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2 years ago

What waves need molecules in order to transfer energy

Physics

1 answer:

nevsk [136]2 years ago

4 0

Answer:

Mechanical waves

Explanation:

Waves are periodic oscillations, that carry energy, but not matter.

Waves are classified into two types:

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, which can oscillate along the direction of propagation of the wave (longitudinal wave) or perpendicular to the direction of motion of the wave (transverse wave). These waves can only propagate in a medium, so they cannot travel in a vacuum. Examples of mechanical waves are sound waves.

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field. They are transverse waves. They are the only type of wave able to propagate through a vacuum (so, through space).

Therefore, the waves that need molecules in order to transfer energy are mechanical waves.

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A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c

Sonbull [250]

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

$F=\frac {m(v-u)}{t}$

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

$F=\frac {0.3(2--5)}{0.0015}=1400N$

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2 years ago

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: $Vf^{2} = Vo^{2} - 2gh$ and $h=Vo*t - \frac{g*t^{2} }{2}$ , First we need to find the height to time iqual to 3.04 s using the formula: $h=Vo*t - \frac{g*t^{2} }{2}$ and remember that golf ball was released from the rest (Vo= 0 (m/s)) so $h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}$ , we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using $Vf^{2} = Vo^{2} - 2gh$ solving for Vf, we get: $Vf = \sqrt{Vo^{2}-2*g*h }$ replacing the values given $Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }$ , so we get: Vf = 29.79 (m/s).

5 0

3 years ago

Read 2 more answers

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$