PHYSICS HIGH SCHOOL

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 8.08 N/m. When the cannon is fired, the ball moves 15.8 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 0 N on the ball.With what speed does the projectile leave the barrel of the cannon?

Answers

Answer 1

Answer:

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=(1)/(2)kx^2=(1)/(2)* 8.08* 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031* 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$(1)/(2)mv^2=0.0052$

$(1)/(2)* 0.00524* v^2=0.0052$

v = 1.40 m/sec

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A dwarf planet is a small celestial body resembling a planet, but lacks necessary things to be a planet

Hey, Sverlnexlcasatyr!
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A shot-putter throws an 8.00 kg shot at a speed of 10.0 m/s. What is the kinetic energy of the shot?

Answers

Answer:

Kinetic energy of the shot is 400 Joules

Explanation:

It is given that,

Mass of the shot, m = 8 kg

Speed of thrown, v = 10 m/s

We have to find the kinetic energy of the shot. It is given by the product of mass of an object and the square of velocity divided by 2 i.e.

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KE = 400 Joules

Hence, the kinetic energy of the shot is 400 joules.

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Answers

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Answers

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