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3 years ago

The work done when a force moves a body through a distance of 15m is 1800j. What is the value of the force applied

Physics

1 answer:

Leviafan [203]3 years ago

7 0

Answer:

120

Work :

W = Fd (work = force x distance)

Force :

F = W/d

Distance :

d = W/F

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Apply a large force over a short time

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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-

Kipish [7]

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg

Take downward direction as positive.

Fg = m g

E = 670 N/C

Fe = q E

Fe + Fg = 0

q E + m g = 0

q = -m g/E

= -1.49 × 10^-3 × 9.81/670

= -2.18 × 10^-5 C

= -2.18 × 10^-5 C

6 0

3 years ago

A sail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 N. how much w

erma4kov [3.2K]

<h3><u>Answer;</u></h3>

a) 5.00 x 10^8 J

<h3><u>Explanation;</u></h3>

The work done to move the sailboat is calculated through the equation;

W = F x d

where F is force and d is the distance.

Substituting the known values from the given above,

W = (5.00 x 10⁴ N)(10 km)(1000 m/ 1km)

= 5.00 x 10⁸ J

Thus, the work done is <u>5.00 x 10⁸Joules</u>

7 0

3 years ago