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Charra [1.4K]
3 years ago
12

A 10.1 g bullet leaves the muzzle of a rifle with a speed of 425 m/s. What constant force is exerted on the bullet while it is t

raveling down the 0.6 m length of the barrel of the rifle?
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer

Force will be 1520.2604 N

Explanation:

We have given mass of the bullet m = 10.1 gram = 0.0101 kg

Distance S= 0.6 m

Final velocity v = 425 m/sec

Initial velocity u will be 0 m/sec

From third equation of motion v^2=u^2+2as

425^2=0^2+2\times a\times 0.6

a=150520.833m/sec^2

We know that force is given by F = ma

So force = 0.0101×150520.833 = 1520.2604 N

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Question #14
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The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

The hydrogen molecule is a form that two hydrogen atoms share their electrons decreasing the total energy of the molecule, this bond has a covalent and hydrogen bonding characteristic.

In a stellar explosion, the energy released increases the energy of the hydrogen atom, for which we have two possibilities:

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When the atoms are thrown into space, the separation between them is so high that it does not allow electrons to be shared and molecules cannot be formed either.

In conclusion, the decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot join another atom, which is why the hydrogen has only one atom.

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2 years ago
A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

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