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ad-work [718]
4 years ago
15

How does climate change affect the ocean??

Physics
1 answer:
Debora [2.8K]4 years ago
6 0
It raises the water level because of the melting glaciers. 
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Physical values in the real world have two componnts: magnitude and ____
Dahasolnce [82]

Answer:

DIRECTION

Explanation:

3 0
3 years ago
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.
Black_prince [1.1K]
<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.
7 0
3 years ago
How many electrons are in 204 C of charge?
zubka84 [21]

Answer:

The mass number 204 – 82 protons = 122 neutrons

Explanation:

Hope this helps!

5 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are l
MariettaO [177]

Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed,v=4\times 10^{3}\ m/s

Time of collision,t=6\times 10^{-8}\ s

We know that,

Force, F = ma

Put all the values,

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N

So, the required force is 6666.7 N.

3 0
3 years ago
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