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3 years ago

The table below shows the acceleration of gravity on different bodies in the

Physics

2 answers:

Aloiza [94]3 years ago

8 0

Answer:

Pluto

because the the gravitation strength is small, so it will be accelerating slow

ololo11 [35]3 years ago

3 0

Answer:

i think is mercury, im not 100% sure

Explanation:

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How many wavelenghts of a wave pass a point if the frequency of the wave is 4 hertz?

sasho [114]

8. Hope this helps.....a little

4 0

4 years ago

Can someone give me four examples of how energy has the ability to cause motion or create change.

tamaranim1 [39]

The answer is that it can involve changing the shape of matter, or changing the speed or direction of an object.

5 0

3 years ago

A 3 kg block is attached to a vertical spring. Initially, you exert a 50 N downwards force on the block, holding it in place, at

Kryger [21]

As you were holding the block down and in place, the spring exerted an upward force that balanced the downward push by your hand and its own weight. So this restoring force has a magnitude of R such that

R - 50 N - (3 kg) g = 0 => R = 79.4 N

As soon as you remove your hand, the block has acceleration a such that, by Newton's second law,

R - (3 kg) g = (3 kg) a => a = (79.4 N - (3 kg) g) / (3 kg) ≈ 16.7 m/s^2

pointing upward.

4 0

3 years ago

Using a radar gun, you emit radar waves at a frequency of 6.2 GHz that bounce off of a moving tennis ball and recombine with the

Keith_Richards [23]

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

$f_{app}$ = frequency observed as the ball approach the radar

$f_{rec}$ = frequency observed as the ball recede away from the radar

V = speed of light

$v$ = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

$f_{app}=\frac{f(V+v)}{V}$ eq-1

frequency observed as the ball recede the radar is given as

$f_{rec}=\frac{f(V-v)}{V}$ eq-2

Beat frequency is given as

$B = f_{app} - f_{rec}$

Using eq-2 and eq-1

$B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}$

inserting the values

$969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}$

$v$ = 23.4 m/s

8 0

3 years ago

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic

shepuryov [24]

Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

$F=m\frac{v^2}{r}$ (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$

6 0

3 years ago