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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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Which best describes the runners

qaws [65]

Answer:

The answer is

Explanation:

Daniela had a 5 meter head start, and Leonard caught up to her at 25 meters.

brainiest would be cool

3 0

3 years ago

What is the velocity of a quarter drooped from a towards after 10secounds

ArbitrLikvidat [17]

U=0
t=10 
a=9.8m/s/s 
v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) 

you work out the result now

6 0

3 years ago

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

2 years ago

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

timofeeve [1]

The percent difference between two numbers $x$ and $y$ is given by

$\dfrac{|y-x|}x \times 100\%$

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from $x$ to $y$ or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

$\dfrac{|6-3|}3 \times 100\% = 100\%$

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

$\dfrac{3-6}3\times100\%=-50\%$

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take $x$ from the left column and $y$ from the right column.

$\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%$

$\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%$

$\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%$

3 0

2 years ago

What is the difference between current and voltage besides their units of measurement

kondor19780726 [428]

"Voltage" is the "pressure" that makes electrons want to leave where they are
and head in some direction, if there's conducting material in that direction.

"Current" is the rate at which they all migrate in that direction.

8 0

3 years ago