How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have
2 answers:
1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
So the power output is
And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.
2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.
So the power output is
And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.
2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.
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Answer:
T² ∝ R³
Explanation:
Given data,
The period of revolution of the planet around the sun, T
The mean distance of the planet from the sun, R
According to the III law of Kepler, " Law of Periods' states that the square of the orbital period to go around the sun once is directly proportional to the cube of the mean distance between the sun and the planet.
T² ∝ R³
From the above equation it is clear that T² varies directly as the R³.
Answer:
The acceleration of the bag of sand is .
Explanation:
We have,
Mass of a bag of sand is 100 kg
Weight of the bag of sand is 100 N
It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :
When it is dropped, a = g
So, the acceleration of the bag of sand is .
Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by
where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so
And so, the work done is also zero.