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Alex
3 years ago
11

The pressure of an ideal gas is 4 atm. If the only change is to increase the temperature by a factor of 4, what will be the new

pressure of the gas. Use (Pv=NkT)
Physics
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

The new pressure is 4 times of the initial pressure.

Explanation:

Given that,

The pressure of an ideal gas is 4 atm.

We need to find the change in the pressure of the gas if the temperature by a factor of 4.

We know that,

PV = nKT ....(1)

P = pressure, V = volume, T = temperature

Let P' is the new pressure of the gas. So,

P'V= nK(4T) ......(2)

Dividing equation (1) and (2) we get :

\dfrac{P}{P'}=\dfrac{T}{4T}\\\\\dfrac{P}{P'}=\dfrac{1}{4}\\\\P'=4P

It means that the new pressure is 4 times of the initial pressure.

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A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
4 years ago
A ball was rolling downhill at 2 m/s. After 5s, it was rolling at 90 m/s. What is its acceleration?
olga55 [171]

Answer:

17.6 m/s²

Explanation:

Given:

v_{f} = 90 m/s (final velocity)

v_{i} = 2 m/s (initial velocity)

Δt = 5s (change in time)

The formula for acceleration is:

a_{avg} = Δv / Δt

We can find Δv by doing

Δv = v_{f} - v_{i}

Replace the values

Δv = 90m/s - 2m/s

Δv= 88m/s

Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.

a_{avg} = Δv / Δt

a_{avg} = \frac{88m/s}{5/s}

acceleration = 17.6 m/s^{2}

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3 years ago
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Natural gas is the answer.
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What is a lenticular (S0) galaxy? A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms A galaxy
RSB [31]

Answer:

A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms

Explanation:

A Lenticular galaxy is a kind of galaxy intermediate between elliptical galaxy and a spiral galaxy in the Morphological classification system of galaxies. They have a central bulge or disc just like a Spiral galaxy but lacks the arms of spiral galaxy. If looked edge on they appear to be spiral and if looked face on they appear to be elliptical.

The absence of spiral arms can be attributed to the absence of star formation. They mainly consists of ageing stars.

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