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zmey [24]
3 years ago
12

How do driver of cars use galvanometer

Physics
1 answer:
adoni [48]3 years ago
6 0
<span>Electricians use them and drivers of cars use them to know when to stop for fuel.. If a wire with a current moved upward in a magnetic field, how would it move when the direction of ... What energy transformation occurs in a galvanometer?</span><span>
</span>
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A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
1 year ago
Help me and you will meet your favorite singer at your door.
Goshia [24]
Sunlight, because it provides a source of energy is the answer because plants also provide a food source (please put as brainliest answer)
4 0
3 years ago
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A party shop delivers helium-filled balloons to homes and businesses. The owners realize from experience that on hot summer days
kvasek [131]

Answer:As the temperature increases, the helium in the ballon expands.

Explanation:

I took the quiz already

6 0
2 years ago
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What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

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3 years ago
Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
photoshop1234 [79]

Your answer would be A):Organize a laboratory in Germany.

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