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Makovka662 [10]
3 years ago
11

While rowing in a race, John does 132 J of work while pulling the oar 0.800 m.

Physics
2 answers:
HACTEHA [7]3 years ago
6 0

Answer:

i got u!!! john's arms exert a force of 165 newtons.

Nastasia [14]3 years ago
5 0

Hi there!

\large\boxed{F = 165 N}

We know the equation:

W = F · d, where:

W = work (J)

d = displacement (m)

F = force (N)

We can rearrange to solve for F:

W/d = F

Thus:

132/0.8 = F

F = 165 N

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HACTEHA [7]
UV Radiation since it has a higher frequency than the others. The higher the frequency the shorter the wavelength.
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3 years ago
Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 n, and it doesn't budge. what is
maks197457 [2]

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N

This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).

When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.

Learn more about frictional force here

brainly.com/question/1714663

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8 0
1 year ago
The force that acts on an object lying on a surface, acting in a direction perpendicular to the surface is called ____________​
DanielleElmas [232]
If it is on land gravitational force
If it is on water thrust
8 0
2 years ago
Is sand a dependent or independent variable
anastassius [24]

Answer:

dependent variable

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6 0
3 years ago
Read 2 more answers
A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300
Illusion [34]
<h2>Answer:</h2>

\boxed{P=96.09MW}

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega

So we can calculate the power loss as:

P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}

Finally, the power loss due to resistance in the line is 96.09MW

5 0
3 years ago
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