Ask question

Ask question

All categories

3 years ago

7

Explain the role that heat plays in phase change. You will need to describe heat’s role in the phase change using terms such as

endothermic or exothermic and kinetic energy

1 answer:

Mekhanik [1.2K]3 years ago

4 0

Answer:

it will change to exothermic and kinetic energy

Explanation:

You might be interested in

Uploaded imageTwo long, parallel wires each carry the same current I in the same direction (see figure below). The total magneti

LekaFEV [45]

Answer:

Zero

Explanation:

Two long parallel wires each carry the same current I in the same direction. The magnetic field in wire 1 is given by :

$B_1=\dfrac{\mu_oI_1}{2\pi r}$

Magnetic force acting in wire 2 due to 1 is given by :

$F_{2}=I_2lB_1$

$F_2=\dfrac{\mu_oI_1I_2 l}{2\pi r}$

Similarly, force acting in wire 1 is given by :

$F_1=\dfrac{\mu_oI_1I_2 l}{2\pi r}$ According to third law of motion, the force acting in wire 1 will be in opposite direction to wire 2 as :

$F_2=-F_1$

So, the total magnetic field at the point P midway between the wires is in what direction will be zero as the the direction of forces are in opposite direction.

7 0

3 years ago

Indicate whether the statement is true or false. an astronaut weighs the same on earth as in space.

tino4ka555 [31]

Weight is different (but mass is the same)

6 0

3 years ago

Read 2 more answers

How does Mr. Anderson define solubility in the video?

VikaD [51]

Answer:

7.3

Explanation:

4 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

Suppose the polar bear was running on land instead of swimming. If the polar bear runs at a speed of about 8.3 m/s, how far will

olga_2 [115]

Answer:

298800 m

Explanation:

$v = \frac{d}{t}$

V=speed

d=distance

t=time

$d = vt$

Change hours to seconds

It will be 36000s

8.3m/s * 36000s

=298800m

8 0

3 years ago