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2 years ago

Easy one here pls help thanks so much

Physics

2 answers:

Kisachek [45]2 years ago

8 0

The answer is B. Nutrients.

Alexus [3.1K]2 years ago

7 0

The answer is B. Nutrients

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A star with a mass like the Sun which will soon die is observed to be surrounded by a large amount of dust and gas -- all materi

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Answer: D. Infrared

Infrared is the best way to observe it.

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2 years ago

The bodies of many cars are designed to compress or crumple during an accident. Why are cars built with a crumple zone?

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By collapsing and crumpling, the force and energy is absorbed by the crumple zone and it is not transferred to the passengers inside. This is a safety feature for the passengers.

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2 years ago

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

pav-90 [236]

Answer:

The permittivity of rubber is $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

The magnitude of the point charge is $q_1 = 70 \ nC = 70 *10^{-9} \ C$

The diameter of the rubber shell is $d = 32 \ cm = 0.32 \ m$

The Electric field inside the rubber shell is $E = 2500 \ N/ C$

The radius of the rubber is mathematically evaluated as

$r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

$E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

=> $E * \epsilon * 4 * \pi * r^2 = Q$

=> $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

$\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

$\epsilon = 8.703 *10^{-11}$

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3 years ago

How much power is needed to move a 2000 kg mass from the bottom of the 150 m talk great pyramid in Egypt up a ramp if the ramp h

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Answer:

How Heavy? More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

Explanation:

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2 years ago

When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas

Bingel [31]

<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure $\Delta P$ is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

$\Delta U=\Delta Q-\Delta W$ (1)

Where:

$\Delta U$ is the internal energy

$\Delta Q$ is the heat transferred

$\Delta W$ is the work

Now, for an isobaric process:

$\Delta W=P\Delta V$ (2)

Where:

$P$ is the pressure (<u>always positive</u>)

$\Delta V$ is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive $\Delta V$ ), the work is positive.

-If the gas compresses (negative $\Delta V$ ), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

$\Delta U=\Delta Q - P\Delta V$ (3)

Clearing $\Delta Q$ :

$\Delta Q=\Delta U+P \Delta V$ (4)

Then, for an ideal gas in an isobaric process, part of the heat ( $Q$ ) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

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3 years ago