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MrRa [10]
3 years ago
13

1. A stationary 50 tonne submarine fires a missile of mass 40kg at

Physics
1 answer:
White raven [17]3 years ago
8 0

The velocity of the submarine immediately after firing the missile is 0.0104 m/s

Explanation:

Mass of the submarine M=50 tonne=50\times 1000=50000kg

Mass of the missile m=40  kg

velocity of the missile v= 13m/s

we have to calculate the velocity of the submarine after firing

This is the recoil velocity and its expression is derived from the law of conservation of momentum

recoil velocity of the submarine

V=-mv/M\\=-40\times 13/50000\\=0.0104 m/s

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frosja888 [35]

Answer:

Acceleration of the object will be 1.05m/sec^2

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by F=kx=95\times 0.1=9.5N

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So 9.5=9\times a

a=1.05m/sec^2

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3 years ago
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2 years ago
Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a
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The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

s_2=\frac{1}{2} g(4t)^2=8gt^2\\  s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2

The length of the window s is given by,

s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2

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The height of the roof h is the distance traveled by the first drop and is given by,

h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m

the height of the roof is 3.57 m



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Answer:

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Explanation:

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