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1 year ago

How does your power output in climbing the stairs compared to the power output of a 100 watt

Physics

1 answer:

alexdok [17]1 year ago

8 0

Answer:

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.

Explanation:

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Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass

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2 years ago

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what english word does the following- the first two signify a male, the 1st 3 letters support a female, the first four support a

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Answer:

lol i know - i was rushing -_-

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2 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$