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soldi70 [24.7K]
3 years ago
6

Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an ampli

tude of 7.00 cm. Find the amplitude of the resultant wave.
Physics
1 answer:
horsena [70]3 years ago
5 0

Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

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Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>Q</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

Mk ×DTk

=<u>2</u><u>5</u><u>0</u><u> </u><u>×</u><u> </u><u>9</u><u> </u><u>×</u><u> </u><u>5</u><u> </u><u> </u>

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = <u>2</u><u>5</u><u>×</u><u>1</u><u>0</u> ×100

1.082

=<u>2</u><u>5</u><u>0</u><u>0</u>

1.082

= 23105.360 g/kj.

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