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3 years ago

How does energy travel through a wave in a medium? Plz help!

Physics

1 answer:

melamori03 [73]3 years ago

7 0

Answer:

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles

Explanation:

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The highest frequency radio waves are which type ?

Zolol [24]

Answer: Gamma rays

Explanation: Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies.

7 0

4 years ago

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$

So, the kinetic energy of the ball at the point of maximum speed is

$K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s$

7 0

3 years ago

A pendulum of length l=5.0m attached to the ceiling carries a ball of mass 10.0 kg. The ball (a massive bob) is moved from its s

never [62]

Answer:

Em₀ = 245 J

Explanation:

We can solve this problem with the concepts of energy conservation, we assume that there is no friction with the air.

Initial energy the highest point

Em₀ = U

Em₀ = m g h

The height can be found with trigonometry

The length of the pendulum is L and the length for the angle of 60 ° is L ’, therefore the height from the lowest point is

h = L - L’

cos θ = L ’/ L

L ’= L cos θ

h = L (1 - cos θ)

We replace

Em₀ = m g L (1- cos θ)

Let's calculate

Em₀ = 10 9.8 5.0 (1 - cos 60)

Em₀ = 245 J

3 0

3 years ago

The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your

Marina CMI [18]

Explanation:

It is given that, the force of gravity pulls down on your school with a total force of 400,000 N.

The force of gravity is given by the formula as follows :

F = mg

m is mass and g is the acceleration due to gravity

Mass of an object remains same always. But if the mass of your school becomes twice of the initial mass, the force of gravity pulling down on your school would be exactly twice.

4 0

3 years ago

A point on a wheel of radius 40 cm that is rotating at a constant 5.0 revolutions per second is located 0.20 m from the axis of

Margarita [4]

Answer:

$197.2 m/s^2$