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OleMash [197]
3 years ago
9

Why are some substances dissolve both polar and nonpolar substances?

Physics
1 answer:
dlinn [17]3 years ago
8 0

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other.<em> polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.</em>

<em></em>

<em></em>

<em>please mark me as the brainliest :) </em>

<em></em>

<em>use your brain :)</em>

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What is true of the moon's orbital and rotational periods?
Eddi Din [679]
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
4 0
3 years ago
4 facts about relative dating?
sasho [114]
Relative dating is used to arrange geological events….

Relative dating puts geologic events in chronological order without requiring that a specific numerical age be assigned to each event….

Relative Dating uses the half life of isotopes to get the exact age of a rock or mineral.
6 0
2 years ago
Which has a greater momentum? A 1200kg car traveling at 30m/s or a 2000kg truck traveling at 20m/s
Fittoniya [83]

the formula for momentum is velocity times mass

car :

1200 x 30 = 36000

truck:

2000 x 20 = 40000

Ans : truck has a greater momentum

7 0
3 years ago
An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only
Gnesinka [82]

Answer:

15.75 m/s

Explanation:

v = Velocity of the combined mass of astronaut and tools = 1.8 m/s

m_1 = Mass of astronaut = 124 kg

m_2 = Mass of tools = 16 kg

v_1 = Velocity of astronaut = 0

v_2 = Velocity of tools

As linear momentum is conserved

m_1v_1 + m_2v_2 =(m_1 + m_2)v\\\Rightarrow v_2=\frac{(m_1 + m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(124+16)\times 1.8-124\times 0}{16}\\\Rightarrow v_2=15.75\ m/s

The velocity of the tools is 15.75 m/s

3 0
3 years ago
How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un
Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

W=K.E

W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2

Put the value into the formula

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=1.122\times10^{-9}\ J

W=7001\ MeV

(b). We need to calculate  the momentum of this proton

Using formula of momentum

p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value into the formula

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}

p=1.404\times10^{-26}c

p=4.20\times10^{8}\ kg-m/s

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

4 0
3 years ago
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