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Karolina [17]
3 years ago
8

)) What do these two changes have in common?

Physics
1 answer:
zloy xaker [14]3 years ago
7 0

Answer:

Both are only physical changes

Explanation:

A physical change is a change that does not involve or alter the chemical composition of the substances involved. Physical changes form no new substance and can be easily separated into individual constituents. Example of physical changes are change in state, boiling, melting etc.

According to this question, two processes were given as follows:

1. mixing chocolate syrup into milk

2. rain forming in a cloud

These two processes are similar in the sense that they are both examples of physical changes.

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Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?
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Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity. 
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Calculate the root mean square velocity of nitrogen molecules at 25 C.
iragen [17]
 root mean square<span>= square root of ( 3RT/M) 

 R = 8.314 J/K/mole 
T = 25 + 273 = 298 K 
M = molecular mas of N2 in kg = 28 X 10^-3 kg 

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3 years ago
What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?In meters
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What two factors affect the rate of acceleration in an object
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Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the
Assoli18 [71]

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

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3 years ago
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