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inessss [21]
3 years ago
8

Use Hooke's Law and Newton's 2nd Law to derive a formula for the acceleration of an object of mass m on a frictionless surface w

hich is attached to a spring with a constant k which is stretched a distance L. The acceleration should be in terms of m, k, and L.
Physics
1 answer:
kramer3 years ago
4 0

Answer:

a = kL/m

Explanation:

Here we can use Hooke's Law to find out the force applied on the system. Hooke's Law states that when a spring is stretched by some force, the force applied is directly proportional to the displacement of spring. The formula is given as:

F = kL

Now, the Newton's Second Law of motion states that whenever an unbalanced force is applied to a body it produces an acceleration in the body, in its own direction. So, the force is given by the formula:

F = ma

Comparing both the forces, we get:

kL = ma

<u>a = kL/m</u>

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Jack is playing with a Newton's cradle. As he lifts one ball to position A and drops it, it impacts the other balls at position
Volgvan

Answer: D

Explanation:

Because the energy from the first ball immediately impacts the other balls.

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3 years ago
you cover 10 meters in a time of 1 second .Is your speed the same if you cover 20 meters in 2 seconds?
IrinaVladis [17]

Definition:
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Speed in the second example = (20 meters) / (2 seconds) = 10 m/s.

Apparently, it is.  This little exercise demonstrates it.

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Which weather event is capable of destroying homes and uprooting trees due to a low pressure area at its center
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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
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