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3 years ago

How do particles move when a transverse wave passes through a medium?

1 answer:

4 0

In transverse waves, particles of themedium vibrate to and from in a direction perpendicular to the direction of energy transport.

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A bowling ball is rolling in a straight line toward the pins at a speed of 2.2 m/s. Its momentum is 4.75 kg*m/s. What is the

Thepotemich [5.8K]

Answer: m= 2.16 kg

Explanation: Momentum is expressed in the following formula:

p = mv

Derive to find m:

m = p / v

= 4.75 kg.m/s / 2.2 m/s

= 2.16 kg

Cancel out m/s and the remaining unit is in kg.

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4 years ago

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Which action might lead scientists to develop new explanations about the

VikaD [51]

Answer:

vrzjxuxuxjcckhcouvou

Explanation:

gghhjjokjjhghhrjxjxkyxygbjv

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3 years ago

a square loop of wire side 3.0 cm carries 3.0 A of current. A uniform magnetic field of magnitude 0.67 T makes an angle of 37 de

Varvara68 [4.7K]

Explanation:

Below is an attachment containing the solution.

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3 years ago

Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an

borishaifa [10]

Answer:a

Explanation:

Given

two bodies are falling with negligible air Resistance and one of the body is given a slight horizontal acceleration

since there is no change in vertical acceleration therefore time taken by both bodies is same irrespective of change in horizontal direction

suppose both start from rest from a height of h

time taken

$t=\sqrt{\frac{2h}{g}}$

The only difference is that body with horizontal acceleration will be some distance away from first

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3 years ago

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A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What

almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.

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3 years ago