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Degger [83]
3 years ago
9

What are two ways you can increase power by climbing the stairs?

Physics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

change the

  • time
  • distance
  • force.

Explanation:

P = W/time

W = F*d

You have control over how fast you go up the stairs.

You also have control over how far up the stairs you go.

Therefore the answer is

  • time
  • distance

If you don't like distance as an answer, you can carry something up the stairs -- anything that increases F will do.

You might be interested in
Which three factors are used to calculate gravitational potential energy?
NeX [460]

Answer:

Height, mass, acceleration.

Explanation:

I hope it helps u dear! ^_^

7 0
2 years ago
If an automobile engine delivers 42.0 hp of power, how much time will it take for the engine to do 6.20 â 105 j of work? (hint:
Elena L [17]
To be able to answer this item, we are to calculate the power that the machine could deliver from hp to kW. 

      (45 hp)(746 W/1 hp) = 33570 W

Power is the amount of energy delivered at a certain period. 

             t = (6.20 x 10^2 J)/ (33570 kJ/s)

             t = 0.01845 s
7 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
Explain how the message is transmitted from the receiving neuron’s dendrites down the axon all the way to the axon terminal (ele
Vikki [24]

The vesicles release neurotransmitters. These cross the synapse and are accepted by the receptors in the dendrites of the next neuron.

Explanation:

An axon, or nerve fiber, is a long slender projection of a nerve cell, or neuron, that conducts electrical impulses away from the neuron's cell body. Axons are in effect the primary transmission lines of the nervous system, and as bundles they help make up nerves.

When an action potential reaches the axon terminal, it depolarizes the membrane and opens voltage-gated Na+ channels. Na+ ions enter the cell, further depolarizing the presynaptic membrane.

6 0
3 years ago
a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The
forsale [732]
The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
6 0
3 years ago
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