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3 years ago

What is the momentum of a 5 kg object that has a velocity of 1.2 m/s? 3.8 kg • m/s 4.2 kg • m/s 6.0 kg • m/s 6.2 kg • m/s

Physics

2 answers:

Gnesinka [82]3 years ago

4 0

Answer:

Your answer will be 6.0kg•m/s

Explanation:

In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity = 5 * 1.2 = 6So the momentum of the object is 6 Newton.

Julli [10]3 years ago

3 0

Answer:

Momentum of an object, p = 6 kg-m/s

Explanation:

It is given that,

Mass of an object, m = 5 kg

Velocity of that object, v = 1.2 m/s

We need to find the momentum of an object. It is equal to the product of mass and its velocity. Mathematically, it is given by :

p = m v

$p=5\ kg\times 1.2\ m/s$

p = 6 kg-m/s

So, the momentum of an object is 6 kg-m/s. Hence, this is the required solution.

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Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

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compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

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$F=m.a$

where:

a= acceleration

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$a\approx 1835\,m.s^{-2}\\$

we have:

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Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

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(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

$v^2=u^2-2g.h$

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

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