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3 years ago

What does a mechanical wave always travel through?

1 answer:

6 0

A mechanical wave<span> requires an initial energy input. Once this initial energy is added, the </span>wave travels through<span> the medium until all its energy is transferred.</span>

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The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$

v is the maximum speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s$

Hence, the maximum speed of the ball is 5.86 m/s.

3 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

3 years ago

Weather maps use isobar lines to show weather conditions. Which of the following information is indicated by these lines?

marishachu [46]

A its is high and low pressure areas

4 0

3 years ago

Read 2 more answers

Atoms contain empty space true or false

olga nikolaevna [1]

The answer would be false

5 0

3 years ago

yvonne van gennip of the netherlands ice skated 10.0 km with an average speed of 10.8 m/s. suppose vang ennip crosses the finish

Sophie [7]

By conservation of momentum, we will find that the mass is 4.97 kg.

So in the original system, we have two objects, the bouquet of flowers of mass M that is not moving and Yvonne, which has a mass of 63 kg and a speed of 10.8 m/s.

Then the total momentum of this system is:

P = (63kg)*(10.8m/s) + M*(0 m/s) = 680.4 kg*m/s.

Remember the conservation of momentum, thus, the final momentum must be equal to the above one.

In the final situation, Yvonne and the bouquet move together with a speed of 10.01 m/s,

Then the final momentum is:

P' = (63kg + M)*(10.01 m/s)

And that must be equal to the initial momentum, then we have the equation:

(63kg + M)*(10.01 m/s) = 680.4 kg*m/s.

630.63kg*m/s + M*10.01 m/s = 680.4 kg*m/s.

M*10.01 m/s = 680.4 kg*m/s - 630.63kg*m/s = 49.77 kg*m/s

M = (49.77 kg*m/s)/(10.01 m/s) = 4.97 kg

So the mass of the bouquet is 4.97 kg

If you want to learn more about momentum, you can read:

brainly.com/question/19636349

3 0

2 years ago