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nikitadnepr [17]
4 years ago
12

The evaporation of the oceans surface water as part of the water cycle is caused by

Physics
1 answer:
Blababa [14]4 years ago
3 0

Answer: The Sun

Explanation:

Water cycle describes the evaporation of water from Earth's surface, formation of clouds and precipitation water back onto the surface of Earth. Seventy percent of the Earth is covered with water in the form of seas and oceans. <u>The heat of the Sun </u>causes the evaporation of water from the surface of the ocean. The water vapors rise and condense to form water droplets. The collection of these water droplets form clouds which rain down again.

The evaporation of the oceans surface water as part of the water cycle is caused by <u>the Sun</u>

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Two identical 0.25 kg balls are involved in a head-on collision. Ball A is initially travelling at 3.5 m/s, and ball B is inital
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Answer:

a) mv(final):<0,0,0> minus mv(initial):<25,0,0> = <-25,0,0>

b) mv(final):<25,0,0> minus mv(initial):<0,0,0> = <25,0,0>

c) conservation of momentum makes it <0,0,0>

for a-b-c, momentum_system + momentum_surroundings = 0

Explanation:

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The kinetic energy of the attracted magnet and iron rod came from potential energy in the system.
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The amount of potential energy stored in the magnetic field depends on both the strength of the magnet and the distance that another magnet or magnetic object is from it. ...

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Thermodynamics is the study of the _____ of heat and other forms of energy. A)conversion
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What is the IMA of an inclined plane that is 5m long and 2m high?
IrinaK [193]

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3 years ago
A beam of protons moves in a circle of radius 0.20 m. The protons move perpendicular to a 0.36-T magnetic field. (a) What is the
gregori [183]

Answer:

a) v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b) F_c=3.97\times 10^{-13}\ N

Explanation:

Given:

radius of path of motion, r=0.2\ m

we know charge on protons, q=1.6\times 10^{-19}\ C

magnetic field strength, B=0.36\ T

we've mass of proton, m=1.67\times 10^{-27}\ kg

a)

From the equivalence of magnetic force and the centripetal force on the proton:

F_B=F_C

q.v.B=\frac{m.v^2}{r}

q.B=\frac{m.v}{r}

where:

v = speed of the proton

(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}

v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b)

Now the centripetal force on each proton:

F_c=m.\frac{v^2}{r}

F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}

F_c=3.97\times 10^{-13}\ N

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