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3 years ago

How is hydrogen separated from other nearby substances

Physics

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:

Membranes

Explanation:

Hydrogen separation uses membranes to separate the hydrogen from other gases, leaving it in its purest and most valuable form. The separated hydrogen that is captured can be used for many different applications including: LIFTING: As a lifting agent similar to helium.

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Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

mart [117]

1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.

7 0

3 years ago

Atoms that have the same number of outer electrons

Varvara68 [4.7K]

B they belong to the same family of elements

8 0

3 years ago

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

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4 years ago

A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distancer1

Jobisdone [24]

Answer:

The magnitude of the electric field in the second case will be $\frac{1}{8}$ of the electric field in the first case

Explanation:

The electric field at a distance $r_1$ < R from the center of the sphere is given by

the formula $E_1$ = $\frac{kQr_1}{R^3}$

where R is the radius of the sphere, and, Q is the charge uniformly distributed on the sphere.

It is given that the same charge Q is distributed uniformly throughout a sphere of radius 2R and we have to find the electric field at same distance $r_1$ from the center.

In the second case the electric field will be given by

$E_2$ = $\frac{kQr_1}{(2R)^3} = \frac{kQr_1}{8R^3} =\frac{1}{8} \frac{kQr_1}{R^3} = \frac{1}{8} E_1$