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3 years ago

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

Physics

1 answer:

nordsb [41]3 years ago

4 0

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

$5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s$

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

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3 years ago

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3 years ago

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frez [133]

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2 years ago

A wave with a short wave length will also have...?

Masja [62]

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3 years ago

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp

dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A) What is its speed after falling for 2.00s?

from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

from the equation of motion $s=ut+0.5at^{2}$ we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

$v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}$

v = 15.4 m/s

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2 years ago