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Marysya12 [62]
2 years ago
15

A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it

Physics
2 answers:
aalyn [17]2 years ago
4 0

Answer:

20 cm

Explanation:

Force = mass * acceleration  F = ma (1)

Spring equation is given as Force = Spring constant*Distance (2)

combine 1 & 2

F = ma = Spring constant*Distance

70 * 9.81 = 3430 * d

d = 0.2 m or 20 cm

Dominik [7]2 years ago
3 0
F = ma = -kx

a = 9.81 m/s²
k = 3430 N/m
m = 70 kg

x = - ma/ k = 0.2m
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5 0
2 years ago
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

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distance, s = 7 m

coefficient of friction = 0.26

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F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

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(c) Work done by the friction

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(d) Work done by the normal force

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Work done by the gravity

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3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
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The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

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α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

#SPJ1

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