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Elden [556K]
3 years ago
5

/How much kinetic energy does an 80 kg man have while running at 1.5 m/s

Physics
1 answer:
igomit [66]3 years ago
7 0
K.E = 0.5 * m * v^2 
  = 0.5 * 80 * (1.5)^2 
K.E = 90 joule 
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DIA [1.3K]
<h3>Answer: <u><em>Kinetic energy is the energy of motion. This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four.</em></u></h3><h3><u><em /></u></h3><h2>Explanation: <u><em>Hope this helps</em></u></h2><h2><u><em /></u></h2>
5 0
3 years ago
Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to complet
notsponge [240]

Answer:

The amount of energy required is 152.68\times 10^{3}Joules

Explanation:

The energy required to convert the ice to steam is the sum of:

1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.

2) Latent heat required to convert the ice into water.

3) Energy required to raise the temperature of water from 0 degrees to 100 degrees

4) Latent heat required to convert the water at 100 degrees to steam.

The amount of energy required in each process is as under

1) Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules

where

'S.heat_{ice}' is specific heat of ice =2.05J/^{o}C\cdot gm

2) Amount of heat required in phase 2 equals

Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules

3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals

Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules

where

'S.heat_{water}' is specific heat of water=4.186J/^{o}C\cdot gm

4) Amount of heat required in phase 4 equals

Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\\\\\\\\\Thus the total heat required equals Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules

5 0
3 years ago
Technician A says that the radiator should always be inspected for leaks and proper flow before installing a rebuilt engine. Tec
Slav-nsk [51]

Answer:

Both the technicians are right so option C is correct

Explanation:

Both the technicians are right so option C is correct

while installing a rebuilt engine, it is necessary to inspect the leaks and proper flow because if there is leak then cooling oil will flow out of the engine which results in engine heating. Secondly flow should be proper otherwise the oil will not reach to all parts and may cause heating of some parts of engine.

If the cooling fan is defective it won't be suck enough heat from engine and it will also cause heating.

8 0
3 years ago
A 5 kg
kotykmax [81]

Answer:

<em>The object-Earth system is open</em>

F_n=ma=5\ (7.2)=36\ N

Explanation:

<u>Accelerated Motion </u>

When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.

By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.

If the object starts from rest, its initial speed is zero, and

v_f=a\ t

where a is the acceleration and t is the time. The distance traveled is given by :

\displaystyle y=\frac{a\ t^2}{2}

From the two above equations, we find that:

v_f^2=2ay

Solving for a

\displaystyle a=\frac{v_f^2}{2y}

\displaystyle a=\frac{12^2}{2\ (10)}

a=7.2\ m/s^2

It means the net force is

F_n=ma=5\ (7.2)=36\ N

The object's weight is

W=5\ (9.8)=49 N

This means there is some external force acting upwards delaying the object's fall of a magnitude of

F_e=49-36=13\ N

3 0
3 years ago
What are 3 examples of elastic collisions?
Andre45 [30]
QUICK ANSWER

The collision between two gas molecules or billiard balls can be approximated as elastic collisions. Elastic collisions are exchanges of kinetic energy between two bodies having different reference frames in which the total kinetic energy of the two bodies after collision is equal to the energy before collision.'

3 0
3 years ago
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