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Arturiano [62]
3 years ago
7

What can accurately be said about a resultant wave that displays both reinforcement and interference? A. The component waves hav

e different frequencies. B. Maximum interference occurs where the troughs of the two component waves are in phase. C. Molecules remain in their normal positions at spots of reinforcement. D. The crests of the two component waves are in phase where interference occurs in the resultant wave. Mark for review (Will be highlighted on the review page)
Physics
2 answers:
Agata [3.3K]3 years ago
7 0
<h3><u>Answer;</u></h3>

A. The component waves have different frequencies.

<h3><u>Explanation;</u></h3>
  • <em><u>interference is one of the property of waves that occurs when  two waves superpose to form a resultant wave of greater, lower, or the same amplitude.</u></em>
  • Interference occurs when two waves travelling in the same medium at the same time. It may be either a constructive or a destructive interference.
  • <em><u>Constructive interference occurs when the two waves are in phase and have the same frequency, the amplitudes are therefore, reinforced resulting to a constructive interference.</u></em> However, if the two waves are out phase the result will be a destructive interference.
Tanya [424]3 years ago
4 0
The answer is A. <span>The component waves have different frequencies.
The magnitudes of reinforcement usually really dependent on the number of frequencies and interference is usually caused due to the difference in frequencies. So, we can conclude that if the frequencies are different and causing interference, the reinforcement will also different
</span>
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S A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected
Flura [38]

The power of is series combination is Vn^2 times that of a parallel combination.

For series combination :

Req = R + R + R + ............... n times = nR

I = Δv/nr

Power = (Δv/nr)^2 × nr = Δv^2/nr

For parallel combination

1/req = 1/R + 1/R + 1/R +................(n times) = n/R

Req = R/n

Power = Δv/(R/n) = nΔv^2/R

Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2

Hence, power of is series combination is Vn^2 times that of a parallel.

Learn more about parallel combination here:

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3 0
1 year ago
Bromine, a liquid at room temperature, has a boiling point
lukranit [14]

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

3 0
3 years ago
. Since the man is walking at a constant velocity, he has what acceleration
docker41 [41]
If he’s walking at a constant velocity there is no acceleration.
6 0
3 years ago
A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
2 years ago
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