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Umnica [9.8K]
2 years ago
7

What is the similarity between pushing and lifting? What is the difference?

Physics
1 answer:
MA_775_DIABLO [31]2 years ago
7 0

Answer:

the similarity is that hands are applied and force

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A care package is dropped from an airplane flying at 1000 m above sea level with a velocity of 200 m/s to an island. If it takes
ikadub [295]

Answer:3000

Explanation:

4 0
3 years ago
Is it possible for a car to be accelerating to the west while it is driving to the east?
Ber [7]

Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

8 0
3 years ago
I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to
Vladimir79 [104]

Answer:

293k

Explanation:

In this question, we are asked to calculate the temperature to which the reaction must be heated to double the equilibrium constant.

To find this value, we will need to use the Van’t Hoff equation.

Please check attachment for complete solution

7 0
3 years ago
Calculate the velocity of the bicyclist between 0 and 3 seconds.
hoa [83]

Answer:

is there an equasion it gives you?

Explanation:

need a little more info.

4 0
3 years ago
Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d
Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

8 0
3 years ago
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