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olga2289 [7]
3 years ago
14

Place where warm air rises and pushes down on the earths surface with less pressure

Physics
1 answer:
BartSMP [9]3 years ago
3 0
 A high pressure area
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Can you please tell me what this is
Anna [14]

Answer:

200000 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE => is the kinetic energy.

m =>is the mass of the object

V => it the velocity of the object.

With the above formula, we can obtain the kinetic energy of the roller coaster as follow:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 1000 × 20²

KE = 500 × 400

KE = 200000 J

Therefore, the kinetic energy of the roller coaster is 200000 J.

4 0
3 years ago
Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rath
Sophie [7]

Answer:

Change/ Potential

Explanation:

Work is the amount of energy required to perform an action that is for a force to cause a displacement.

From work-energy theorem, work done by body is equal to change in its kinetic energy.

Work of gravity is basically the potential energy stored in the body due to gravity. From the law of conservation of mechanical energy, increased kinetic energy comes from the change of the potential energy of the stone.

8 0
3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe
iren [92.7K]
Second law is it:))))))))
4 0
3 years ago
Can someone answer this 2 correctly please
ahrayia [7]

Answer:

I know the first one is C.) 4J. I don't know of the answer for the second oneis suppose to be in N/m form? but I got

2,500N/m

5 0
3 years ago
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