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2 years ago

A good refrigerant must have a high boiling point. Question 5 options: True False

Physics

2 answers:

Liono4ka [1.6K]2 years ago

5 0

Answer:

true a good refrigiant needs a high boiling point

NeTakaya2 years ago

3 0

A good refrigerant must have a high boiling point is FALSE.

Explanation:

It has the low boiling point. It should have the ability to turn into gas in absorption of heat.

If the refrigerant need to be good and ideal must have the boiling point low and freezing point also low. The low boiling point makes the refrigerant to be good as it needs to boil when the temperature crosses the limit.

It has the high critical pressure and temperature in order to avoid the large power requirements of the compressor in the refrigerator.

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= A/k x [ 0 + e^(-k * 0.1) ]

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3 years ago

Q. A mass of 300g is lifted to a height of 10m 205 by a person. Calculate his work done

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3 0

2 years ago

Pluto has been reassigned and is now a dwarf planet. Why did scientists think this reassignment was necessary? If you were a sci

kap26 [50]

Answer:

Explanation:

we humans have our own ify classification for celestial objects, most people are saddened that pluto is not a planet anymore altho it hasn't changed at all.

scientist say that if an object is going to be considered a planet it must fill in these three checkboxes:

You must be spherical, you must orbit a star, and you must have already cleared your path or debris.

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6 0

2 years ago

Read 2 more answers

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

7 0

3 years ago

Heat is applied to an ice cube in a closed container until only steam is present. draw a representation of this process, assumin

olga55 [171]

The solid, liquid and gas phases of water would have the same structure of the molecules since they are same substance. The only difference would be the distances of the molecules in the container. For a ice, the molecules are close to each other where the molecules vibrate only in place. For liquid, the molecules are freely moving and are at some distance with each other but not that far away with each other. Steam, on the other hand, would have molecules that are very far from each other and are freely moving in the whole container. As the container is heated, the size of the molecules would not change. It is only the volume that has changed. Also, the mass is the same since there is no outflow of the substances.

6 0

3 years ago