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ratelena [41]
3 years ago
8

A ball took 0.45s to hit the ground 0.72m from the table. What was the horizontal velocity of the ball as it rolled off the tabl

e?
Physics
1 answer:
dusya [7]3 years ago
8 0

the ball hits the ground a very high velocity


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What type of phase change occurs when water vapor in the atmosphere changes to liquid water in clouds?
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Answer:

condensation, the process of changing from a gas to a liquid

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Which of the following is an example of kinetic energy?
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B. Sound, because everything else sits still and sound waves move
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The table below describes some features of methods used to generate electricity. Name method 4.
alina1380 [7]

Answer:

Hydroelectricity

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4 0
3 years ago
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s
nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where:

F=1.1 \cdot 10^5 N is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

I=\Delta p = m \Delta v

where

m = 68 kg is the mass of the man

\Delta v = 0-55 m/s = -55 m/s is the change in velocity of the man

Substituting,

I=(68 kg)(-55 m/s)=-3740 kg m/s

7 0
3 years ago
A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
3 years ago
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