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Olegator [25]
3 years ago
7

An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl

ide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of n mol of the gas is raised from T1 to T2? (Use any variable or symbol stated above along with the following as necessary: R.)
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

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A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
nignag [31]
<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

3 0
3 years ago
This is my question. I need the answer stat.
Kipish [7]

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

5 0
2 years ago
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Another athlete using a different spring exerts an average force of 400 N to enable her
babymother [125]

Answer:84Nm

Explanation:

force=400N

Distance=0.210m

Workdone=force x distance

Workdone=400 x 0.210

Workdone=84Nm

6 0
3 years ago
1. The horizontal and vertical components of a projectile's velocity are
Anni [7]

The horizontal and vertical components of a projectile's velocity are independent of each other.

Answer: Option C

<u>Explanation:</u>

The path of a projectile is determined by two components of motion. They are termed as horizontal and the vertical components. Since both components velocity are perpendicular to each other, so it can stated that they are independent of each other.

Even it can seen that when the horizontal components of velocity is constant, then there will be change in the vertical components of velocity leading to free fall projectile path.

And in the absence of gravity, there will be change in the horizontal components of velocity with zero vertical component of velocity. Thus, the horizontal and the vertical components of a projectile’s velocity are seemed to be independent of each other.

5 0
3 years ago
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