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3 years ago

Which gas law states that the pressure of a gas decreases when volume is increased and the temperature is unchanged?

1 answer:

6 0

<span>Boyle's Law, in short, states that the volume of a gas is inversely proportional to its pressure AS LONG AS THE TEMPERATURE REMAINS UNCHAnged </span>

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What is te leading cause of type 2 diabetes

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The insulin levels lead to the cause of type 2 diabetes

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3 years ago

Read 2 more answers

Encontrar la distancia y desplazamiento de las dos trayectorias si se mueve el móvil Desde A hasta B

Lerok [7]

Answer:

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Explanation:

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3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

3 years ago

calculate how much charge passes through the LED in 1 hour if the current flowing in the circuit below is 350mA

Morgarella [4.7K]

Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

<h3>What is Current?</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = Q/t

Where Q is the charge and t is time elapsed.

Given the data in the question;

Time elapsed t = 1hr = 3600s

Current I = 350mA = 0.35A

Charge Q = ?

We substitute our given values into the expression above to determine the charge.

I = Q/t

Q = I × t

Q = 0.35A × 3600s

Q = 1260C

Therefore, given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

Learn more about current here: brainly.com/question/3192435

#SPJ1

3 0

2 years ago