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charle [14.2K]
3 years ago
10

Heat from the surface is transferred underground by _______.

Physics
2 answers:
gulaghasi [49]3 years ago
7 0

Answer: Option (D) is the correct answer.

Explanation:

  • When we place two objects of different temperature adjacent to each other then there will occur transfer of heat from hotter object to colder object till their temperature becomes equal.  

This process is known as conduction.

Hence, when heat is being transferred from surface to the underground then it is due to the process of conduction.

  • Convection is defined as the process in which a fluid (liquid or gas) when heated then more dense (cooler) substance sinks ate the bottom whereas less dense (hotter) substance will rises at the top.
  • Radiation is defined as a process in which an electromagnetic energy will travel through space in the form of waves or particles.
  • Condensation is a process in which vapor state of a substance changes into liquid state.

Thus, we can conclude that heat from the surface is transferred underground by conduction.

Ostrovityanka [42]3 years ago
5 0
A, convection, is your answer
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Explanation:

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3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

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Answer:

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