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N76 [4]
3 years ago
10

When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At th

is pressure, 1.00kg of water has a volume of 1.00×10−3m3, and 1.00 kg of steam has a volume of 0.824m3.
a)Compute the work done when 1.00kg of steam is formed at this temperature

b)Compute the increase in internal energy of the water.
Physics
1 answer:
Vikki [24]3 years ago
6 0

Answer:

2033219.05 J

Explanation:

V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

L_v = Heat of vaporization = 2.2\times 10^6\ J/kg

Work done in an isobaric system is given by

W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J

Work done is 166780.95 J

Change in internal energy is given by

\Delta U=Q-W

Heat is given by

Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J

\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J

The increase in internal energy of the water is 2033219.05 J

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As a box is pushed 30 meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the b
irakobra [83]

Answer:

1,050 Joules

Explanation:

<u>Step 1:</u> work done in moving the box 30 meters

work done = force X distance

                  = 25N X 30 = 750 Joules

<u>Step 2: </u>calculate total internal energy

Total internal energy = work done + kinetic energy

                                   = 750 Joules + 300 Joules

                                   = 1,050 Joules = 1.05 KJ

5 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
The following force diagram represents Newton’s Third Law of Motion:
Julli [10]

Answer:

<u>FALSE.</u>

Explanation:

Newton's third law states that :

  • <em>Every action has equal and opposite reaction</em>
  • <em>That is , the magnitude is the same but the directions are opposite</em>
  • <em>The action reaction forces DONOT operate on the same body.</em>

For example ,

If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>

Thus,

we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>

4 0
3 years ago
A 20 N unbalanced force causes an object to accelerate at 1.5 m/s2. What is the mass of the object?
Ket [755]

<u>Answer:</u>

Force = 20N

acceleration (a) = 1.5 m/s²

Mass of object (m) = ?

<u>From Newtons II law</u>

                   <em>    F = m. a N</em>

                        m =  F/a

                        m = 20/1.5

                      <em>  m = 13.34 Kg</em>

<em>Mass of an object is 13.34 Kg</em>

8 0
4 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
guajiro [1.7K]

Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          f_{e} = 1.7 cm

6 0
3 years ago
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