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belka [17]
3 years ago
8

Which is the correct equation for the force applied by a spring?

Physics
1 answer:
slava [35]3 years ago
3 0

Answer:

B. F = -k*x

Explanation:

The force of a spring can be easily calculated using Hooke's law which tells us that the force applied on a spring is equal to the product of the constant of a spring by the compressed or stretched distance of the spring.

That is:

F = k*x

k = spring constant [N/m]

x = distance [m]

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At what time did time begin???
blsea [12.9K]

Answer:

approximately 14 billion years ago

Explanation:

brainliest please

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2 years ago
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

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6 0
2 years ago
What name is given to the turning effect of a force?
Alex787 [66]

Answer:

The turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. When undoing a nut fastened to a screw by hand one realises that the amount of force required is a lot greater than when undoing the same nut using a spanner.

Explanation:

6 0
3 years ago
Read the passage.
nevsk [136]

Answer:

d in my opinion is the most opitmal answer

Explanation:

aka a graph comparing distances traveled by objects thrown on Earth and the moon

7 0
3 years ago
Read 2 more answers
The carnival ride has a 2.0m radius and rotates 1.1 times persecond.
Lady_Fox [76]

Answer:

a) v = 13.8 m / s , b) a = 95.49 m / s² , c) a force that goes to the center of the carnival ride  and d)   μ = 0.10

Explanation:

For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics

a) reduce the magnitudes to the SI system

     w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s

The equation that relates linear and angular velocity is

     v = w r

     v = 6.91  2

     v = 13.8 m / s

b) centripetal acceleration is given by

     a = v² / r = w² r

     a = 6.91² 2

     a = 95.49 m / s²

c) this acceleration is produced by a force that goes to the center of the carnival ride

d) Here we use Newton's second law

     fr -W = 0

     fr = W

     μ N = mg

Radial shaft

      N = m a

      N = m w² r

      μ m w²  r = m g

      μ = g / w² r

      μ = 9.8 / 6.91² 2

      μ = 0.10

3 0
3 years ago
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