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kiruha [24]
3 years ago
10

Newton’s Laws of Motion are absolute in classical physics. One example that uses all three laws simultaneously is the firing of

a rocket in space. Hollywood uses this idea when they use fire extinguishers, or a rocket, in space to move about. Write a brief essay describing how Newton’s Laws explain how a rocket in space can move objects. Be sure to touch on each law within your essay.
Physics
2 answers:
inysia [295]3 years ago
5 0

Explanation:

Since there is no medium in space we need rockets as they can travel without any medium. They release the the thrust from the nozzle (which is action) and an equal and opposite force acts on the nozzle (which is reaction) to move it forward. This is as per the third law of motion (<em>Action Reaction Principle</em>) which states that every action has an equal and opposite reaction.

As per the second law of motion, the acceleration of an object depends on its mass and force applied on it. As the rocket moves it discards its unused parts which makes it lighter. Also the decreasing amount of fuel further reduces the mass. Both of these factors increases the acceleration of the rocket.

As per the first law of motion, <em>law of Inertia</em>, an object will not change its state of motion until an unbalanced force is applied on it. So, if the rocket is moving at a speed of x Km/h and we shut down the engine, it will keep moving at the same speed without using any extra fuel.

Ksju [112]3 years ago
4 0
Action and reaction are the rocket fuel going backwards and the rocket going forward. 3
f=ma 2
the rocket was at rest, 1, until there was a resultant external force 1 at which point 2 came in, with 3 alongside
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Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
Dennis_Churaev [7]

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

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where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

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3 years ago
Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d
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Answer:

The width of the slit is 0.4 mm (0.00040 m).

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From the Young's interference expression, we have;

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d = (3.30 × 563 ×10^{-9} ) ÷ (0.0047)

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d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

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