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1 year ago

Which of these phase changes requires energy?

Physics

1 answer:

frutty [35]1 year ago

7 0

Answer:

Explanation:

You have to give energy away for B. You have to think about that carefully. The CO2 starts out with a great deal of energy (mostly KE) and has to slow down to go from gas to a solid. Not B

In general C is the same way. Water has more energy than ice. Not C

Same principle in D. Not D.

So it's A

You might be interested in

What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?

Anuta_ua [19.1K]

The potential energy of the rock is 30,000 J

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):

$E=PE+KE$

where

PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field

KE is the kinetic energy, which is the energy possessed by the object due to its motion

In this problem, the rock is at rest, so its kinetic energy is zero:

KE = 0

Therefore, the energy of the rock is just equal to its potential energy, which is:

$E=PE=mgh$

where

m = 10.2 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

h = 300 m is the height of the rock above the ground

Substituting and solving, we find

$PE=(10.2)(9.8)(300)=30,000 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

A bus driver heads south with a steady speed of

Nutka1998 [239]

ANS

{ A }

Magnitude

$5400.097 \: ✓ \: m$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

{ B }

Average speed in ( m/s )

$23.43 \: ✓ \: m/s$

{ C }

Magnitude

$14.06 \: ✓ \: m/s$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

5 0

2 years ago

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$

the distance is given by:

$d=\theta*r$

$d=381rad*(35.5*10^{-2}m)\\d=135m$

4 0

3 years ago

What is championship game of baseball called? The World Series The Super Bowl The World Cup

emmasim [6.3K]

Answer:

The World Series

Explanation:

The Super Bowl is the championship American Football game, and the World Cup is the Soccer/Football game.

American Football and Football are different things. The first is what Americans call football, while the other is what Americans call soccer. It is confusing.

6 0

2 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

t= 1hr

4 0

3 years ago

Read 2 more answers