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11111nata11111 [884]
3 years ago
5

A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d

one on the hoop to stop it
Physics
1 answer:
Softa [21]3 years ago
6 0

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

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Impulse=force*time
impluse=120N*2.0s
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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t
castortr0y [4]

Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

T=\frac{2uSin\theta }{g}

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

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So, it takes 4.86 s more time of flight on moon than the earth.  

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3 years ago
A pendulum is made by letting a 2.0-kg object swing at the end of a string that has a length of 2.1 m. The maximum angle the str
just olya [345]

Answer:

The speed of the object at the lowest point in its trajectory is:

v=2.34\: m/s

Explanation:

We can use the conservation of energy between the maximum point of swing and the lowest point of the pendulum.

P=K

mgh=\frac{1}{2}mv^{2} (1)

Where:

  • h is the height of the object at 30° with the vertical.
  • v is the speed at the lowest point.

We can find h using trigonometry.

h=L-Lcos(30)=L(1-cos(30))

h=2.1(1-cos(30))=0.28\: m

Now, using equation (1) we can find v.

gh=\frac{1}{2}v^{2}

2gh=v^{2}

v=\sqrt{2gh}

v=\sqrt{2(9.81)(0.28)}

v=2.34\: m/s

I hope it helps you!

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A forest is most sustainable when it has-
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I think it is D
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