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3 years ago

What happens to the energy of a wave as it moves away from its source?

Physics

1 answer:

Eddi Din [679]3 years ago

3 0

Answer:

As the particles move further away from their normal position (up towards the wave crest or down towards the trough), they slow down.

Explanation:

This means that some of their kinetic energy has been converted into potential energy – the energy of particles in a wave oscillates between kinetic and potential energy. Hope that this helps you and have a great day :)

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A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a

IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

$\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}$

Put the value into the formula

$\dfrac{dM}{dt}=53.3+3.63$

$\dfrac{dM}{dt}=56.93\ kg/s$

We need to calculate the thrust of the jet engine

Using formula of thrust

$T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v$

Put the value into the formula

$T=56.93\times317-53.3\times260$

$T=4188.81\ N$

Hence, The thrust of the jet engine is 4188.81 N.

7 0

3 years ago

A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to

Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

$\tau = I\alpha$

Where $\tau$ is the torque

$I$ is the moment of inertia

$\alpha$ is the angular acceleration

But, the angular acceleration is given by

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

Now, to determine the angular speed, we first determine the Torque $\tau$ and the moment of inertia $I$ .

Here, The torque is given by,

$\tau = rF$

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴ $I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0

3 years ago

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b

Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is $n_2 = 1.333$ , and for air $n_1 = 1.00$ : the angle of light with the normal is $90^o-60^o = 30^o$ ; therefore Snell's law gives