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3 years ago

5

A 10-meter rope is lying on the floor and has a mass force of 20 N. How much work is required to raise one end of the rope to a

height of 5 meters?

1 answer:

VMariaS [17]3 years ago

7 0

Answer:10N

Explanation: I think

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A wire, of length L = 4.1 mm, on a circuit board carries a current of I = 1.96 μA in the j direction. A nearby circuit element g

tamaranim1 [39]

Answer:

The magnitude of the magnetic field B is 5.921 T.

Explanation:

Given that,

Length = 4.1 mm

$B_{x}=4.9\ G$

$B_{y}=2.3\ G$

$B_{z}=2.4\ G$

Current $I = 1.96\ mu A$

We need to calculate the magnetic field

Using formula of magnetic field

$B=\sqrt{B_{x}^2+B_{y}^2+B_{z}^2}$

Put the value into the formula

$B=\sqrt{(4.9)^2+(2.3)^2+(2.4)^2}$

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3 years ago

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hope this helps:)

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3 years ago

1. A student moves a box of books down the hall by pulling on a rope attached to the box. The student

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Explanation:

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2 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

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3 years ago