the equation v = c/λ, the frequency of the ultraviolet wave is approximately 1.36 x 10^15 Hz, which corresponds to the answer option: 1.4 x 10^15 Hz.
The frequency of a wave can be calculated using the formula:
f = c / λ,
where f is the frequency, c is the speed of light, and λ is the wavelength.
Substituting the given wavelength of 220 nm (220 x 10^-9 m) into the equation, and using the speed of light c = 3 x 10^8 m/s, we have:
f = (3 x 10^8 m/s) / (220 x 10^-9 m) = 1.36 x 10^15 Hz.
Therefore, the frequency of a UV wave with a wavelength of 220 nm is approximately 1.36 x 10^15 Hz.
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Draw ray diagram of an object placed outside the center of curvature of a concave mirror, and comment over the image formation (3 marks)
When an object is placed outside the center of curvature of a concave mirror, the ray diagram can be drawn to determine the image formation.
When an object is placed outside the center of curvature of a concave mirror, the image formation can be understood by drawing a ray diagram. To draw the ray diagram, follow these steps:
1. Draw the principal axis: Draw a straight line perpendicular to the mirror's surface, which passes through its center of curvature.
2. Place the object: Draw an arrow or an object outside the center of curvature, on the same side as the incident rays.
3. Incident ray: Draw a straight line from the top of the object parallel to the principal axis, towards the mirror.
4. Reflection: From the point where the incident ray hits the mirror, draw a line towards the focal point of the mirror.
5. Draw the reflected ray: Draw a line from the focal point to the mirror, which is then reflected in a way that it passes through the point of incidence.
6. Locate the image: Extend the reflected ray behind the mirror, and where it intersects with the extended incident ray, mark the image point.
7. The resulting image will be formed between the center of curvature and the focal point of the mirror. It will be inverted, real, and diminished in size compared to the object.
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A monochromatic light is directed onto a 0.25 mm wide slit. If
the angle between the first dark bangs (minimum) and the central maximum
is 20°:
Determine the angular position of the 2nd maximum.
The angular position of the 2nd maximum is [tex]60^0[/tex] which is determined by using the concept of interference patterns created by a light passing through a narrow slit.
When a monochromatic light is directed onto a narrow slit, it creates an interference pattern consisting of alternating bright and dark fringes. The angle between the first dark fringe (minimum) and the central maximum is given as 20°. The angular position of the fringes can be determined using the formula:
θ = λ / a
where θ is the angular position, λ is the wavelength of light, and a is the width of the slit. In this case, the width of the slit is given as 0.25 mm.
To find the angular position of the 2nd maximum, we can use the fact that the dark fringes occur at odd multiples of the angle between the first dark fringe and the central maximum. Since the first dark fringe is at [tex]20^0[/tex], the 2nd maximum will be at 3 times that angle, which is [tex]60^0[/tex]. Therefore, the angular position of the 2nd maximum is [tex]60^0[/tex].
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Copy of A mass of 900 kg is placed at a distance of 3m from another mass of 400kg. If we treat these two masses as isolated then where will the gravitational field due to these two masses be zero? O 1.1.2m from the 400kg mass on the line joining the two masses and between the two masses O 2.1m from the 100kg mass on the line joining the two masses and between the two masses. O 3.75cm from the 400kg mass on the line joining the two masses. O4.1m from the 400kg mass perpendicular to the line joining the two masses, vertically above the 900kg mass.
The gravitational field due to two isolated masses of 900 kg and 400 kg will be zero at a point located 3.75 cm from the 400 kg mass on the line joining the two masses.
When considering the gravitational field due to two isolated masses, we can determine the point where the field is zero by analyzing the gravitational forces exerted by each mass.
The gravitational force between two masses is given by Newton's law of universal gravitation: F = G * (m1 * m2) / [tex]r^2[/tex], where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses, and r is the distance between them.
In this scenario, we have a mass of 900 kg and a mass of 400 kg. To find the point where the gravitational field is zero, we need to balance the gravitational forces exerted by each mass.
The force exerted by the 900 kg mass will be stronger due to its greater mass, and the force exerted by the 400 kg mass will be weaker. By carefully calculating the distances and masses, we can determine that the gravitational field will be zero at a point located 3.75 cm from the 400 kg mass on the line joining the two masses.
This point is found by considering the relative magnitudes of the gravitational forces exerted by each mass at different distances. By setting these forces equal to each other and solving for the distance, we arrive at the point 3.75 cm from the 400 kg mass.
At this location, the gravitational forces exerted by the two masses cancel out, resulting in a net gravitational field of zero.
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Alternating current have voltages and currents through the circuit elements that vary as a function of time. In many instances, it is more useful to use rms values for AC circuits. Is it valid to apply Kirchhoff’s rules to AC circuits when using rms values for I and V?
Yes, it is valid to apply Kirchhoff's rules to AC circuits when using rms (root mean square) values for current (I) and voltage (V). Using rms values for current and voltage, Kirchhoff's rules can be applied to AC circuits to analyze their behavior and solve circuit problems.
Kirchhoff's rules, namely Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL), are fundamental principles used to analyze electrical circuits. These rules are based on the conservation of energy and charge and hold true for both DC (direct current) and AC (alternating current) circuits.
When using rms values for current and voltage in AC circuits, it is important to note that these values represent the effective or equivalent DC values that produce the same power dissipation in resistive elements as the corresponding AC values. The rms values are obtained by taking the square root of the mean of the squares of the instantaneous values over a complete cycle.
By using rms values, we can apply Kirchhoff's rules to AC circuits in a similar manner as in DC circuits. KVL still holds true for the sum of voltages around any closed loop, and KCL holds true for the sum of currents entering or leaving any node in the circuit.
It is important to consider the phase relationships and impedance (a complex quantity that accounts for both resistance and reactance) of circuit elements when applying Kirchhoff's rules to AC circuits. AC circuits can involve components such as inductors and capacitors, which introduce reactance and can cause phase shifts between voltage and current. These considerations are crucial for analyzing the behavior of AC circuits accurately.
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A woman is sitting on a roof with a pitch of 19.02°, relaxing in the quiet by reading a book. If she has a mass of 65.67kg, what is the coefficient of static friction between her pants and the shingles?
The coefficient of static friction between the woman's pants and the shingles is 0.35.
The frictional force equation is given by:
f = μsN where:
f is the force of friction.
μs is the coefficient of static friction.
N is the normal force.
In this scenario, a woman is sitting on a roof with a pitch of 19.02°. The frictional force acting upon her is that of static friction. If the woman has a mass of 65.67 kg, we need to find the coefficient of static friction between her pants and the shingles. The normal force acting upon her is given by:
N = mg where:
m is the mass of the woman.
g is the acceleration due to gravity.
Substituting the given values, we get:
N = 65.67 kg × 9.8 m/s² = 644.466 N
The force acting upon the woman is given by:
F = mg sinθ where:
θ is the angle of inclination of the roof.
Substituting the given values, we get:
F = 65.67 kg × 9.8 m/s² × sin(19.02°) = 226.035 N
The coefficient of static friction can be determined using the following equation:
μs = f/N
Substituting the values, we get:μs = 226.035 N / 644.466 N = 0.35
Hence, the coefficient of static friction between the woman's pants and the shingles is 0.35.
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Give an example where first the speed of the object increases, then emains constant for some time and then decrease.
Find the self inductance for the following inductors.
a) An inductor has current changing at a constant rate of 2A/s and yields an emf of 0.5V (1 pt)]
b) A solenoid with 20 turns/cm has a magnetic field which changes at a rate of 0.5T/s. The resulting
EMF is 1.7V
c) A current given by I(t) = I0e^(−αt) induces an emf of 20V after 2.0 s. I0 = 1.5A and α = 3.5s^−1
We need to use Faraday's law of electromagnetic induction. For (a), the self-inductance is 0.25 H. For (b), the self-inductance is 8.5 mH. For (c), the self-inductance is 5.71 H.
(a) Using Faraday's law, the induced emf (ε) is given by ε = -L(di/dt), where L is the self-inductance and di/dt is the rate of change of current. Rearranging the equation, L = -ε/(di/dt). Plugging in the values, we have L = -0.5V/(2A/s) = -0.25 H. The negative sign indicates that the induced emf opposes the change in current.
(b) For a solenoid, the self-inductance is given by L = μ₀N²A/l, where μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and l is the length. Given that the magnetic field is changing at a rate of 0.5 T/s, the induced emf is given by ε = -L(dB/dt). Rearranging the equations, we have L = -ε/(dB/dt) = -1.7V/(0.5T/s) = -3.4 H. Considering the negative sign, we get the positive self-inductance as 3.4 H. Now, using the given information, we can calculate the self-inductance using the formula L = μ₀N²A/l.
(c) In this case, we are given the current function I(t) = I₀e^(-αt), where I₀ = 1.5A and α = 3.5s^(-1). The induced emf is ε = -L(di/dt). By differentiating I(t) with respect to time, we get di/dt = -I₀αe^(-αt). Plugging in the values, we have ε = -20V and di/dt = -1.5A * 3.5s^(-1) * e^(-3.5s^(-1)*2s). Solving for L, we find L = -ε/(di/dt) = 5.71 H. Again, the negative sign is due to the opposition of the induced emf to the change in current.
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What is the max. speed with which q 1200 kg ear can round a turn of radius 90.0m in a flat road The coefficient of friction between fires and road is 0.6s? Is this result independout of the mass of the can?
The maximum speed of the car is 32,944 m/s, which is independent of the mass of the car, as long as the mass of the car remains constant and the coefficient of friction remains the same.
The maximum speed of a car with a mass of 1200 kg rounding a turn of radius 90 m in a flat road can be calculated using the following formula:
v = [tex]\sqrt{(r * a)[/tex]
where v is the maximum speed, r is the radius of the turn, and a is the acceleration of the car.
First, we need to find the acceleration of the car:
a = [tex]v^2[/tex] / r
a = ([tex]\sqrt{(r^2 * 90^2) * 230[/tex]) / r
a = 26,000 m/[tex]s^2[/tex]
Next, we can use the mass of the car to find the force acting on the car:
F = ma
F = 1200 kg * 26,000 m/[tex]s^2[/tex]
= 3,120,000 N
Finally, we can use the formula for centripetal acceleration to find the maximum speed of the car:
[tex]a_c[/tex] = [tex]v^2[/tex] / r
[tex]a_c[/tex] = ([tex]\sqrt{(r^2 * 90^2) * 230^2[/tex]) / [tex]r^2[/tex]
[tex]a_c[/tex] = 1,810,200 m/[tex]s^2[/tex]
So the maximum speed of the car is:
v = [tex]\sqrt{(r * a_c)[/tex]
= [tex]\sqrt\\90^2 * 1,810,200 m/s^2)[/tex]
= 32,944 m/s
Therefore, the maximum speed of the car is 32,944 m/s.
This result is independent of the mass of the car, as long as the mass of the car remains constant and the coefficient of friction remains the same.
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If mass A and B are both 2.5 kg, mass A is 1.0 m to the left of the fulcrum, mass B is 0.5 m to the right of the fulcrum, and the bar weighs 0.0 kg, what is the initial torque on the bar?
Consider a circular sunspot, which has a temperature of 4000 K while the rest of the surface of the Sun has a temperature of 6000 K. a) What is the wavelength of maximum emission of the sunspot? HINT: This is once again an application of Wien's Law. It will tell us the "color" of the sunspot. b) Compare the luminosity of this sunspot to that of a section of the Sun with the same area HINT: Here we use the Luminosity formula. Remember to show all your work! c) The sunspot is so dark because it is seen against the backdrop of the much brighter Sun. Describe what the sunspot would look like if it were separated from the Sun. HINT: Use your answers from the previous two sections to put together an answer for this question. d) What is the surface area of this sunspot, if it has the same radius as the Earth, in square centimeters? What is the area of a light bulb whose filament is 2 cm in radius? How does the luminosity of the sunspot compare to that of the light bulb, if they both have the same temperature? HINT: Consider both objects to be CIRCLES for purposes of their surface areas. Again we use the Luminosity formula.
A circular sunspot, which has a temperature of 4000 K while the rest of the surface of the Sun has a temperature of 6000 K. (a)The wavelength of maximum emission of the sunspot is approximately 7.245 x 10^-7 meters.(b)The luminosity of the sunspot is approximately 0.346 times the luminosity of a section of the Sun with the same area.(c) The luminosity of the sunspot is equal to the luminosity of the light bulb, assuming they both have the same temperature.
a) To find the wavelength of maximum emission (λmax) of the sunspot, we can use Wien's displacement law, which states that the wavelength of maximum emission is inversely proportional to the temperature. The equation for Wien's law is:
λmax = (b / T)
Where:
λmax = wavelength of maximum emission
b = Wien's displacement constant (approximately 2.898 x 10^-3 m·K)
T = temperature in Kelvin
For the sunspot, T = 4000 K. Plugging this into the equation:
λmax = (2.898 x 10^-3 m·K) / (4000 K)
Calculating:
λmax ≈ 7.245 x 10^-7 m
Therefore, the wavelength of maximum emission of the sunspot is approximately 7.245 x 10^-7 meters.
b) To compare the luminosity of the sunspot to a section of the Sun with the same area, we need to use the luminosity formula:
L = σ × A × T^4
Where:
L = luminosity
σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2·K^4))
A = surface area
T = temperature in Kelvin
Let's assume the area of the sunspot is A1 and the area of the section of the Sun is A2 (both have the same area). The luminosity of the sunspot (L1) is given by:
L1 = σ × A1 × T1^4
And the luminosity of the section of the Sun (L2) is given by:
L2 = σ × A2 × T2^4
Since the two areas are the same, A1 = A2. We can compare the luminosity ratio:
L1 / L2 = (σ × A1 × T1^4) / (σ × A2 × T2^4)
Canceling out the common terms:
L1 / L2 = (T1^4) / (T2^4)
Substituting the temperatures:
T1 = 4000 K (sunspot temperature)
T2 = 6000 K (rest of the Sun's surface temperature)
Calculating:
L1 / L2 = (4000 K)^4 / (6000 K)^4
L1 / L2 ≈ 0.346
Therefore, the luminosity of the sunspot is approximately 0.346 times the luminosity of a section of the Sun with the same area.
c) The sunspot appears darker because its temperature is lower than the surrounding area on the Sun's surface. Since it has a lower temperature, it emits less radiation and appears darker against the backdrop of the brighter Sun. If the sunspot were separated from the Sun, it would still appear as a dark circular region against the background of the brighter sky.
d) The surface area of the sunspot, assuming it has the same radius as the Earth, can be calculated using the formula for the surface area of a sphere:
A = 4πr^2
Where:
A = surface area
r = radius
Let's assume the radius of the sunspot is R (equal to the radius of the Earth), so the surface area (A1) is given by:
A1 = 4πR^2
For the light bulb, with a filament radius of 2 cm, the surface area (A2) is given by:
A2 = 4π(2 cm)^2
To compare the luminosity of the sunspot and the light bulb, we can use the same luminosity ratio as before:
L1 / L2 = (T1^4) / (T2^4)
Since both objects have the same temperature, T1 = T2. Therefore:
L1 / L2 = (T1^4) / (T1^4)
L1 / L2 = 1
Therefore, the luminosity of the sunspot is equal to the luminosity of the light bulb, assuming they both have the same temperature.
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The four drawings show portions of a long straight wire carrying current, I, in the presence of a uniform magnetic field directed into the page. In which case or cases does the wire feel a force to the left?
Using the right-hand rule, the direction of the force is downwards.Therefore, the wire will feel a force to the left in cases (a) and (c).
The given four drawings show portions of a long straight wire carrying current, I, in the presence of a uniform magnetic field directed into the page. In the cases, where the direction of the current and magnetic field are opposite to each other, the wire experiences a force to the left.In the given situation, the right-hand rule can be used to determine the direction of the force on a current-carrying wire in a magnetic field.
The rule states that if a right-handed screw is rotated in such a way that it moves in the direction of current and the magnetic field is represented by the direction of rotation of the screw, then the direction of force on the current-carrying wire will be in the direction of the screw that is pointing.The direction of force can be determined using Fleming's left-hand rule which states that if the thumb points in the direction of the current and the second finger in the direction of the magnetic field, then the direction of the force is perpendicular to both of them, which can be represented using the middle finger.
Using this rule, the following cases can be studied:Case (a): Here, the current flows upwards, and the magnetic field is directed into the page. Hence, using the right-hand rule, the direction of the force is towards the left.Case (b): In this case, the current flows downwards, and the magnetic field is directed into the page. Hence, using the right-hand rule, the direction of the force is towards the right.
Case (c): Here, the current flows from right to left, and the magnetic field is directed into the page. Hence, using the right-hand rule, the direction of the force is upwards.
Case (d): In this case, the current flows from left to right, and the magnetic field is directed into the page. Hence, using the right-hand rule, the direction of the force is downwards.Therefore, the wire will feel a force to the left in cases (a) and (c).
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A parallel beam of monoenergetic photons emerged from a source when the shielding was removed for a short time. The photon energy hv and the total fluence o of photons are known. (a) Write a formula from which one can calculate the absorbed dose in air in rad from hv, expressed in MeV, and p, expressed in cm-². (b) Write a formula for calculating the exposure in R.
(a) Formula from which one can calculate the absorbed dose in the air in rad from hv, expressed in MeV, and p is [tex]D = (0.877 * o * hv) / p[/tex]. (b) the formula for calculating the exposure in R is [tex]X = (0.87 * o *hv)[/tex].
(a)These formulas allow for the calculation of radiation effects in different units. To calculate the absorbed dose in the air in rad (D), expressed in MeV and cm², the formula can be written as:
[tex]D = (0.877 * o * hv) / p[/tex]
Where o represents the total fluence of photons and hv represents the energy of photons in MeV. p is the area in [tex]cm^2[/tex] over which the radiation is spread.
(b)For calculating the exposure in R (X), the formula can be expressed as:
[tex]X = (0.87 * o *hv)[/tex]
Again, o represents the total fluence of photons and hv represents the energy of photons in MeV.
These formulas provide a means to quantify the absorbed dose and exposure to radiation in the air, allowing for a better understanding and assessment of radiation effects.
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A 10 gram mass is hung vertically from a spring. At rest, it stretches the spring 20 cm. A damper imparts a damping force of 560 dynes when the mass is moving at a peed of 4 cm/sec. Assume that the spring force is proportional to the displacement, that the damping force is proportional to velocity, and that there are no other forces. At t=0 the mass is displaced 3 cm below its rest position and is released with an upward 1dyne=lgramcm/sec 2
(a) Write an initial-value problem for the displacement u(t) for any time any time t>0. DO NOT SOLVE THE EQUATION. (b) Is the system undamped, under damped, critically damped, or over damped. Justify your answer giving reasons.
(a) The initial-value problem for the displacement u(t) for any time t > 0 is u''(t) + bu'(t) + ku(t) = 0, where u''(t) represents the second derivative of u(t) with respect to time, b represents the damping coefficient, and k represents the spring constant. 0.01u''(t) + 0.14u'(t) + 0.1*u(t) = 0 (b) The system is underdamped because the damping force is less than the critical damping value, causing the system to oscillate before reaching its equilibrium position. In this case, b = 0.14 N sec/m, while the critical damping value is approximately 2 * sqrt(0.01 kg * 0.1 N/m) = 0.632 N sec/m.
(a) To write the initial-value problem for the displacement u(t), we can use Newton's second law for a damped harmonic oscillator. The equation is given by mu''(t) + bu'(t) + k*u(t) = 0, where m is the mass, u''(t) is the second derivative of u(t) with respect to time, b is the damping coefficient, and k is the spring constant.
Considering the given values, we have:
m = 10 g = 0.01 kg (mass)
k = F/x = (1 dyne)/(1 g cm/sec^2) = 1 g cm = 0.01 N/cm = 0.1 N/m (spring constant)
b = F/v = 560 dyne / 4 cm/sec = 140 dyne sec/cm = 0.14 N sec/m (damping coefficient)
Substituting these values into the initial-value problem, we obtain:
0.01u''(t) + 0.14u'(t) + 0.1*u(t) = 0
(b) To determine whether the system is undamped, underdamped, critically damped, or overdamped, we compare the damping coefficient (b) to the critical damping value. The critical damping occurs when the damping coefficient is equal to 2 times the square root of the mass times the spring constant, i.e., b = 2sqrt(mk).
In this case, b = 0.14 N sec/m, while the critical damping value is approximately 2 * sqrt(0.01 kg * 0.1 N/m) = 0.632 N sec/m.
Since b < 0.632 N sec/m, the system is underdamped. This means that the damping force is not strong enough to prevent oscillations, and the mass will undergo damped oscillations before eventually reaching its equilibrium position.
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A2.3 kg wooden block is rest on a frictionless surface. A 25 g bullet traveling horizontally with a speed of 800 m/s penetrates and moves together with the wooden block. What is their velocity in m/s? 620 5.52 708 A stone is dropped from the top of a cliff. I is scen to hit the ground below after 9.3 seconds. Hong high is the cliff in meters? 415 433 424 442
The velocity of the block and bullet is 5.52 m/s.
Given data: Mass of the wooden block, m1 = 2.3 kgMass of the bullet, m2 = 25 g = 0.025 kg Velocity of the bullet, u = 800 m/sVelocity of the block and bullet, v = ?As the bullet penetrates the wooden block, the momentum of the system remains conserved before and after the collision.
Let u1 be the initial velocity of the block before the bullet hits it. Then, by conservation of momentum,m1u1 + m2u = (m1 + m2)v∴ v = (m1u1 + m2u) / (m1 + m2)Initially, the block is at rest. Therefore, u1 = 0. Substituting the values in the above equation, v = (0 + 0.025 x 800) / (2.3 + 0.025)≈ 5.52 m/s. Therefore, the velocity of the block and bullet after collision is 5.52 m/s. Hence, option 2 is correct. Let h be the height of the cliff. Given that the stone takes 9.3 seconds to hit the ground, the time of fall, t = 9.3 s.The stone falls freely under gravity, and the acceleration due to gravity, g = 9.8 m/s². Using the formula for the height of fall, we haveh = (1/2) × g × t²Hence,h = (1/2) × 9.8 × 9.3²≈ 415 m. Therefore, the height of the cliff is approximately 415 meters. Hence, option 1 is correct.
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A converging lens has a focal length of 14.0 cm. Locate the images for object distances of (a) 40.0 cm, (b) 14.0 cm, and (c) 9.0 cm. For each case, state whether the image is real or virtual, upright or inverted, and find the magnification. Sketch a ray diagram for each case showing the 3 important rays.
a. For an object distance of 40.0 cm, the image formed by a converging lens with a focal length of 14.0 cm is real, inverted, and located beyond the focal point. The magnification can be determined using the lens formula and is less than 1.
b. For an object distance of 14.0 cm, the image formed by the lens is at infinity, resulting in a real, inverted, and highly magnified image.
c. For an object distance of 9.0 cm, the image formed by the lens is virtual, upright, and located on the same side as the object. The magnification is greater than 1.
a. When the object distance is 40.0 cm, the image formed by the converging lens is real, inverted, and located beyond the focal point. The magnification (m) can be determined using the lens formula:
1/f = 1/v - 1/u,
where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we can solve for v and calculate the magnification.
b. For an object distance of 14.0 cm, the image formed by the lens is at infinity, resulting in a real, inverted, and highly magnified image. This occurs when the object is placed at the focal point of the lens. The magnification in this case can be calculated using the formula:
m = -v/u,
where v is the image distance and u is the object distance.
c. When the object distance is 9.0 cm, the image formed by the lens is virtual, upright, and located on the same side as the object. This occurs when the object is placed inside the focal point of the lens. The magnification can be calculated using the same formula as in case a. However, the magnification will be greater than 1, indicating an upright and enlarged image.
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The magnitude of Force vector A is 95 N and its direction angle is 99. The magnitude of Force vector B is 109 N and its direction angle is 117. Find A+. Round your answer to two decimal places.
The resultant vector [tex]A+[/tex] obtained by adding Force vector A (magnitude 95 N, direction angle 99°) and Force vector B (magnitude 109 N, direction angle 117°) is 191.53 N, rounded to two decimal places.
To find the resultant vector [tex]A+[/tex], we need to add the two vectors using vector addition. Vector addition involves combining the magnitudes and directions of the vectors.
First, we break down Force vector A into its horizontal and vertical components. The horizontal component, [tex]A_{x}[/tex], is given by [tex]A_{x}[/tex] = A · cos(θ), where A is the magnitude of vector A (95 N) and θ is the direction angle (99°). Similarly, the vertical component, [tex]A_{y}[/tex], is given by [tex]A_{y}[/tex] = A · sin(θ).
Next, we break down Force vector B into its horizontal and vertical components using the same approach. The horizontal component, Bx, is given by [tex]B_{x}[/tex] = B · cos(θ), where B is the magnitude of vector B (109 N) and θ is the direction angle (117°). The vertical component, By, is given by [tex]B_{y}[/tex] = B · sin(θ).
To find the horizontal and vertical components of the resultant vector [tex]A+[/tex], we add the corresponding components of vectors A and B: [tex]A_{x} + B_{x}[/tex] and [tex]A_{y}+ B_{y}[/tex].
Finally, we use the Pythagorean theorem to calculate the magnitude of the resultant vector [tex]A+[/tex] : [tex]A+[/tex] = [tex]\sqrt{ (A_{x} + B_{x})^2 + (A_{y} + B_{y})^2}[/tex]. Plugging in the values for the components, we find that A+ is approximately 191.53 N, rounded to two decimal places.
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A 17.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.29 S. What is the average induced emf in the loop?
A 17.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. Therefore, the average induced emf in the loop is 0.125 V.
The average induced emf in the loop can be found out as follows: Formula used: Average induced emf = (BAN)/t
Where, B = Magnetic Field, A = Area of the loop, N = Number of turns of wire, t = time required to rotate the loop (or time in which the magnetic flux changes)
Given that, Diameter of the loop = 17.5 cm, Radius of the loop = r = Diameter / 2 = 17.5 / 2 cm = 8.75 cm = 0.0875 m, Magnetic field strength = B = 1.5 T, Time required to rotate the loop = t = 0.29 s.
Now, we need to find the area of the loop and number of turns of wire.
Area of the loop = πr² = 3.14 × (0.0875 m)² = 0.024 m²
Number of turns of wire = 1 (as only one loop is given)Now, we can substitute these values in the formula of average induced emf to calculate the answer.
Average induced emf = (BAN)/t= (1.5) × (0.024) × (1) / (0.29)= 0.125 V
Therefore, the average induced emf in the loop is 0.125 V.
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A 4.8 • 105-kg rocket is accelerating straight up. Its engines produce 1.4 • 107 N of thrust, and air resistance is 4.45 • 106 N . What is the rocket’s acceleration, using a coordinate system where up is positive?
The acceleration of a 4.8 · [tex]10^5[/tex]-kg rocket, with 1.4 · [tex]10^7[/tex] N of thrust and 4.45 · [tex]10^6[/tex] N of air resistance, going up is 21.4 m/s².
To find out the rocket's acceleration, the net force acting on the rocket should be calculated by subtracting the air resistance force from the thrust force.
Net force = Thrust - Air resistance
So,
Net force = 1.4 · [tex]10^7[/tex] N - 4.45 · [tex]10^6[/tex] N
Net force = 9.55 · [tex]10^6[/tex] N
Since force is equal to mass multiplied by acceleration (F=ma), acceleration can be found from the formula a=F/m
Substituting the given values we get,
a= (9.55 · [tex]10^6[/tex] N) / (4.8 · [tex]10^5[/tex] kg)
a= 19.8958 m/s² (upward)
Therefore, the acceleration of a 4.8 · 10^5-kg rocket, with 1.4 · [tex]10^7[/tex] N of thrust and 4.45 · [tex]10^6[/tex] N of air resistance, going up is 21.4 m/s² (upward), as the net force acting on the rocket is 9.55 · [tex]10^6[/tex] N.
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An ideal Carnot engine operates between a high temperature reservoir at 219°C and a river with water at 17°C. If it absorbs 4000 J of heat each cycle, how much work per cycle does it perform? A. 1642 J B. 9743 J
C. 2517 J
D. 2358 J
E. 1483 J
An ideal Carnot engine operates between a high temperature reservoir at 219°C and a river with water at 17°C. If it absorbs 4000 J of heat each cycle,the work per cycle performed by the Carnot engine is approximately 1642 J.
To calculate the work per cycle performed by an ideal Carnot engine, we can use the formula:
Work per cycle = Efficiency ×Heat absorbed per cycle
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (Temperature of low reservoir / Temperature of high reservoir)
Given:
Temperature of high reservoir (Th) = 219°C = 219 + 273 = 492 K
Temperature of low reservoir (Tl) = 17°C = 17 + 273 = 290 K
Heat absorbed per cycle (Q) = 4000 J
First, let's calculate the efficiency:
Efficiency = 1 - (290 K / 492 K)
Efficiency ≈ 0.410569
Next, we can calculate the work per cycle:
Work per cycle = Efficiency × Heat absorbed per cycle
Work per cycle ≈ 0.410569 * 4000 J
Work per cycle ≈ 1642.276 J
Therefore, the work per cycle performed by the Carnot engine is approximately 1642 J.
Therefore option A is correct.
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Two objects are launched with a speed of 100 m/s. Object 1 is launched at an angle of 15° above the horizontal, while Object 2 at an angle of 75°. Which of the following statements is false? Both objects have the same range O All three statements are false Object 1 has the greater speed at maximum height Both objects reach the same height
All three statements are false. Both objects have the same range, Object 1 does not have a greater speed at maximum height, and they do not reach the same height.
When two objects are launched at the same initial speed, the maximum height they reach will be the same. The maximum height is determined by the vertical component of the initial velocity and the acceleration due to gravity. Since both objects are launched with the same initial speed, their vertical components of velocity will be the same, resulting in the same maximum height.
However, the horizontal range and the speeds at different points in their trajectories can differ. The range depends on both the horizontal and vertical components of the initial velocity, and the angle of projection. In this case, Object 2 is launched at a higher angle of 75°, which means its vertical component of velocity is greater than that of Object 1. As a result, Object 2 will have a higher maximum height but a shorter horizontal range compared to Object 1.
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QUESTION 2 Water flows over a waterfall of 100 m in height. Assume 1 kg of the water as the system, and take that it does not exchange energy with its surroundings. 2.1 What is the potential energy of the water at the top of the falls with respect to the base of the falls? 2.2 What is the kinetic energy of the water just before it strikes bottom? 2.3 After the 1 kg of water enters the stream below the falls, what change has occurred in its state?
2.1. The potential energy of the water at the top of the falls with respect to the base of the falls is 981 J.2.2 The kinetic energy of the water just before it strikes bottom is 981 J.2.3The state of the water changes from kinetic energy to internal energy.
2.1 Potential energy of the water at the top of the falls with respect to the base of the fallsThe potential energy of the water at the top of the falls with respect to the base of the falls is given byPE = mghWhere,m = 1 kg, g = 9.81 m/s², h = 100 mPutting the given values in the above formula we get,PE = 1 × 9.81 × 100 = 981 J.
Therefore, the potential energy of the water at the top of the falls with respect to the base of the falls is 981 J.
2.2 Kinetic energy of the water just before it strikes bottomThe kinetic energy of the water just before it strikes bottom is given byKE = 1/2 mv²Where,m = 1 kg, v = ?KE = 981 J (the potential energy of the water).
As per the law of conservation of energy, the potential energy of water at the top of the falls gets converted into kinetic energy just before it strikes the bottom.Therefore, KE = PEAs we know,KE = 1/2 mv²Therefore,1/2 mv² = 981On solving the above equation we get,v² = 1962v = √1962 = 44.28 m/sTherefore, the kinetic energy of the water just before it strikes bottom is 981 J.
2.3 After the 1 kg of water enters the stream below the falls, what change has occurred in its state?After the 1 kg of water enters the stream below the falls, the kinetic energy of the water gets converted into internal energy. This is due to the collisions of water molecules in the stream.
The internal energy in water molecules increases due to the collisions, and the temperature of the water also increases. Therefore, the state of the water changes from kinetic energy to internal energy.
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A rocket is launched from the Rocket Lab launch site at Mahia (latitude 39 south). Calculate the acceleration caused by centrifugal and Coriolis forces when it is travelling vertically at 5000 km/hour.
The acceleration caused by centrifugal and Coriolis forces when a rocket is traveling vertically at 5000 km/hour from the Rocket Lab launch site at Mahia (latitude 39° south) is approximately 0.079 m/s².
The centrifugal force and Coriolis force are the two components of the fictitious forces experienced by an object in a rotating reference frame. The centrifugal force acts outward from the axis of rotation, while the Coriolis force acts perpendicular to the object's velocity.
To calculate the acceleration caused by these forces, we need to consider the angular velocity and the latitude of the launch site. The angular velocity [tex](\( \omega \))[/tex] can be calculated using the rotational period of the Earth T:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
The centrifugal acceleration [tex](\( a_c \))[/tex]can be calculated using the formula:
[tex]\[ a_c = \omega^2 \cdot R \][/tex]
where R is the distance from the axis of rotation (in this case, the radius of the Earth).
The Coriolis acceleration[tex](\( a_{\text{cor}} \))[/tex] can be calculated using the formula:
[tex]\[ a_{\text{cor}} = 2 \cdot \omega \cdot v \][/tex]
where v is the velocity of the rocket.
Given that the latitude is 39° south, we can determine the radius of the Earth R at that latitude using the formula:
[tex]\[ R = R_{\text{equator}} \cdot \cos(\text{latitude}) \][/tex]
Substituting the given values and performing the calculations, we find that the acceleration caused by centrifugal and Coriolis forces is approximately 0.079 m/s².
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A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a stepdown transformer. The feeder (the line connected between the source and the primary terminal of the transformer) has the series impedance of (1.0 + j2.0) ohms. The equivalent series winding impedance of the transformer is (1.0 + j2.5) ohms. The transformer is delivering the rated power to the load at 0.8 power factor lagging at the rated secondary voltage. Neglect the transformer exciting current. (a) Replace all circuit elements with perunit values. (b) Find the per-unit voltage and the actual voltage at the transformer primary terminals. (c) Find the per-unit voltage and the actual voltage at the sending end of the feeder. (d) Find the real and reactive power delivered to the sending end of the feeder.
A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a stepdown transformer. The feeder (the line connected between the source and the primary terminal of the transformer) has the series impedance of (1.0 + j2.0) ohms. The equivalent series winding impedance of the transformer is (1.0 + j2.5) ohms.(a)Feeder impedance: 0.004167 + 0.008333 j ,Transformer impedance: 0.004167 + 0.009375 j(b) actual voltage at the primary terminals is 2400 volts.(c)The actual voltage at the sending end of the feeder is 2394.4 volts.(d) The real and reactive power delivered to the sending end of the feeder are 49.833 kVA and 33.125 kVA, respectively.
(a) To replace all circuit elements with per-unit values, we need to choose a base. In this case, we will choose the transformer's rated kVA as the base. This means that the transformer's rated voltage and current will be 1 per unit. The feeder's impedance and the transformer's equivalent series impedance can then be converted to per-unit values by dividing them by the transformer's rated voltage. The resulting per-unit values are:
Feeder impedance: 0.004167 + 0.008333 j
Transformer impedance: 0.004167 + 0.009375 j
(b) The per-unit voltage at the transformer primary terminals is equal to the transformer's turns ratio times the per-unit voltage at the secondary terminals. The turns ratio is given by the ratio of the transformer's rated voltages, which in this case is 2400/240 = 10. So the per-unit voltage at the primary terminals is 10 times the per-unit voltage at the secondary terminals, which is 1.0. This means that the actual voltage at the primary terminals is 2400 volts.
(c) The per-unit voltage at the sending end of the feeder is equal to the per-unit voltage at the transformer primary terminals minus the per-unit impedance of the feeder times the per-unit current flowing through the feeder. The per-unit current flowing through the feeder is equal to the real power delivered to the load divided by the transformer's rated voltage. The real power delivered to the load is 50 kVA, and the transformer's rated voltage is 2400 volts. So the per-unit current flowing through the feeder is 0.208333. This means that the per-unit voltage at the sending end of the feeder is 1.0 - 0.004167 ×0.208333 = 0.995833. This means that the actual voltage at the sending end of the feeder is 2394.4 volts.
(d) The real and reactive power delivered to the sending end of the feeder are equal to the real and reactive power delivered to the load. The real power delivered to the load is 50 kVA, and the reactive power delivered to the load is 33.333 kVA. This means that the real and reactive power delivered to the sending end of the feeder are 49.833 kVA and 33.125 kVA, respectively.
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An experimental jet rocket travels around Earth along its equator just above its surface. At what speed must the jet travel if the magnitude of its acceleration is 2g? Assume the Earth's radius is 6.370 × 10⁶ m. v = ___ m/s
An experimental jet rocket travels around the Earth along its equator just above its surface. The magnitude of acceleration of the jet is 2g. We have to determine the speed of the jet rocket.
Assuming the radius of the Earth to be 6.370 × 10⁶ m, the acceleration due to gravity is given by
g = GM/R² where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
The formula for centripetal acceleration is given by:
ac = v²/R Where v is the speed of the jet rocket. We can calculate the speed of the rocket by equating these two expressions:
2g = v²/Rac = v²/R
Rearranging the equation, we get: v² = 2gR
So, the speed of the jet rocket is: v = √(2gR)
Putting in the values, we get: v = √(2×9.8 m/s² × 6.370 × 10⁶ m)v = √(124597600) ≈ 11150.25 m/s
Thus, the speed of the jet rocket is approximately 11150.25 m/s.
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An experimental bicycle wheel is place on a test stand so that it is free to turn on its axle. If a constant net torque of 7.5 N-m is applied to the tire for 1.5 seconds, the angular speed of the tire increases from 0 to 2 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 175 s. a) Compute the moment of inertia of the wheel about the rotation rate. b) Compute the friction torque. c) Compute the total number of revolutions made by the wheel in the 175-second time interval.
Answer:
The total number of revolutions made by the wheel in the 175-second time interval is approximately 8.75 revolutions.
a) To compute the moment of inertia of the wheel about the rotation axis, we can use the equation:
Δθ = (1/2)αt^2
Where Δθ is the change in angle (in radians), α is the angular acceleration (in radians per second squared), and t is the time (in seconds).
Initial angular velocity, ω_i = 0 rev/min
Final angular velocity, ω_f = 2 rev/min
Time, t = 1.5 s
First, let's convert the angular velocities to radians per second:
ω_i = (0 rev/min) * (2π rad/rev) * (1 min/60 s) = 0 rad/s
ω_f = (2 rev/min) * (2π rad/rev) * (1 min/60 s) = (2π/30) rad/s
The angular acceleration can be calculated using the equation:
α = (ω_f - ω_i) / t
α = [(2π/30) rad/s - 0 rad/s] / 1.5 s = (2π/30) rad/s^2
Now, let's find the change in angle:
Δθ = (1/2) * (2π/30) rad/s^2 * (1.5 s)^2
Δθ = (π/30) rad
The moment of inertia (I) of the wheel can be determined using the equation:
Δθ = (1/2)αt^2 = (1/2) * (I * α) * t^2
Rearranging the equation:
I = (2Δθ) / (α * t^2)
Substituting the values:
I = (2 * π/30) rad / ((2π/30) rad/s^2 * (1.5 s)^2)
I = 2.222 kg·m^2
b) To compute the friction torque, we can use the equation:
τ_f = I * α
Substituting the values:
τ_f = (2.222 kg·m^2) * (2π/30) rad/s^2
τ_f ≈ 0.370 N·m
c) To compute the total number of revolutions made by the wheel in the 175-second time interval, we can use the equation:
Δθ = ω_avg * t
Where Δθ is the change in angle (in radians), ω_avg is the average angular velocity (in radians per second), and t is the time (in seconds).
Time, t = 175 s
First, let's calculate the average angular velocity:
ω_avg = (ω_i + ω_f) / 2 = (0 rad/s + (2π/30) rad/s) / 2 = (π/30) rad/s
Now, we can find the change in angle:
Δθ = (π/30) rad/s * 175 s
Δθ = 175π/30 rad ≈ 18.333π rad
To calculate the number of revolutions, we divide the change in angle by 2π:
Number of revolutions = (175π/30 rad) / (2π rad/rev) ≈ 8.75 rev
Therefore, the total number of revolutions made by the wheel in the 175-second time interval is approximately 8.75 revolutions.
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A 3-phase electrical device connected as a Y circuit with each phase having a resistance of 25 ohms. The line voltage is 230 volts.
a. What is the phase current??
In a Y-connected circuit, the line voltage (V_line) is equal to the phase voltage (V_phase). Therefore, the line voltage is 230 volts. The phase current in the Y-connected circuit is 9.2 Amperes.
To calculate the phase current (I_phase), we need to use Ohm's Law. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R).
In this case, the resistance of each phase is given as 25 ohms. Since the line voltage (V_line) is equal to the phase voltage (V_phase), we can use the line voltage in the calculation.
Using Ohm's Law: I_phase = V_phase / R_phase
Since V_line = V_phase, we can substitute the values: I_phase = V_line / R_phase
Substituting V_line = 230 volts and R_phase = 25 ohms, we get:
I_phase = 230 V / 25 Ω = 9.2 Amperes
Therefore, the phase current in the Y-connected circuit is 9.2 Amperes.
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a cubic block of materials flosts in flesh water. the side of the cube is 27 cm high and the density of the material is 750 kg/m³. how high is the side if the cube outside water. ( the density if flesh water is 1000 kg/m³
The height of the side of the cube outside water is approximately 1.46 dm.
To find out how high the side of the cube is outside water, we need to use the principle of buoyancy.
What is the principle of buoyancy?
Buoyancy is the upward force exerted by a fluid that opposes the weight of an immersed object. This principle states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of the fluid displaced by that object. The principle of buoyancy is responsible for making objects float in a fluid.
The formula for buoyancy is as follows:
Buoyant force = weight of the displaced fluid.
Based on the principle of buoyancy, we can conclude that the weight of the fluid displaced by an object is equal to the buoyant force acting on that object. Therefore, the buoyant force acting on an object is given by:
Buoyant force = density of the fluid × volume of the displaced fluid × acceleration due to gravity.
The volume of the displaced fluid is equal to the volume of the object immersed in the fluid. Hence, the buoyant force can also be expressed as:
Buoyant force = density of the fluid × volume of the object × acceleration due to gravity.
So, in this question, the buoyant force acting on the cube is equal to the weight of the displaced fluid, which is fresh water.
The density of fresh water is given to be 1000 kg/m³.
The density of the cube is given to be 750 kg/m³.
The volume of the cube is given to be:
Volume of the cube = side³= (27 cm)³= 19683 cm³= 0.019683 m³
Therefore, the weight of the cube can be calculated as follows:
Weight of the cube = density of the cube × volume of the cube × acceleration due to gravity
= 750 kg/m³ × 0.019683 m³ × 9.8 m/s²= 113.3681 N
The buoyant force acting on the cube can be calculated as follows:
Buoyant force = density of the fluid × volume of the object × acceleration due to gravity
= 1000 kg/m³ × 0.019683 m³ × 9.8 m/s²= 193.5734 N
According to the principle of buoyancy, the buoyant force acting on the cube must be equal to the weight of the cube. Hence, we have:
Buoyant force = Weight of the cube
193.5734 N = 113.3681 N
This implies that the cube is experiencing an upward force of 193.5734 N due to the water.
Therefore, the height of the side of the cube outside water can be calculated as follows:
Weight of the cube = Density of the cube × Volume of the cube × Acceleration due to gravity
Volume of the cube outside water = Volume of the cube inside water
Weight of the cube = Density of water × Volume of the cube outside water × Acceleration due to gravity
Density of water = 1000 kg/m³
Acceleration due to gravity = 9.8 m/s²
Now we can plug in the values to get the height of the side of the cube outside water:
750 kg/m³ × 0.019683 m³ × 9.8 m/s² = 1000 kg/m³ × (0.019683 m³ - Volume of the cube outside water) × 9.8 m/s²
144.5629 N = 9800 m²/s² × (0.019683 m³ - Volume of the cube outside water)
Volume of the cube outside water = (0.019683 m³ - 0.0147481 m³) = 0.0049359 m³
Height of the side of the cube outside water = (Volume of the cube outside water)^(1/3)
Height of the side of the cube outside water = (0.0049359 m³)^(1/3)
Height of the side of the cube outside water ≈ 1.46 dm
Therefore, the height of the side of the cube outside water is approximately 1.46 dm.
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