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Vlad1618 [11]
3 years ago
14

When a refrigerant enters the compressor, it is a ____ and when it leaves the compressor, it is a ____. A. low pressure low temp

erature superheated vapor, high pressure high temperature superheated vapor B. low pressure low temperature superheated vapor, low pressure low temperature subcooled vapor C. high pressure superheated vapor, low pressure superheated vapor D. low pressure low temperature subcooled liquid, high pressure high temperature superheated vapor
Physics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

A

Explanation:

The compressor receive hot refrigerant and raises the pressure and temperature even further as it is send to the condenser.

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Non-native species that have rapidly increasing populations that spread
hjlf

Answer:

Invasive Species

Explanation:

Plz give brainliest

6 0
4 years ago
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil
Oksanka [162]

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

p_f=p_i

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}    (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

6 0
4 years ago
A car traveling in a straight line has a velocity of 3.29 m/s at some instant. After 4.99 s, its velocity is 8.29 m/s.
seraphim [82]

Use the definition of average acceleration:

<em>a</em> = ∆<em>v</em> / ∆<em>t</em>

We have

<em>a</em> = (8.29 m/s - 3.29 m/s) / (4.99 s)

<em>a</em> = 1.002 m/s²

4 0
3 years ago
Which of the following is an example of an even population distribution?
icang [17]
Even population distribution refers to a type of population distribution in which the arrangement is done in such a way that the distance between neighboring individual is maximized and uniform. This type of population distribution is usually found on the farm land, where the space in which the crops are planted have been carefully measured out. Thus an example of even population distribution is corn planted in a field.
8 0
4 years ago
1.5 x 10^3 standard notation
poizon [28]

Answer:

1500

Explanation:

3 0
3 years ago
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