Answer:
17.92 Ω
Explanation:
R₀ = Initial resistance of the aluminum wire at 30.0°C = 7.00 Ω
R = resistance of the aluminum wire at 430.0°C = ?
α = temperature coefficient of resistivity for aluminum = 3.90 x 10⁻³ °C⁻¹
ΔT = Change in temperature = 430 - 30 = 400 °C
Resistance of the wire is given as
R = R₀ (1 + α ΔT)
R = (7) (1 + (3.90 x 10⁻³) (400))
R = 17.92 Ω
A windmill is turning at 12.5 rad/s in a clockwise direction. The wind suddenly shifts, producing an angular acceleration of 0.75 rad/s2 in a counterclockwise direction. After 12.0 s, how many revolutions has the windmill blades turned?
Answer:
Windmill will complete 0.27 revolutions.
Explanation:
We have equation of motion s = ut + 0.5at²
Initial speed, u = 12.5 rad/s
Time, t = 12 seconds.
Angular acceleration, a = -0.75 rad/s²
Substituting
s = 12.5 x 12 - 0.5 x 0.75 x 12² = 96 rad
1 revolution = 360 rad
[tex]96rad=\frac{96}{360}=0.27revolution[/tex]
So, windmill will complete 0.27 revolutions.
Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?
Answer:
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
At the time of collision velocity of ball one is descending.
Explanation:
Velocity of ball 1 = 146 ft/sec = 44.50m/s
The balls are to collide at an altitude of 234 ft
H = 234 ft = 71.32 m
We have equation of motion
v² = u² + 2as
v² = 44.50² + 2 x (-9.81) x 71.32
v = ±24.10 m/s.
Time for each velocity can be calculated using equation of motion
v = u + at
24.10 = 44.50 - 9.81 t , t = 2.07 s
-24.10 = 44.50 - 9.81 t , t = 6.99 s
Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.
So at the time of collision velocity of ball one is descending.
The collision occurs at t = 6.99 s.
Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.
Height traveled by ball 2 = 71.32 m
We need to find velocity
We have
s = ut + 0.5 at²
71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²
u = 38.21 m/s = 125.36 ft/s
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
Answer:
v2=139 ft
Explanation:
First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.
h_1=v_1*t-(a*t/2)
We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.
t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s
Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.
Solving a similar equation, but this time for v2 to obtain the result.
h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft
The momentum of an isolated system is conserved A) only in inelastic collisions. B) only in elastic collisions. C) in both elastic and inelastic collisions
Answer:
It is actually in both.
Explanation:
The momentum of an isolated system is conserved for both elastic and inelastic collision. OPtion C is correct
Collision is the process by which two bodies come into contact with the release of energy.
Collision can be elastic or inelastic.
Foe elastic collision, both momentum, and energy is conserved while for inelastic collision only momentum is conserved.
From the explanation above, we can see that momentum of an isolated system is conserved for both elastic and inelastic collision
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A gun with a muzzle velocity of 100 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange will the bullet be 8 seconds later? g
Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
[tex]y(t)-h = \frac{1}{2}gt^2[/tex]
where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]
So the bullet has travelled 313.6 m downward.
f a vector has magnitude V_1 = 46 N, and direction theta_1 = 30 degree Another vector V_2 = 63 N, and direction theta_2 = 151 degree. Calculate the reference angle of the vector V_1 + V_2. Round your answer such that it has no decimal.
Answer:
105.8 degree from + X axis
Explanation:
V1 = 46 N at 30 degree
V2 = 63 N at 151 degree
Write the vector form
V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j
V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j
The resultant of V1 and V2 is given by
V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j
The angle made by resultant of V1 and V2 with X axis is
tan∅ = 53.54 / (- 15.16) = - 3.53
∅ = 105.8 degree from + X axis.
A wire carrying a current is shaped in the form of a circular loop of radius 3.0mm If the magnetic field strength that this current produces at the center of the loop is 1.1mT, what is the magnitude of the current that flows through the wire? (μo = 4π x10^-7 T. m/A) A) 5.3A B) 16A C) 23 A D) 9.1A
Final answer:
The magnitude of the current that flows through the wire is approximately 16A.
Explanation:
To calculate the magnitude of the current that flows through the wire, we can use the formula for the magnetic field produced by a current-carrying circular loop: B = μ0 * I / (2 * R), where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and R is the radius of the loop.
Given that the radius of the loop is 3.0 mm and the magnetic field strength at the center of the loop is 1.1 mT, we can rearrange the formula to solve for I:
I = (2 * B * R) / μ0
Substituting the given values, we get:
I = (2 * 1.1 x 10^-3 T * 3.0 x 10^-3 m) / (4π x 10^-7 T.m/A)
Simplifying the expression, we find that the magnitude of the current is approximately 16 A. Therefore, the correct answer is B) 16A.
The bar of Prob. 5/82 is repeated here. The ends of the 0.4-m bar remain in contact with their re- spective support surfaces. End B has a velocity of 0.5 m/s and an acceleration of 0.3 m/s2 in the di- rections shown. Determine the angular accelera- tion of the bar and the acceleration of end A.
Answer:
0.24
Explanation:
w=vt
then
w1/v1=w2/v2
0.4/0.5=w2/0.3
w2= 0.3*0.4/0.5=0.24//
The angular acceleration of the bar is 0.75 rad/s², and the acceleration of end A is 0.3 m/s² in the opposite direction to end B. These values were determined using the relationships between linear and angular velocities and accelerations.
To determine the angular acceleration of the bar and the acceleration of end A, let's go through the step-by-step process:
Determine the Angular Velocity
Given that the bar has a length of 0.4 meters and the velocity of end B is 0.5 m/s, we start by calculating the angular velocity ω. The formula relating linear velocity v and angular velocity ω is:
[tex]v = \omega * r[/tex]
where r is the length of the bar. Rearranging for ω :
[tex]\omega = v / r = 0.5 m/s / 0.4 m = 1.25 rad/s[/tex]
Determine the Angular Acceleration
Next, we use the given acceleration of end B, which is 0.3 m/s². To find the angular acceleration α, we use the formula:
[tex]a = \alpha * r[/tex]
Rearranging for α:
[tex]\alpha = a / r = 0.3 m/s^2 / 0.4 m = 0.75 rad/s^2[/tex]
Determine the Acceleration of End A
The acceleration of end A, which is located at the opposite end of the bar, can be computed using:
[tex]a_A = \alpha * r[/tex]
Since end A is on the same bar, r here will be the same:
[tex]a_A = 0.75 rad/s^2* 0.4 m = 0.3 m/s^2[/tex]
Therefore, the acceleration of end A is also 0.3 m/s², but in the opposite direction to end B.
40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ) is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as given by B = 50 sin(10 t) mT where t is measured in s. What is the magnitude of the induced current in the coil at 0.10 s?
To find the magnitude of the induced current in the coil at 0.10 s, we need to calculate the magnitude of the magnetic flux through the coil and use Faraday's law of electromagnetic induction.
Explanation:To find the magnitude of the induced current in the coil at 0.10 s, we need to first calculate the magnitude of the magnetic flux through the coil. The magnetic flux is given by Φ = B * A, where B is the magnetic field and A is the area of the coil. The area of the coil is calculated as A = π * r^2, where r is the radius of the coil.
Next, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. Mathematically, this can be written as emf = -dΦ/dt. Since the coil has resistance, we can use Ohm's Law to find the magnitude of the induced current, which is given by I = emf/R. Plugging in the values and calculating, we can find the magnitude of the induced current at 0.10 s.
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To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction. The magnetic field B is given as B = 50sin(10πt) mT. At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.
Explanation:To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction, which states that the induced emf (voltage) is equal to the negative rate of change of magnetic flux through the coil.
The magnetic field B is given as B = 50sin(10πt) mT, where t is measured in seconds.
The magnetic flux through the coil is given by Φ = B.A, where A is the area of the coil.
Substituting the given values, we have A = πr^2 = π(0.04^2) = 0.005 cm^2.
At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.
We can now calculate the magnetic flux through the coil using Φ = B.A.
Finally, we can use Faraday's Law to find the induced emf and divide it by the resistance of the coil to find the magnitude of the induced current.
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The angular velocity of the rear wheel of a stationary exercise bike is 5.40 rad/s at time t = 0.000 sec, and its angular acceleration is constant and equal to 1.50 rad/s^2. A particular spoke coincides with the +x axis at time t = 0.000 sec. (a) What angle (in rev) does this spoke make with the +x axis at time t = 4.00 s? (b) What is the angular velocity (rev/s) at this time?
Answer:
The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.
Explanation:
Given that,
Angular velocity = 5.40 rad/s
Time t = 0.000 sec
Angular acceleration = 1.50 rad/s^2
(a). We need to calculate the angle at time t = 4.00 s
Using formula for angle
[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t^2[/tex]
Where, [tex]\omega_{0}[/tex]=angular velocity
[tex]\alpha[/tex]=angular acceleration
t = time
Put the value into the formula
[tex]\theta=5.40\times4.00+\dfrac{1}{2}\times1.50\times(4.00)^2[/tex]
[tex]\theta=33.6\ rad[/tex]
[tex]\theta=\dfrac{33.6}{2\pi}\ rad[/tex]
[tex]\theta=5.35\ rev[/tex]
(b). We nee to calculate the angular velocity at 4.00 s
Using formula of angular velocity
[tex]\omega=\omega_{0}+\alpha t[/tex]
[tex]\omega =5.40+1.50\times4.00[/tex]
[tex]\omega=11.4\ rad/s[/tex]
[tex]\omega=\dfrac{11.4}{2\pi}\ rad/s[/tex]
[tex]\omega=1.81\ rev/s[/tex]
Hence, The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.
A metal wheel 50 cm in radius rotates on a desk anti-clockwise at a constant rate of 800 rev/min about its central axis. Determine (a) its angular velocity (not speed!!!),
(b) its linear speed at a point 30 cm from its center of rotation,
(c) the radial acceleration of a point on the rim, and
(d) the total distance a point on the rim moves in 120 seconds.
Note: You must draw a diagram and pick appropriate axes to aid your calculation.
Write down the detail process.
Answer:
(a) 83.73 rad/s
(b) 251.2 m/s
(c) 3505.4 rad/s^2
(d) 5024 m
Explanation:
R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps
(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s
(b) The relation between linear speed and the angular speed is
V = r w
Here, r = 30 cm = 0.3 m
V = 0.3 x 83.73 = 25.12 m/s
(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2
(d) Time period T = 2 x 3.14 / w
T = 2 x 3.14 / 83.73 = 0.075 sec
In 0.075 second, angle turn = 360 degree
In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree
In 360 degree, the distance traveled = 2 x pi x R
In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360
= 5024 m
Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.
Answer:
Inductive reactance is 125.7 Ω
Explanation:
It is given that,
Inductance, [tex]L=50\ mH=0.05\ H[/tex]
Voltage source, V = 15 volt
Frequency, f = 400 Hz
The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :
[tex]X_L=2\pi fL[/tex]
[tex]X_L=2\pi \times 400\ Hz\times 0.05\ H[/tex]
[tex]X_L=125.66\ \Omega[/tex]
[tex]X_L=125.7\ \Omega[/tex]
So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.
To calculate the inductive reactance for a 50 mH inductor with a 400 Hz source, apply the formula XL = 2πfL, resulting in an inductive reactance of 125.6 ohms.
To determine the inductive reactance of a 50 mH inductor across a 15 volt, 400 Hz source, we use the formula for inductive reactance, which is:
L = 2πfL
where:
XL is the inductive reactance
f is the frequency of the AC source (400 Hz in this case)
L is the inductance of the coil (50 mH or 0.050 H)
Substituting the given values:
XL = 2π x 400 Hz x 0.050 H
XL = 125.6 Ω
Therefore, the inductive reactance is 125.6 ohms (Ω).
A 8.0 N force acts on a 0.30-kg object for 0.20 seconds. By how much does the object's momentum change (in kg-m/s)? (NEVER include units in the answer to a numerical question.)
Answer:
The change in momentum of an object is 1.6 kg-m/s.
Explanation:
It is given that,
Mass of the object, m = 0.3 kg
Force acting on it, F = 8 N
Time taken, t = 0.2 s
We need to find the change in momentum of an object. The change in momentum is equal to the impulse imparted on an object i.e.
[tex]\Delta p=J=F.\Delta t[/tex]
[tex]\Delta p=8\ N\times 0.2\ s[/tex]
[tex]\Delta p=1.6\ kg-m/s[/tex]
So, the change in momentum of an object is 1.6 kg-m/s. Hence, this is the required solution.
The change in the momentum of the object is 1.6 kg-m/s.
What is change in momentum?Change in momentum is the product of force acting on a body and time
To calculate the change in momentum of the object, we use the formula below
Formula:
ΔM = Ft...............Equation 1Where:
ΔM = change in momentum of the objectm = Force acting on the objectt = time.From the question,
Given:
F = 8 Nt = 0.20 secondsSubstitute these values into equation 1
ΔM = (8×0.2)ΔM = 1.6 kg-m/sHence, The change in the momentum of the object is 1.6 kg-m/s.
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The differential equation below models the temperature of a 91°C cup of coffee in a 21°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 71°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 91°C.) dy/dt =(− 1/50)(y − 21)
Answer:
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
Explanation:
We start from:
[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]
Separating variables:
[tex]-50dT=(T-21)dt[/tex]
[tex]-50\frac{dT}{T-21}=dt[/tex]
Integrating with initial conditions:
[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]
[tex]-50ln(\frac{T-21}{91-21})=t[/tex]
[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]
Isolating T:
[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)
The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.
Explanation:In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).
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A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.
Answer:
a) 0.1832 A
b) 11.91 Volts
c) 2.18 Watt , 0.0168 Watt
Explanation:
(a)
R = external resistor connected to the terminals of the battery = 65 Ω
E = Emf of the battery = 12.0 Volts
r = internal resistance of the battery = 0.5 Ω
i = current flowing in the circuit
Using ohm's law
E = i (R + r)
12 = i (65 + 0.5)
i = 0.1832 A
(b)
Terminal voltage is given as
[tex]V_{ab}[/tex] = i R
[tex]V_{ab}[/tex] = (0.1832) (65)
[tex]V_{ab}[/tex] = 11.91 Volts
(c)
Power dissipated in the resister R is given as
[tex]P_{R}[/tex] = i²R
[tex]P_{R}[/tex] = (0.1832)²(65)
[tex]P_{R}[/tex] = 2.18 Watt
Power dissipated in the internal resistance is given as
[tex]P_{r}[/tex] = i²r
[tex]P_{r}[/tex] = (0.1832)²(0.5)
[tex]P_{r}[/tex] = 0.0168 Watt
To compute the values in the question, one can utilize Ohm’s law and the power dissipation formula. The current in the circuit is 0.183 A, the terminal voltage of the battery is 11.908 V, the power dissipated in the 65.0 Ω resistor is 2.18 W, and the energy wasted in the battery's internal resistance is 0.0167 W.
Explanation:To calculate the current in the circuit, we can use Ohm's law which states that the current is equal to the voltage divided by the resistance. This gives us: I = emf/(R+r) where I is the current, emf is the electromotive force of the battery, R is the load resistance and r is the internal resistance of the battery.
Substituting the values given in the question, we get: I = 12.0 V/(65.0 Ω + 0.5 Ω) = 0.183 A. Therefore, the current flowing through the circuit is 0.183 Amperes.
Now that we have the current, we can calculate the terminal voltage of the battery. The terminal voltage (Vab) is given by the equation: Vab = emf - Ir. Substituting the values, we get: Vab = 12.0V - (0.183A * 0.5Ω) = 11.908 V. Therefore, the terminal voltage of the battery is 11.908 Volts.
Lastly, we can calculate the power dissipated in the resistor R and in the battery’s internal resistance r using the formula: Power = I²R. For the resistor, power dissipation = (0.183A)² * 65.0Ω = 2.18 Watts, and for the battery's internal resistance, power dissipation = (0.183A)² * 0.5Ω = 0.0167 Watts.
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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.9 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?
By using similar triangles concept and rates of change in Mathematics, we can find the man's shadow length is decreasing at a rate of 0.5 m/s when he is 4 meters away from the building.
Explanation:The problem you're describing is typically solved using similar triangles concept in Mathematics and rates of change. As the man walks towards the building, his shadow becomes shorter. We can set up a ratio between the man's height and his distance from the building, and the length of the shadow and the wall. When the man is 4 meters away from the building, a right-triangle is formed with the man's height (2m), his distance from the building (4m), and his shadow's length.
Let s gives the length of the shadow, then the similar triangles imply 12/s = 2/4 which gives us s = 6 meters.
To find the rate at which the man's shadow is decreasing, we can differentiate the similar triangles ratio with respect to time to get -12/s^2 ds/dt =-1/2, sub in s = 6 into the equation to find ds/dt = -0.5 m/s.
So the length of the man's shadow on the building is decreasing at a rate of 0.5 m/s when he is 4 m from the building.
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Final answer:
To find how fast the man's shadow is decreasing, we use similar triangles and related rates in calculus. When the man is 4 m from the building, his 2 m tall shadow is shortening at 3.8 m/s, twice his walking speed of 1.9 m/s.
Explanation:
The question involves determining how fast the length of a man's shadow on a building is decreasing when he walks toward the building and away from the spotlight. This requires the application of similar triangles and the concept of related rates in calculus to solve. If we let the distance between the man and the building be x, and the length of the shadow be y, as the man walks towards the building, both x and y are changing with respect to time.
Given that the man is 2 m tall and the spotlight is 12 m away from the building, we can set up a proportion between the heights of the man and his shadow, and their respective distances from the spotlight. When he is 4 m from the building, the similar triangles can be represented as 2/y = (12-x)/(12). Differentiating both sides with respect to time t yields the related rate of change for the length of the shadow y when x is 4 m and the speed of the man is known, which is 1.9 m/s.
By solving this proportion and applying the derivative, we can find that the rate at which the length of his shadow decreases is exactly double the speed at which the man is walking, because in this scenario, the man and the tip of his shadow form a straight line with the light source. Thus, the length of the shadow decreases at a rate of 3.8 m/s when the man is 4 m from the building.
The primary coil of a transformer contains100 turns; the secondary has 200 turns. The primary coil isconnected to a size-AA battery that supplies a constant voltage of1.5 volts. What voltage would be measured across the secondarycoil? zero
Answer:
There will be no voltage across secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.
Explanation:
In this question we have given
AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.
Therefore, no EMF will induce in the secondary coil
and we know that in a transformer,
[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }[/tex]...........(1)
hence from above equation it is clear that,
[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}[/tex]............(2)
Here ,
[tex]E_{2}=0[/tex]
Put [tex]E_{2}=0[/tex] in equation (2)
We got
[tex]V_{2}=0[/tex]
There will be no voltage across secondary coil because there is no electromotive force induced in secondary coil.
Acharged particle moving through a magnetic field at right angles to the field with a speed of 36.2 m/s experiences a magnetic force of 7.38x104 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 12.1 m/s at an angle of 17° relative to the magnetic field. Express your answer in microNewtons.
Answer:
7212.3 N
Explanation:
F = 7.38 x 10^4 N, v = 36.2 m/s
Let b be the strength of magnetic field and charge on the particle is q.
F = q v B Sin theta
Here theta = 90 degree
7.38 x 10^4 = q x 36.2 x B x Sin 90
q B = 2038.7 .....(1)
Now, theta = 17 degree, v = 12.1 m/s
F = q v B Sin theta
F = 2038.7 x 12.1 x Sin 17 ( q v = 2038.7 from equation (1)
F = 7212.3 N
The angle of elevation to the top of a very tall Building is found to be 6° from the ground at a distance of 1 mi from the base of the building. Using this information, find the height of the building. (Round your answer to the nearest foot.)
Answer:
555
Explanation:
The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.
Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.
Tangent ratio relates the opposite and adjacent side by following formula:
[tex]tan(\theta)=\frac{Opposite}{Adjacent}[/tex]
Using the given values, we get:
[tex]tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555[/tex]
Thus, rounded to nearest foot, the height of the building is 555 feet
To find the height of the building, we can use trigonometry and the given angle of elevation. The height of the building is approximately 555 feet.
Explanation:To find the height of the building, we can use trigonometry and the given angle of elevation. Let x be the height of the building. From the given information, we have the opposite side (x) and the adjacent side (1 mile = 5280 feet). The tangent function is used to relate these sides: tan(6°) = x/5280 feet. Solving for x, we get x = 5280 feet × tan(6°).
Using a calculator, we find that tan(6°) is approximately 0.1051. Multiplying this by 5280 feet gives us the height of the building in feet: 0.1051 × 5280 = 554.928 feet.
Rounding to the nearest foot, the height of the building is approximately 555 feet.
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Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium
Answer:
A
Explanation:
Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but
Aluminum by itself is not able to be magnetized.
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is connected to a 2.00-V battery, what charge is stored?
Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb
As a radioactive material undergoes nuclear disintegration, a) b) c) d) e) Its atomic number always remains the same Its mass number never increases; Its mass number never decreases; It always emits an alpha particle; None of the preceding statements is correct.
A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static frictional force of 560 N. Find the mass of the worker.
Answer:
99.63 kg
Explanation:
From the force diagram
N = normal force on the worker from the surface of the roof
f = static frictional force = 560 N
θ = angle of the slope = 35
m = mass of the worker
W = weight of the worker = mg
W Cosθ = Component of the weight of worker perpendicular to the surface of roof
W Sinθ = Component of the weight of worker parallel to the surface of roof
From the force diagram, for the worker not to slip, force equation must be
W Sinθ = f
mg Sinθ = f
m (9.8) Sin35 = 560
m = 99.63 kg
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T s. (a) What emf is induced in this loop? (b) If the loop has a resistance of find the current induced in the loop
Answer:
(a) 0.0171 V
Explanation:
A = 0.09 m^2, dB/dt = 0.190 T/s
(a) According to the law of electromagntic induction
e = dФ / dt
e = A dB / dt
e = 0.09 x 0.190 = 0.0171 V
(b)
as we know
i = e / R
we can find induced current by dividing induced emf by resistance
(a) The emf induced in the loop is 0.0171 V.
(b) The current in the loop is determined from the ratio of induced emf to resistance in the loop.
emf induced in the loopThe emf induced in the loop is calculated by applying Faradays law as follows;
emf = dФ/dt
emf = A dB/dt
where;
A is the area of the loopdB/dt is the change in magnetic field with timeemf = 0.09 x 0.19
emf = 0.0171 V
Current in the loopThe current in the loop is determined by applying Ohm's law;
I = emf/R
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When a charged particle moves at an angle of 25° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2F?
Answer:57.7
Explanation:
Force on a moving charge in a magnetic field is given by
[tex]F=q\times V\times Bsin\theta [/tex]
Where F=Force experienced by charge
q=charge of particle
V=velocity of particle
B=magnetic field
F=qVBsin(25)-------1
[tex]2F=qVBsin(\theta )------2[/tex]
Divide (1)&(2)
[tex]\frac{1}{2}[/tex]=[tex]\frac{sin(25)}{sin(\theta)}[/tex]
[tex]Sin(\theta )=2\times sin(25)[/tex]
[tex]\theta =57.7 ^{\circ}[/tex]
Given the constraints of the question and the formula for magnetic force on a moving charge, the situation described isn't possible.
The magnetic force experienced by a moving charged particle is given by the formula F = qvBsin(θ), where q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the magnetic field. When the force is F at 25°, to experience a force of 2F, the sin of the new angle must be double that of sin(25°) because all other factors (charge, speed, magnetic field strength) are constant.
In the case given, sin(θ) would need to be 2sin(25°), but since the maximimum value sin(θ) can have is 1 (at 90°), this situation isn't possible within the constraints of the question (angle less than 90°). Hence, the student might have misunderstood the problem or there might be an error the question itself. However, if it is assumed that the question intends to find when the component of the force perpendicular to the field is double, the answer would be when θ is 90 degrees because the sin(90°) = 1. This is because when a charged particle moves perpendicular to a magnetic field, the sin(θ) factor in the force equation is maximized.
A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the object and (b) the magnitude of the resultant force.
Answer:
(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).
(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .
Explanation:
m= 3kg
a= 2 i + 5 j = 5 .38 < 68.19 º
F= m * a
F= 3* ( 5.38 < 68.19º )
F= 16.4 N < 68.19º
Fx= F * cos(68.19º)
Fx= 5.99
Fy= F* sin(68.19º)
Fy= 14.98
The resultant force acting on the object is (6.00 i + 15.00 j) N, and the magnitude of the resultant force is 16.16 N.
To find the resultant force acting on a 3.00-kg object with an acceleration given by a = (2.00 i + 5.00 j) m/s2, we use Newton's second law of motion (Force = mass imes acceleration).
Calculate the resultant force: F = m imes a where m is the mass and a is the acceleration vector.Substitute the given values into the equation: F = 3.00 kg imes (2.00 i + 5.00 j) m/s2.This gives F = (6.00 i + 15.00 j) N.For the magnitude of the resultant force, we calculate it using the Pythagorean theorem:
|F| = \\(6.002 + 15.002)\sqrt N.Which simplifies to |F| = \\(36 + 225)\sqrt N = \\(261)\sqrt N.Thus, the magnitude of the resultant force is |F| = 16.16 N.
What is the value of (8 104)2, written with the correct mumber of significant figures? A) 206.324 B) 206.323 C)2063 D) 206 E) 200
Answer:
B) 206.323
Explanation:
The value of (8 104)2, written with the correct mumber of significant figures is 206.323.
Answer:
B. 206.323
Explanation:
A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result in all of the light reflected back into the diamond? The index of refraction for air is 1.000.
Answer:
24.2 degree
Explanation:
n = 2.419
Let the minimum angle of incidence is i.
The value of minimum angle of incidence so that all the light reflects back into the diamond is the critical angle for air diamond interface.
The relation for the critical angle and the refractive index of diamond is given by
Sin ic = 1 / n
where, ic is the critical angle for air diamond interface and n be the refractive index for diamond.
Sin ic = 1 / 2.419 = 0.4133
ic = 24.4 degree
Thus, the value of minimum value of angle of incidence for which all the light reflects back into diamond is 24.4 degree.
Explanation:
A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result ...
Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.
Final answer:
To calculate the power converted to thermal energy due to frictional effects, subtract the ideal mechanical power required to lift water to the higher reservoir (13.2435 kW) from the actual pump power (20 kW), yielding 6.7565 kW.
Explanation:
The student is asking to determine the mechanical power that is converted to thermal energy due to frictional effects when water is pumped from a lower reservoir to a higher reservoir. Given the shaft power of the pump is 20 kW, height difference is 45 m, and flow rate is 0.03 m³/s, we can first calculate the ideal mechanical power required to lift the water to that height.
The gravitational potential energy given to the water per second (which is the power) by pumping it to the height is calculated using the formula P = ρghQ, where ρ (rho) is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the height difference (45 m), and Q is the flow rate (0.03 m³/s). This results in P = 1000 kg/m³ * 9.81 m/s² * 45 m * 0.03 m³/s = 13243.5 W or 13.2435 kW.
The difference between the actual power supplied by the pump (20 kW) and the ideal power required (13.2435 kW) is the power lost to thermal energy due to friction. Therefore, the power converted to thermal energy is 20 kW - 13.2435 kW = 6.7565 kW.
Suppose a 50 turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.325 m^2. It is squished to have no area in 0.225 s. What is the magnitude of the average induced emf in volts if the uniform magnetic field has a strength of 1.5 T?
Answer:
EMF = 108.3 Volts
Explanation:
As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
[tex]EMF = N\frac{d\phi}{dt}[/tex]
[tex]EMF = N\frac{BA - 0}{\Delta t}[/tex]
now we know that
N = 50 turns
B = 1.5 T
A = 0.325 m^2
[tex]\Delta t = 0.225 s[/tex]
now we have
[tex]EMF = (50)(\frac{1.5\times 0.325}{0.225})[/tex]
[tex]EMF = 108.3 Volts[/tex]
The magnitude of the average induced emf is 110.12 volts.
The magnitude of the average induced emf can be calculated using Faraday's law. The formula is given as:
emf = -N * A * (dB/dt)
Where:
N is the number of turns in the coilA is the area of the coildB/dt is the rate of change of the magnetic fieldIn this case, the number of turns is 50, the area is 0.325 m², and
the rate of change of the magnetic field is (0 T - 1.5 T) / 0.225 s = -6.67 T/s.
Substituting these values into the formula, we get:
emf = -(50)(0.325)(-6.67) = 110.12 V
The magnitude of the average induced emf is 110.12 volts.
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We are given a piece of copper of mass m=0.82 g to form a cylindrical wire of resistance R=0.87.2. What should be the length / of this wire?: 2.16 m. What would be the diameter d of this wire?: d 0.232 mm. To make things certain, let us use copper's mass density of 8.96 g/cm3 and resistivity p= 1.7 x 10-8 2.m.
Answer:
Length of wire=2.16 m
Diameter of wire=0.232 mm
Explanation:
m= mass of copper wire= 0.82 g
R= Resistance of copper wire= 0.87 ohms
D= Density of copper= 8.96 g/cm^3
ρ= Resistivity= 1.7×10^-8 Ωm
[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]
[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]
[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]