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Studentka2010 [4]
2 years ago
14

If the amplitude of the resultant wave is twice as great as the amplitude of either component wave, and the wave exhibits reinfo

rcement, the component waves must
A. Have a different wave length than the resultant wave
B. Be in phase with each other
C. Have a different frequency than the resultant wave
D. Be traveling in the opposite direction of the resultant wave
Physics
2 answers:
shutvik [7]2 years ago
8 0

Correct answer choice is:

B. Be in phase with each other.

____________________________________________________________

Explanation:

Components waves mix to produce a resultant with the equivalent wavelength but the amplitude which is bigger than the amplitude of both of the unique component waves, and this results in constructive interference, which states that the interference of two or more waves of identical frequency and phase, constructing in their respective reinforcement and constructing a single amplitude equivalent to the total of the amplitudes of the unique waves.

Lunna [17]2 years ago
5 0
Be in phase with each other! otherWise B
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Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
2 years ago
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using 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 time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0
3 years ago
Read 2 more answers
approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos
Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as:  v_{av}  = \frac{x - x_{0} }{t - t_{0} }

where t = final time

           t₀ = initial time

           v = final speed

           v₀ = initial time

where x denotes the position, then

v_{ave} = \frac{v+v_{0} }{2}

where v = \frac{dx}{dt} and dx = change in distance with respect to time.

6 0
3 years ago
A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

v = u + at

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0
2 years ago
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