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Marat540 [252]
3 years ago
9

This is what occurs when matter transitions between solid, liquid and gas.

Physics
2 answers:
Annette [7]3 years ago
5 0

Answer:

phase change

Explanation i did the usatest prep

vodomira [7]3 years ago
4 0

Answer:

The answer is Phase Change

Explanation:

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I’ve done this before the answer is B
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1 year ago
If a persons skin is dry she will experience more of a shock
NeX [460]

Answer:

add lotion

Explanation:

7 0
3 years ago
Read 2 more answers
PE=30J, m=?, g=10m/s2, h=10m
OleMash [197]
Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

PEgrav=mgh
30J=m(10)(10[tex] \frac{30}{100} =m)[/tex]

Do operations that you can with what is given first.

30J=m(100m)

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

\frac{30}{100} =m

m = 0.30kg

5 0
3 years ago
One way in which the atmosphere helps us is by absorbing solar _____?
Damm [24]
The atmosphere absorbs most of the solar RADIATION in the atmosphere, but not all which is how we still get sun burns. :)
8 0
3 years ago
A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge
Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

\tau=RC

where

R=50,000 \Omega is the resistance

C=2.0\mu F=2.0\cdot 10^{-6}F is the capacitance

Substituting,

\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s

The charge on a charging capacitor is given by

Q(t)=Q_0 (1-e^{-t/\tau} ) (1)

where

Q_0 is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Q(t)=0.90 Q_0

Substituting this into eq.(1) we find

0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s

4 0
3 years ago
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