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Nimfa-mama [501]
4 years ago
9

Amir observes Wave 1 and Wave 2 crashing into each other at two different intervals. His experiments produce Wave 3 and Wave 4.

Amir records his data in a table.
What is the best statement about the data collected in Amir’s table?


Wave 3 resulted from destructive interference, and Wave 4 resulted from constructive interference.


Waves 3 and 4 resulted from constructive interference.


Waves 3 and 4 resulted from destructive interference.


Wave 3 resulted from constructive interference, and Wave 4 resulted from destructive interference.

Physics
2 answers:
topjm [15]4 years ago
6 0

Answer:

"Wave 3 resulted from constructive interference and Wave 4 resulted from destructive interference."

Explanation:

Constructive Interference:

When two waves meet in such a way that their crests line up together, then it's called constructive interference. The resulting wave has a higher amplitude.

Destructive Interference:

In destructive interference, the crest of one wave meets the trough of another, and the result is a lower total amplitude.

The amplitude of Wave 3 is higher than both the Waves, 1 & 2, so it must be a result of constructive interference.

The amplitude of Wave 4 is less than both the Waves, 1 & 2, so it must be a result of destructive interference.

Leokris [45]4 years ago
6 0

Answer:

Wave 3 resulted from constructive interference and Wave 4 resulted from destructive interference."

Explanation:

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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Gnesinka [82]

Answer:

a).a_p=-2.39x10^{-12} rad/s^2

b).t=1016298.8 years

c).T_i=80.58x10^{-3}s

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

a_p=\frac{dw}{dt}

w=\frac{2\pi}{t}

a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}

dT=0.0808s

dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s

Replacing

a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2

b).

If the pulsar will continue to decelerate at this rate, it will  stop rotating at time:

t=\frac{w}{a_p}

w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s

t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s

t=1016298.8 years

c).

582 years ago to 2019

1437

T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s

5 0
3 years ago
At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan
Elina [12.6K]

Answer:

2.67kg

Explanation:

The maximum velocity, v _ {max} of a body experiencing simple harmonic motion is given by equation (1);

v_{max}=\omega A............(1)

where \omega is the angular velocity and A is the amplitude.

The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);

\omega=\sqrt{\frac{k}{m}}.................(2)

where k is the force constant of the spring and m is the loaded mass.

We can make \omega the subject of formula in equation (1) as follows;

\omega=\frac{v_{max}}{A}.................(3)

We then combine equations (2) and (3) as follows;

\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)

According to the problem, the following  are given;

v_ {max }=1.92m/s\\A=0.21m\\k=223N/m

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}

9.143=\sqrt{\frac{223}{m}}

Squaring both sides, we obtain the following;

9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg

8 0
3 years ago
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