What happens to molecules when heat is absorbed in a substance

A constant net force acts on an object. which of the following best describes the object's motion?

Anton [14]

The object is moving with a decreasing acceleration. The object is moving with a constant velocity.

4 0

3 years ago

PLEASE HELP!!! 20 PIONTS!!!

timama [110]

Answer:

Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.

Explanation:

The movement of the medium is different. In the longitudinal wave, the medium moves left to right, while in thee transverse wave, the medium moves vertically up and down. Longitudinal waves have a compression and rarefaction, while the transverse wave has a crest and a trough. Longitudinal waves have a pressure variation, transverse waves don't have pressure variation. Longitudinal waves can be propagated in solids, liquids and gases, transverse waves can only be propagated in solids and on the surfaces of liquids. Longitudinal waves have a change in density throughout the medium, transverse waves don't.

4 0

2 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when th

iogann1982 [59]

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

$F_{e} = F_{m}$

q E = qv B

v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

6 0

3 years ago

A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?

dusya [7]

The cliff is 319. kilograms above the water

5 0

2 years ago