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Vanyuwa [196]
4 years ago
8

Which of the following statement(s) about energy and phase is/are correct? Select all that apply. Choose one or more: A. While o

nly one phase is present, adding or removing energy changes PE but not KE. B. While only one phase is present, adding or removing energy changes KE but not PE. C. During a phase change, adding or removing energy changes KE but not PE. D. During a phase change, adding or removing energy changes PE but not KE.
Physics
1 answer:
Free_Kalibri [48]4 years ago
3 0

Answer: Option (A) and (D) are the correct statements.

Explanation:

Potential energy is defined as the energy obtained by an object due to its position. Whereas kinetic energy is defined as the energy obtained by an object due to  the motion of its molecules.

When there will be only one phase present then addition or removal of energy will lead to change in kinetic energy of the substance but no change in potential energy will take place.

Whereas if change in phase is occurring then adding or removing any energy will lead to change in potential energy of the substance while kinetic energy will remain the same.

Thus, we can conclude that correct statements are as follows.

  • While only one phase is present, adding or removing energy changes PE but not KE.
  • During a phase change, adding or removing energy changes PE but not KE.
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Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

Thrust = mv

where,

m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

Thrust = (0.05\ kg/s)(4000\ m/s)

<u>Thrust = 200 N</u>

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3 years ago
The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries
harkovskaia [24]

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}

E=0.036 V/m

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3 years ago
What is meant by the term 'total internal reflection'? (GCSE Level)
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Explanation:

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In the figure i > © experience total internal reflection

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3 years ago
3730 watts equals about how many horsepower<br><br>A.5<br>B.10<br>C.20<br>D.30
ikadub [295]
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3 years ago
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A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

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             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
4 years ago
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