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3 years ago

A sledgehammer hits a wall How do the hammer and the wall act on each other?

Physics

1 answer:

tigry1 [53]3 years ago

7 0

We want to study the impact of a sledgehammer and a wall.

Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.

When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.

This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.

Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.

If you want to learn more, you can read:

brainly.com/question/13952508

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Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

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3 years ago

If a sled has a mass of 4kg what is the force of gravity on the sled?

ale4655 [162]

Answer:

Explanation:

Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.

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3 years ago

How is momentum conserved in a newtons cradle <br><br> when one steel ball hits the other

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In Newton's Cradle experiment we know that all cradles of same mass and identical to each other

Now we know that when two identical objects collide elastically then they interchange their velocity

So here we have same illustration

When Newton pulls up a cradle and release it will move hit another cradle which is at rest

Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.

So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.

So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.

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3 years ago

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is

tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

$v_x = 4.8 m/s$

now the acceleration is given as

$\vec a = -3.2 \hat i + 4.6 \hat j$

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

$v_f = v_i + at$

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Part a)

the velocity of the particle at this moment in Y direction is given as

$v_f_y = v_i + at$

$v_f_y = 0 + 4.6(1.5)$

$v_f_y = 6.9 m/s$

Part b)

X coordinate of the particle at this time

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

so position is given as (3.6 m, 5.175 m)

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Answer:

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Explanation:

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2 years ago

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