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stepladder [879]
3 years ago
13

A coil spring with a spring rate of 0.28 N / cm is stretched with a force of 1.0 N to a length of 12.0 cm. Calculate the length

of the spring in untensioned condition.
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

2.3

Explanation:

firstly divide 1by12

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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu
madam [21]

The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

<h3>Speed of the satellite</h3>

v = √(GM/r)

where;

  • G is universal gravitation constant
  • M is mass of Earth
  • r is radius of the satellite

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)

v = 1.32 x 10⁵ m/s

Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

Learn more about speed of satellite here: brainly.com/question/22247460

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7 0
2 years ago
Why is there an upward force on objects in water?
Paraphin [41]
If you'r referring to some objects, it means that the mass of the object is less than the water so it floats. If the mass of an object is greater than the mass of the water, it will sink. Compare it to a balloon, helium makes it rise, while normal air makes it sink.
4 0
3 years ago
If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling
ryzh [129]

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after  falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter.  After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

5 0
3 years ago
A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run
Butoxors [25]

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

  • Initial velocity; u = 0
  • Final velocity; v = 13.4m/s
  • Time elapsed; t = 6.5s

Acceleration of the runner; a = \ ?

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

v = u + at

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

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On what criteria is the Fujita scale for Tornadoes based?
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B. Tornado destruction
It is based on the amount of damage
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