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stiks02 [169]
3 years ago
9

When a ball rolls down a hill, what energy conversion occurs?

Physics
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

C

Explanation:

When a ball rolls down a hill, Potential Energy Conversion takes place to Kinetic Energy.

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Which type of neuron stimulates muscles to contract?​
lakkis [162]

Answer:

a motor neuron

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3 years ago
A weightlifter raises a 200-kg barbell through a height of 2 m in 2.2 s. the average power he develops during the lift is
Lunna [17]
Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.
4 0
3 years ago
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Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea
LuckyWell [14K]

Answer:

Explanation:

the weight of the telescope decreases  because the moon attract the body with less force as compared to earth due to less gravity as compared to earth

6 0
3 years ago
A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
Doss [256]

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

5 0
2 years ago
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit
damaskus [11]

Answer:

a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

Explanation:

According to Newton's Third Equation of motion

V_f^2-V_i^2=2as

Where:

V_f is the final velocity

V_i is the initial velocity

a is the acceleration

s is the distance

In our case:

V_f=V_{escape},  V_i=0,s=4 m

So Equation will become:

V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

5 0
3 years ago
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