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Alex73 [517]
2 years ago
7

Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on ex

pansion, what is its temperature when the initial volume has tripled
Physics
1 answer:
tiny-mole [99]2 years ago
3 0

Answer:

Explanation:

It is a case of adiabatic expansion .

T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}

T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.

Given ,

V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .

( 3 )^{\gamma-1}= \frac{292}{ T_2}

( 3 )^{1.4-1}= \frac{292}{ T_2}

1.552 = 292 / T₂

T₂ = 188 K .

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Answer:

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Explanation:

8 0
2 years ago
A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0
2 years ago
An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc
Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0
2 years ago
A hair dryer is a pump for air. Do a battery and a Genecon act like pumps for charge?
lyudmila [28]

Answer:

No

Explanation:

8 0
3 years ago
Read 2 more answers
A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground
Oksi-84 [34.3K]

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

  Using;

                      V²  = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

     V²  = 2gH

    V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

7 0
3 years ago
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