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3 years ago

What's the relationship between work done and force applied

1 answer:

3 0

When a force acts on a body along some path, the work done is W=F*s, where W is the work done, F is the force that is doing the work on the body and s is the path. The force doing the work has to be in the same direction, or parallel, as the path. This is called positive work. If the force and the path are anti-parallel, the work is negative. So the relationship between work and force is W=F*s.

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The velocity of waves in a ripple tank is 10 centimeters per second, and standing waves are produced with nodes spaced 3.0 centi

Vsevolod [243]

<span>node spacing = half of wavelength = 3 cm velocity = 10 cm/s = freq * wavelength hench freq = 10/6 = 5/3 = 1.7 hz</span>

6 0

3 years ago

Read 2 more answers

During which phase of the moon can a lunar eclipse happen?.

Umnica [9.8K]

Full moon!

when Earth is exactly between the Moon and Sun, Earth's shadow falls upon the surface of the Moon, dimming it and sometimes turning the surface red over the course of a few hours.

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2 years ago

(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.

LiRa [457]

Answer:

a) [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

V = k q / r

V = [N m² / C²] C / m

V = [N m / C]

The electric potential is defined as

V = E .s

V = [N / C] [m]

V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

[volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

v = $- \frac{dE}{dx} i^$

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0

3 years ago

A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its

faust18 [17]

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2

6 0

3 years ago

Read 2 more answers

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f

Butoxors [25]

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = $\frac{p}{2l} \sqrt{\frac{T}{m} }$

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 = $\frac{3}{120} \sqrt{\frac{T}{2} }$ I

Now if there is 5 antinodes , the value of p = 5

Thus n = $\frac{5}{120} \sqrt{\frac{T}{2} }$ II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

3 0

2 years ago