3 years ago

Brain

Physics

1 answer:

alina1380 [7]3 years ago

3 0

Chapter name please then I can answer

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Given v_in = 20 m/s and a = 3 m/s2, assuming that the body moves at constant acceleration, the motion is modeled by the equation:

s(t) = (v_in)t + (1/2)a(t^2)

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substituting the given,

s(t) = 20t + (3/2)(t^2)

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A wave on a string is described by D(x,t)= (2.00cm)sin[(12.57rad/m)x−(638rad/s)t], where x is in m and t is in s. The linear den

mamaluj [8]

Answer : The string tension is $T = 12.882 N$

The maximum displacement is 0.02 m

The maximum speed is $v = 12.76\ m/s$

Explanation :

Given that,

D(x,t) = (2.00 cm) sin [(12.57rad/m)x - (638rad/s)t]

Where, x is in m and t is in sec.

Linear density of the string = 5.00 g/m

We know that,

Velocity of the wave

$v = \dfrac{\omega}{k}$

$v = \dfrac{638}{12.57}\ m/s$

$v = 50.76\ m/s$

Now, the string tension

$v = \sqrt\dfrac{T}{m}$

$T = 0.005\times (50.76)^{2}\ N$

$T = 12.882 N$

The maximum displacement is 0.02 m

The maximum speed

$v = a \omega$

$v = 0.02\times638\ m/s$

$v = 12.76\ m/s$

Hence, this is the required solution.

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