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Lynna [10]
2 years ago
15

The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.​

Physics
1 answer:
Natali [406]2 years ago
8 0

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

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Water flows over a section of Niagara Falls at the rate of 1.2×106 kg/s and falls 50.0 m. How much power is generated by the fal
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Answer:

5.88×10⁸ W

Explanation:

Power = energy / time

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P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

7 0
2 years ago
Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0
3 years ago
What two bodily functions are increased by a warm up
drek231 [11]

1) Blood flow: increases during warming improving muscle and joint elasticity. This decreases the possibility of having an injury.


2) Body temperature: This causes the cellular metabolism to increase. It also causes vasodilatation that allows a greater supply of oxygen and nutrients.

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Recycled plastics cannot be used to produce new food containers because ___
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they cannot be sterilized

4 0
3 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

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Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
2 years ago
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