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Step2247 [10]
3 years ago
8

What is the motion of a body when only the force of gravity is acting on it

Physics
1 answer:
Mashcka [7]3 years ago
4 0
When only the force of gravity acts on an object,
the situation is called "free fall".

The motion is uniformly accelerated.  The object falls
straight toward the center of the other object in whose
gravity it shares, and its speed increases at a constant
rate. 

When this happens on or near the surface of the Earth,
the rate is (9.8 meters per second greater) for each
second of falling.

True free-fall is never observed in daily life.  Whenever
we see an object falling, it is always falling through air,
which counteracts the force of gravity to a greater or
lesser degree, depending on the object's size, shape,
weight, surface texture, and speed through the air.
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What is difference between non uniform and uniform circular motion?
statuscvo [17]

Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

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3 years ago
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
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Answer:

I belive it would be ture

Explanation:

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