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3 years ago

8

What is the acceleration of a 10.5 kg mass pushed by a 50.5 n force

1 answer:

Lina20 [59]3 years ago

8 0

Answer:4.8 m/s^2

Explanation:

mass=10.5kg

Force=50.5N

Acceleration =force ➗ mass

Acceleration =50.5 ➗ 10.5

Acceleration =4.8 m/s^2

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now we are given that

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3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

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3 years ago

Newton's third law can be summarized as "every action has an equal and opposite reaction". In this problem, consider the action

scoundrel [369]

Answer:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

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Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

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Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

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And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

Explanation:

For this case we have this:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

$W = mg$

Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:

$N = \mu f_f$

Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

$W=N= mg$

And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

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