If you could find the coldest object on Earth, you would find that it _________.

2 answers:

8 0

B. Has some thermal energy. A and B both refer to absolute zero, which is nearly impossible to reach on earth or elsewhere. D is incorrect because it's impossible to emit cold, since cold isn't a form of energy (rather, it's just the absence of heat).

eduard4 years ago

7 0

If you found the coldest object on Earth, you would find that
it has some thermal energy.

(In fact, this is even true of the coldest object in the solar system, or
in space itself. Space itself is about 3K warmer than absolute zero.)

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What is the wave speed of a wave that has a frequency of 250 Hz and a wavelength of 0.35 m?

azamat

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

$v=\lambda f$

where

v is the wave speed

$\lambda$ is the wavelength

f is the frequency

In this problem, we have

$f=250 Hz$ is the frequency

$\lambda=0.35 m$ is the wavelength

Substituting into the equation, we find

$v=(0.35 m)(250 Hz)=87.5 m/s$

6 0

3 years ago

A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two

AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

$V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\$

k=8.89*10^9

For both cases we have

$k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}$

(b) by replacing this values of r in the expression for V we obtain

$k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}$