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love history [14]
3 years ago
12

Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica

lly compressed spring. Skid of 55 kg is standing on a massless platform attached to a spring of constant 8700 N/m that is compressed 125 cm. Skid is launched from rest. He eventually leaves the platform and reaches a peak high above the platform. (a) What is the velocity of Skid when he leaves the platform
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

        Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

        Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         ½ k x² + m g x ’= ½ m v²

         v² = \frac{k}{m}  x² + g x

let's calculate

         v² = \frac{8700}{55}  1.25² + 9.8 1.25

         v² = 247.159 + 12.25 = 259.409

         v = 16.11 m / s

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1. Which of the following is correct about the sampling distribution of the sample mean
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5 0
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When you describe a liquid as thick, are you saying that it has a high or a low viscosity?
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6 0
2 years ago
A man pushing a crate of mass
marin [14]

(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

<h3>Magnitude of net force on the crate</h3>

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

<h3>Net work done on the crate</h3>

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

<h3>Acceleration of the crate</h3>

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

<h3>Speed of the crate</h3>

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

7 0
2 years ago
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