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denis23 [38]
2 years ago
8

A sinusoidal wave on a string is described by the wave function

Physics
1 answer:
MAXImum [283]2 years ago
5 0

The power transmitted by the wave is 21.1ω.

<h3>What is a wave?</h3>

It refers to the disturbance in a medium that carries energy without a net movement of particles.

<u>Given</u>

y = A.sin(kx−ωt)

A = 15m

k = 8m^−1

ω=  50rad/sec

P = 1/2​μA²ω²v

P = 1/2 ×12×10^−3 × (0.15)^2 ×  (50)^2 × 0.62.5

P = 21.1ω

Therefore, the power transmitted by the wave is 21.1ω.

Read more about wave

brainly.com/question/25699025

#SPJ4

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A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin
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Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

x_{1} = 0.035 m  , v_{1} = 0

x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

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Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

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W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

v_{2} = 0.188 \frac{m}{s}

This is the velocity of block.

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Answer:

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