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2 years ago

Which of these phase changes requires energy?

Physics

1 answer:

frutty [35]2 years 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

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Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0

3 years ago

A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?

MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.

Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.

8 0

3 years ago

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

3 years ago

The talk test can be used to measure the __________. A. knowledge of an activity B. intensity of an activity C. length of an act

marta [7]

Answer:

Explanation:

According to CDC, if you’re doing moderate-intensity activity, you can talk but not sing during the activity.

8 0

3 years ago

Read 2 more answers

41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h

romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0

3 years ago