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almond37 [142]
3 years ago
9

Can power be negative in physics? I am doing a lab in which I run down the stairs as quickly as possible. They are 5.4m tall and

my work is 5321.7 J. After calculating work, I got -5321.7 J. I am supposed to calculate my power used over 3 trials. The time for trial 1 was 4.69 and the seconds for trials 2 and 3 are 4.25. After calculating my power I got -1134.69 W and -1252.16 W. I am not sure if I got my power right. Is my power really negative or is my power positive?
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

Yes.

Explanation:

A negative power would just represent a loss of power. So in your case it lost -1252.16 W

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A cylinder with initial volume V contains a sample of gas at pressure p. On one end of the cylinder, a piston is let free to mov
Elina [12.6K]

Answer:

The work done on the gas = 2PV

Explanation:

In this question, we need to apply the basic gas laws to determine the word done.

For this question, we need to draw a PV diagram (a pressure-volume diagram), which I have made and attached in the attachment. So please refer to that attachment. I will be using this diagram to solve for the work done on the gas.

So, Please refer to the attachment number 1. where x -axis is of volume and y-axis is of pressure.

As we know that, the work done on the gas is equal to the area under the curve.

W = Area of the triangle

W = 0.5 x ( base) x ( height)

W = 0.5 x (BC) x (AC)

W = 0.5 x (3V-V) x (3P-P)

W = 2PV

Hence, the work done on the gas = 2PV

4 0
3 years ago
POSSIBLE POINTS: 1
faust18 [17]

Answer:

Pushes and pulls refer to the force that attracts or repels certain other materials without actually touching them.

Explanation:

Pushes and pulls are the forces exerted by the magnet on certain materials around it without, actually touching them. This push and pull is exerted through a region around the magnet called its magnetic field. The strength of this push and pull force is determined by, the strength of the magnetic field. A strong push or pull force is exerted by a strong magnetic field, and in turn a strong magnet and, a weak push and pull force is exerted by a weak magnetic field and, in turn a weak magnet. A push force is a repulsion while a pull force is an attraction. When a magnetic object is in the region of the magnetic field, it either attracts or is repelled away from the source of the magnetic field.

7 0
3 years ago
A metal sphere, X, has an initial net charge of −6 × 10−6 coulomb and an identical sphere, Y, has an initial net charge of +2 ×
STatiana [176]
When the two spheres touch each other, part of the charge recombine together. The part of charge that recombines is the excess of positive charge on sphere Y, Q_y = +2 \cdot 10^{-6}C, that recombines with an equivalent charge of -2 \cdot 10^{-6}C located on sphere X. As a result, the total charge remained on the two spheres is the excess of negative charge remained on sphere X:
Q=-4 \cdot 10^{-6}C
Since the two spheres are identical, they have same capacity, so this charge will now redistribute equally on the two spheres: therefore, at the end, each sphere will have
\frac{Q}{2}=-2 \cdot 10^{-6} C

8 0
3 years ago
when you run, your feet are pushing you forward. Friction keeps your foot in contact with the ground. According to Newton's thir
Studentka2010 [4]

Answer:

only you are in motion

Explanation:

it is because the ground causes friction to make you stop but you are the object in motion :)

3 0
3 years ago
WILL MARK AS BRAINLIEST!!!!!!
yaroslaw [1]

M1V1 + M2V2 = M1V1' + M2V2'

where:

M1 is the mass of the large marble = 0.05 kg

V1 is the initial velocity of the large marble = 0.6 m/sec

M2 is the mass of the small marble = 0.03 kg

V2 is the initial velocity of the small marble = 0 m/sec (marble is at rest)

V1' is the final velocity of the large marble = -0.2 m/sec

V2' is the final velocity of the small marble that we want to calculate

Substitute with the givens in the above equation to get V2' as follows:

M1V1 + M2V2 = M1V1' + M2V2'

(0.05)(0.6) + (0.03)(0) = (0.05)(-0.2) + 0.03V2'

0.03 = -0.01 + 0.03V2'

0.03V2' = 0.03+0.01 = 0.04

V2' = 0.04/0.03

V2' = 1.334 m/sec

4 0
3 years ago
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