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3 years ago

What happens when you push a spring? How is this different than pulling it? (Hooke’s Law)

2 answers:

7 0

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.

Furkat [3]3 years ago

4 0

When you push a spring you are putting force on it, or rather forcing the spring onto its self. This is different than pulling the spring because when you push it and let go it will spring but if you pull it its would not move as far as it would if you push it. Pulling the spring would most likely deform the spring will pushing it would cause it to spring back into place.

<u>Hooke's law states that the force that extends or compresses a spring by some distance scales linearly with respect to that distance.</u>

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Temperature. The other three dont have anything to do with determining climate

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3 years ago

Read 2 more answers

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

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3 years ago

Frequency, period and wavelength<br> 11th grade high school physics

zzz [600]

Answer:0.38

Explanation:

the formula is f = c / λ

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and that equals 0.38 46 and so on so i just rounded it

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3 years ago

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Answer: please see attached work.

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