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3 years ago

What is unit of pressure

Physics

2 answers:

Amiraneli [1.4K]3 years ago

6 0

Answer:

It should be PCI

Explanation:

My head

REY [17]3 years ago

5 0

Answer:

pascal

Explanation:

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gregori [183]

Answer:

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3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

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3 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

According to the Big Bang Theory, the universe is __________________. A) constant B) depleting C) expanding D) exploding

DiKsa [7]

C) expanding hope this answer helps

6 0

3 years ago

Read 2 more answers

I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow

Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0

3 years ago