PHYSICS COLLEGE

An emf is induced by rotating a 1207 turn, 20.0 cm diameter coil in the Earth's 4.13 10-5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms

Answers

Answer 1

Answer:

Answer:

0.157 V

Explanation:

Parameters given:

Number of turns, N = 1207

Diameter of coil = 20 cm = 0.2 m

Radius of coil, r = 0.2/2 = 0.1 m

Magnetic field strength, B = $4.13 * 10^(-5) T$

Time interval, t = 10 ms = $10 * 10^(-3) = 0.01 s$

The average EMF induced in a coil due to a magnetic field is given as:

EMF = $(- N * A * B)/(t)$

where A = Area of coil

A = π $r^(2)$

Therefore, EMF will be:

$EMF = (- 1207 * 3.142 * 0.1^2 * 4.13 * 10^(-5))/(0.01) \n\n\nEMF = -0.157 V\n$

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At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)

Answers

Answer:

(a) 4.875 m

(b) 5.72 rad/s

Explanation:

(a) Assuming constant angular speed, the angular distance the ball would have traveled after 5s at the rate of 6.5 rad/s is

6.5 * 5 = 32.5 rad

With radius of 0.15m, the linear distance it would have traveled is

32.5 * 0.15 = 4.875 m

(b)The angular velocity of the ball after 0.65s when subjected to an angular acceleration of -1.2 rad/s is

6.5 - 1.2*0.65 = 5.72 rad/s

(c)The linear speed of the ball is the product of the angular speed and radius 0.15 m

5.72 * 0.15 = 0.858 m/s

A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

Answers

Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

Explanation:

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

Learn more about Projectile Motion here:

brainly.com/question/29545516

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

6. As distance increases, gravitational force *
(10 Points)
increases
decreases

Answers

It decreasessssssssss

A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?