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3 years ago

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3. Describe the flow of one molecule of water through the water cycle, beginning in the ocean.

2 answers:

lubasha [3.4K]3 years ago

7 0

Answer:

Water travels in a continuous path through Earth called the water cycle. The journey starts in the Earth's oceans. As the Sun heats the ocean, the water evaporates into a gas. The gas meets cooler air, and the gas changes into droplets of tiny water molecules. This process is called condensation. Little droplets of water fall back down to Earth again as precipitation. The precipitation fills up the rivers and various other things and the cycle begins all over again.

Explanation:

scZoUnD [109]3 years ago

6 0

molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.

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At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?

Ipatiy [6.2K]

The resistance of the sample is $682.5\Omega$

Explanation:

The relationship between resistance of a material and temperature is given by the equation

$R(T)=R_0(1+\alpha (T-T_0))$

where

$R_0$ is the resistance at the temperature $T_0$

$\alpha$ is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

$R_0 = 525 \Omega$ when the temperature is $T_0 = 20^{\circ}C$

While the temperature coefficient of resistance of nickel is

$\alpha = 0.006/^{\circ}C$

Therefore, the resistance of the sample when its temperature is

$T=70^{\circ}C$

is

$R=(525)(1+0.006(70-20))=682.5 \Omega$

Learn more about resistance:

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3 0

3 years ago

Which are base units used in the metric system? Check all that apply.

Shtirlitz [24]

Liters

Grams

Degrees Celsius

The other answer choices are from the imperial system

7 0

3 years ago

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I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!

anyanavicka [17]

Answer:

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2 years ago

What would the velocity of an object be if it was on a 1.75m

JulijaS [17]

Answer:

T = 4 sec / 2 = 2 sec period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

V = S / T = 11 m / 2 sec = 5.5 m/s speed of object

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2 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago