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mars1129 [50]
3 years ago
12

What Kinetic energy is exactly equal to Gravitational Potential Energy why is height halfway between the maximum height?

Physics
1 answer:
prohojiy [21]3 years ago
5 0

Explanation:

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force

Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

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What's the formula to find out power
Julli [10]

In general, 

                 Power = (energy moved) / (time to move the energy) .

If it's mechanical power, then     

                 Power = (work done) / (time to do the work) .

If it's electrical power, then it can be any one of these:

                 Power  =  (volts)  x  (amperes)

                 Power  =  (volts)²  /  (resistance, ohms)

                 Power  =  (amperes)²  x  (resistance, ohms) .

Whatever kind of energy you're dealing with, power always
turns out to be

                  (amount of energy produced, used, or moved)
divided by
                  (time taken to produce, use, or move the energy) .          
3 0
3 years ago
A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?
kupik [55]

Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

3 0
3 years ago
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen
muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass m=0.20kg

Velocity v=4m/s

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 M_{a1}=mV

 M_{a1}=0.2*4

 M_{a1}=0.8

Final Momentum

 M_{a2}=-0.8kgm/s

Therefore

 \triangle M_a=-1.6kgm/s

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 M_{b1}=mV

 M_{b1}=0.2*4

 M_{b1}=0.8

Final Momentum

 M_{b2}=-0 kgm/s

Therefore

 |\triangle M_a|>|\triangle Mb|

Option A

4 0
3 years ago
A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration
babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10


hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0
3 years ago
A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
3 years ago
Read 2 more answers
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