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Blizzard [7]
3 years ago
13

A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

ion) with respect to Earth, and one of the fragments is subsequently moving vertically with speed v, find the velocity ' of the other fragment immediately following the explosion. (Use the following as necessary: v.)
Physics
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e  \dfrac{m}{2} .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v

The velocity of the other fragment immediately following the explosion is v .

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Answer:

<h2>40 kg</h2>

Explanation:

Find the diagram relating to the question for proper explanation of the question below.

Using the principle of moment

Sum of clockwise moments = Sum of anticlockwise moments

Moment = Force * perpendicular distance

For anti-clockwise moment:

Since the 30 kg moves in the anticlockwise direction according to the diagram

ACW moment = 30 * 1 = 30 kgm

For clockwise moment

If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).

Equating both moments we have;

0.75M = 30

M = 30/0.75

M = 40 kg

The second child's mass is 40 kg

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3 years ago
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In a solution,a Is the substance being dissolved
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Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?
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A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

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 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

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    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
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