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goldfiish [28.3K]
3 years ago
6

While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on Earth. If Neil Armstrong jum

ped up on the moon with an initial speed of 1.51 m/s to a height of 0.700 m, what amount of gravitational acceleration did he experience?
Physics
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

The gravitational acceleration experienced was of 1.63m/s².

Explanation:

We know, from the kinematics equations of vertical motion that:

v^{2} =v_0^2-2gy

Solving for g, we get:

g=-\frac{v^2-v_0^2}{2y}

Since the final speed is zero, because Neil Armstrong came to a stop in his maximum height, we obtain:

g=\frac{v_0^2}{2y}

Finally, we plug in the given values of the initial speed and the maximum height:

g=\frac{(1.51m/s)^2}{2(0.700m)}=1.63m/s^2

This means that the gravitational acceleration experienced by Neil Armstrong in the moon, was of 1.63m/s².

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If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
A disk rotates freely on a vertical axis with an angular velocity of 50 rpm . An identical disk rotates above it in the same dir
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Answer:

Final angular velocity is 35rpm

Explanation:

Angular velocity is given by the equation:

I1w1i + I2w2i = I1w1f -I2w2f

But the two disks are identical, so Ii =I2

wf can be calculated using

wf = w1i - w2i/2

Given: w1i =50rpm w2i= 30rpm

wf= (50 + 20) / 2

wf= 70/2 = 35rpm

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3 years ago
Use the concepts of kinetic energy and potential energy to describe the motion of a child on a swing. Why does the child need a
andrew-mc [135]
When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.

The child may need a push from time to time because friction with the air causes loss of energy.
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3 years ago
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Describe how water may be treated before people use it
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Water is treated by purifying it by adding slaked lime or potash alum

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

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At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

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3 years ago
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