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arsen [322]
3 years ago
6

1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper

ature of the air was 300 k, and the air temperature becomes 500 k after the expansion. what is the boundary work done by the air

Physics
1 answer:
rewona [7]3 years ago
4 0

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W =  R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n =  34,53 mol

Finally, in (2):

W =  (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

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Explanation:

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The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty
nydimaria [60]

Answer:

Uncertainty in position of the bullet is \Delta x=1.07\times 10^{-33}\ m

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

p=mv

p=0.035\times 709=24.81\ kg-m/s

Uncertainty in momentum is,

\Delta p=0.2\%\ of\ 24.81

\Delta p=0.049

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

\Delta p.\Delta x\geq \dfrac{h}{4\pi}

\Delta x=\dfrac{h}{4\pi \Delta p}

\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}

\Delta x=1.07\times 10^{-33}\ m

Hence, this is the required solution.

7 0
3 years ago
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0
3 years ago
PLEASE HELP can not find the answer
romanna [79]
I think the answer would be letter B.
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3 years ago
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