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zhuklara [117]
3 years ago
7

if the coolant in a refrigerator was not compressed back into a liquid after it flowed through the refrigerator what would happe

n

Physics
2 answers:
Olin [163]3 years ago
4 0

The system would stop cooling, and simply recycle warm vapor through the refrigeration loop.

Explanation:

From the Mollier diagram we see that one part of the “refrigeration” loop is not complete. The step from 1 to 2 in the diagram would not occur.
This may occur due to a lack of sufficient refrigerant in the system or a faulty compressor. In any case, this results in warm “coolant” vapor passing into the receiver without any condensation (Step 3-4 cannot happen).
Without any liquid coolant available, the cooling effect of expansion of the liquid through the evaporator is lost (step from 5-6), and the system will simply use energy (and generate more heat) while recirculating warm vapor.

Pavlova-9 [17]3 years ago
3 0

Less heat would be removed from the refrigerator.

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Determine the amount of torque
malfutka [58]

Answer:

250Nm

Explanation:

Given parameters:

Length of the long pry bar  = 1m

Force acting on it  = 250N

Angle  = 90°

Unknown:

Amount of torque applied  = ?

Solution:

Torque is the turning force on a body that causes the rotation of the body.

The formula is given as:

 Torque  = Force x r Sin Ф  

r is the distance

 So;

   Torque  = 250 x 1 x sin 90  = 250Nm

7 0
3 years ago
Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ic
Setler [38]

Answer:

final temperature will be 0 degree C

Total amount of ice will be

m_{ice} = 182 g

total amount of water

m_{water} = 17 g

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

Q_1 = m_1s_1 \Delta T_1

Q_1 = 0.030(4186)(50 - 0)

Q_1 = 6279 J

heat absorbed by ice cubes to reach 0 degree

Q_2 = m_2 s_2 \Delta T_2

Q_2 = (0.169)(2100)(30)

Q_2 = 10647 J

so we will have

Q_2 > Q_1

so here we can say that few amount of water will freeze here to balance the heat

10647 - 6279 = mL

m = \frac{10647 - 6279}{335000}

m = 13 g

so final temperature will be 0 degree C

Total amount of ice will be

m_{ice} = 84.5 + 84.5 + 13

m_{ice} = 182 g

total amount of water

m_{water} = 30 - 13

m_{water} = 17 g

4 0
3 years ago
A published hypothesis:
kow [346]

Answer:

should be tested by the scientific community

4 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
What happens when you want to move the boat forward? You pull the oars toward yourself.Explain why you do this.
vampirchik [111]

Answer:

You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.

8 0
3 years ago
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