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den301095 [7]
3 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]3 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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A wheel, starting from rest, rotates with a constant angular acceleration of 2.80 rad/s2. During a certain 5.00 s interval, it t
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Answer:

a) time t1 = 2.14s

b) initial angular speed w1 = 6 rad/s

Explanation:

Given that;

Initial Angular velocity = w1

Angular distance = s = 65 rad

time = t = 5 s

Angular acceleration a = 2.80 rad/s^2

Using the equation of motion;

s = w1t + (at^2)/2

w1 = (s-0.5(at^2))/t

Substituting the values;

w1 = (65 - (0.5×2.8×5^2))/5

w1 = 6rad/s

Time to reach w1 from rest;

w1 = at1

t1 = w1/a = 6/2.8 = 2.14s

a) time t1 = 2.14s

b) initial angular speed w1 = 6 rad/s

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4 years ago
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We know that an electric field contains energy. What causes the electric field itself?
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Electrical charges on one or more particles within the field cause the electric field

Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field

A region of space surrounding an electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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