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2 years ago

6

The gas in a cylinder has a volume of 3 liters at a pressure of 117 kpa. the pressure of the gas is increased to 219 kpa. assumi

ng the temperature remains constant, what would the new volume be?

1 answer:

solniwko [45]2 years ago

8 0

V = (1 / p) × p0×V0 = (1 / 219kPa) × 117kPa × 3L = 1.06L

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640 nanometer setara dengan

lidiya [134]

Answer:

640 nanometer setara dengan 6.4e-7 meter

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1 year ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

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2 years ago

Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur

Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

$\mu_{J,T}=(\frac{dT}{dP})_H$

or,

$\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}$

As per question the formula will be:

$\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}$ .........(1)

where,

$\mu_{J,T}$ = Joule-Thomson coefficient of the gas = $0.13K/atm$

$T_1$ = initial temperature = $19.0^oC=273+19.0=292.0K$

$T_2$ = final temperature = ?

$P_1$ = initial pressure = 200.0 atm

$P_2$ = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

$0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}$

$T_2=266.12K$

Therefore, the final temperature of gas is 266.12 K

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2 years ago

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Inessa05 [86]

Class 1 lever

Explanation:

In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.

The effort is the force input and the load is the force output
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Movement of the effort and load are in opposite directions.
There are other classes of lever like the class 2 and 3.
They all have different load, fulcrum and effort configurations

learn more:

Load related problems brainly.com/question/9202964

Torque brainly.com/question/5352966

#learnwithBrainly

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2 years ago

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olchik [2.2K]

Answer:

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Let the initial displacement of the hiker, = x = 2km

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R' = x' + y'

Check the image uploaded for solution;

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2 years ago