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3 years ago

15

My question is about pressure. Atmospheric pressure on the surface of Venus is 90 times larger than on Earth. On Mars it is 100

times smaller than on Earth. What else does this tell you about the atmospheres of these planets?

1 answer:

Brrunno [24]3 years ago

8 0

It's high because Venus atmospheric mass is about 93 times that of Earth.

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Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the

inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change

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2 years ago

How does the current in a circuit change if the resistance is doubled? The current is halved. The current is doubled. The curren

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Answer:

The current is halved

Explanation:

The relationship between the current and the resistance is given by Ohm's Law, as follows:

$V = IR\\I = \frac{V}{R}$

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I = Current

R = Resistance

Therefore, if we double the resistance:

$I' = \frac{V}{2R}\\\\I' = \frac{1}{2}I$

Hence the correct option is:

<u>The current is halved</u>

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2 years ago

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Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?

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Answer:

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2 years ago

A student is given a 5.4 g mineral sample. She determines that the volume of her sample is 2 cm3. What is the density of her min

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3 years ago

How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J

algol13

Answer:

Work done by the frictional force is $3.41\times 10^5\ J$

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

$W=k_f-k_i$

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)$

W = −340605 J

or

$W=3.41\times 10^5\ J$

Hence, the correct option is (a).

6 0

2 years ago