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IceJOKER [234]
3 years ago
6

A ball is thrown up into the air. Describe the work done on the ball and the energy transformations that take place

Physics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

When a ball is thrown straight up into the air, all its initial kinetic energy is converted into gravitational potential energy when it reaches its maximum height.

Explanation:

hope this helps

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None hsbisneienbjsnsidns ueheijejeie idhdidnis idndidjd idjdi
lys-0071 [83]

Answer:

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7 0
2 years ago
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil.
Nana76 [90]

Answer:

0.32 V

Explanation:

N = 10, A = 0.23 m^2, B = 0.47 T, t = 0.34 s

The average induced emf is given by

e = - N dФ / dt

Where, dФ be the change in magnetic flux in time dt.

dФ / dt = d / dt (B A) = A dB/dt

So,

e = - 10 x 0.23 x 0.047 / 0.34 = - 0.32 V

The negative sign shows the direction of induced emf.

6 0
3 years ago
To stretch a spring 8.00cm from its unstretched length, 16.0J of work must be done.A)What is the force constant of this spring?B
ad-work [718]

A) 5000 N/m

The force constant of the spring can be found by using the expression for the elastic potential energy stored in the spring (which is equal to the work done on it):

W=U=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching/compression of the spring

In this problem, we have

W = 16.0 J is the work done

x = 8.00 cm = 0.08 m is the stretching

Substituting into the formula and re-arranging it, we find

k=\frac{2W}{x^2}=\frac{2(16.0 J)}{(0.08 m)^2}=5000 N/m

B) 400 N

The magnitude of the force needed to stretch the spring by x = 8.00 cm = 0.08 m is given by Hook's law:

F=kx

where k=5000 N/m as we found previously. Substituting x=0.08 m, we find:

F=(5000 N/m)(0.08 m)=400 N

C) 4 J

The work done to compress the spring by x=4.00 cm=0.04 m is given by the same formula used for part A:

W=\frac{1}{2}kx^2

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

W=\frac{1}{2}(5000 N/m)(0.04 m)^2=4 J

D) 200 N

As we did in part B), the force needed to stretch this distance is given by Hook's law:

F=kx

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

F=(5000 N/m)(0.04 m)=200 N

8 0
3 years ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
4 years ago
How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves
Westkost [7]

Answer:

If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.

6 0
2 years ago
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