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den301095 [7]
2 years ago
5

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas

.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
3 0

Answer:

  1. Q=0
  2. U=2820
  3. Energy increases

Explanation:

From the question we are told that

Work done W=2820

a)Generally the heat flow for an adiabatic process is 0 (zero)

Q= U + W =>0

Q=0

b)Generally Change in internal energy of gas is mathematically given by

Since W=-2820J

Therefore

U=2820

Giving

Q= 2820 -2820

Q=O

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

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Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m
Rufina [12.5K]
<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

<em>But;</em>

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0
3 years ago
What is the current when the resistance is 5 ohms and the voltage is 10 volts?
Travka [436]
Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps
5 0
3 years ago
Read 2 more answers
A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3
PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

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3 years ago
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

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Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

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The air pressure is most likely lower.
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