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Elodia [21]
4 years ago
6

A girl weighs 200 newtons. How much work does she do climbing 10 meters of steps?

Physics
2 answers:
lina2011 [118]4 years ago
5 0
W=fscosx
w=fscos0 {as x(angle between force and displacement)=0 because force and displacement is in same direction}
w=200*10
w=2000 joules
Yuliya22 [10]4 years ago
3 0
When she gets to the top, she has (200x10)=2000 more joules of potential energy than she had at the bottom. But unfortunately, she has to do a lot more work than that in order to get there. She has to lift a leg, put it down, and then extend it, for each step, for example. The human body is not 100% efficient.
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rewona [7]

Answer:

The magnitude of the magnetic field is 1.83 x 10^{-5} T.

Explanation:

The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.

In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.

The magnetic field (B) between the given wires can be determined by:

B = \frac{U_{o}I }{2\pi r}

where: I is the current, r is the distance between the wires and U_{0} is the magnetic field constant.

But, I = 11 A, r = 0.12 m and U_{0} = 4\pi x 10^{-7} Tm/A

So that;

B = \frac{4\pi *10^{-7}*11 }{2\pi *0.12}

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3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

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We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

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We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

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3 years ago
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Answer:

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Explanation:

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<h3><u> Explanation: </u></h3>

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