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zepelin [54]
3 years ago
12

Wave C has an amplitude of 1 and wave D has an amplitude of 3 as shown below. What will happen when the trough of wave C meets t

he crest of wave D?
They will interfere to create a crest with an amplitude of 4.
They will interfere to create a crest with an amplitude of 2.
They will interfere to create a crest with an amplitude of 0.
They will bounce off each another.

Physics
2 answers:
Lesechka [4]3 years ago
5 0

Answer:

They will interfere to create a crest with an amplitude of 2.

Explanation:

The two waves will interfere and the resultant amplitude will be given by the principle of superposition, which states that the resultant amplitude is given by the sum (or the difference, if they meet in opposite phase) of the amplitudes of the two waves. In this case, the crest of wave C meets the through of wave D, so they are in opposite phase, therefore the resultant amplitude will be

A = A(C) - A(D) = 3 - 1 = 2

gtnhenbr [62]3 years ago
4 0
They will interfere to create a crest with an amplitude of 2
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What are the main factors that influence the climate of any location on earth?
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Answer:

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8 0
3 years ago
In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 vo
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Answer:

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Explanation:

Given that,

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Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

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6 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

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#LearnwithBrainly

6 0
3 years ago
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