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vaieri [72.5K]
4 years ago
9

How is heat energy transferred from a warm object to a cold object there is direct contact?

Physics
2 answers:
sdas [7]4 years ago
8 0
A is the answer!!!!!!!!!!!!
eduard4 years ago
4 0

<u>Answer:</u> The heat is getting transferred by conduction.

<u>Explanation:</u>

There are 3 modes of heat transfer:

  • <u>Conduction:</u> This type of heat transfer happens when there is direct contact between the two objects.
  • <u>Convection:</u> This type of heat transfer happens when there is a movement of fluid (liquid or gas).
  • <u>Radiation:</u> This type of heat transfer happens when there is direct transfer of energy through space.

We are given:

Heat energy is getting transferred from a warm object to cold object by direct contact.

Hence, the heat is getting transferred by conduction.

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I got the correct answer, but I don't know what I did. Why is the answer E?
MariettaO [177]

Explanation:

Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle.  So her displacement is 100 m.

Person Y walks 3/4 of the way around.  So his displacement is 50√2 m ≈ 70.7 m.

Person Z walks completely around the circle, so their displacement is 0 m.

Therefore:

Z < Y < X

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3 years ago
Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?
Molodets [167]
Taking notes,because you can read back on what you learned
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3 years ago
A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

7 0
4 years ago
The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's
Ivan

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity v>0.

The acceleration a is the rate of change of the velocity v. This means that if v is <em>increasing during</em> time, then a must be positive. But if v is <em>decreasing over</em> time, then a will be negative (even though the velocity is positive).

Mathematically:

a=\frac{dv}{dt}

v decreases ⇒\frac{dv}{dt}

⇒a.

Example:

v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}

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