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gulaghasi [49]
3 years ago
5

Who made the first game

Physics
1 answer:
likoan [24]3 years ago
8 0
I think the answer is D

Hope this helps
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Jason has 13720 J of gravitational potential energy standing at the top of a cliff over the lake. If he jumps off the cliff and
Anna [14]

The conservation of energy and Newton's second law allows us to find the results about Jason's falling motion are;

  • The energy when reaching the water is K = 13720 J
  • The average force of the water to stop it is: F = 2744 N

<h3>Energy conservation.</h3><h3> </h3>

The conservation of energy is one of the most important principles of physics, stable that if there is no friction force, mechanical energy is conserved at all points.

Mechanical energy is the sum of kinetic energy plus potential energy.

Let's look for the energy at two points

Starting point. Get higher.

         Em₀ = U = 13720 J

Final point. Lower down.

         Em_f = K

Friction in the air is negligible, so energy is conserved.

          Em_o= Em_f

          K = 13720J

<h3>Kinematics and Newton's law.</h3><h3> </h3>

They indicate that it stops 5m under the water, if we assume that the water acts with a constant force, we can use kinematics and Newton's second law to find this force.

The kinematics expression to find the acceleration is

            v² =v₀² – 2ay

When it stops the speed is zero.

            a = \frac{v_o^2}{2y}  

 

Newton's second law is:

           F = ma

           F = m ( \frac{v_o^2}{2y} )

The expression for the kinetic energy is:

          K = ½ m v₀²

          v_o^2 = \frac{2K}{m}  

Let's substitute.

           F = m (\frac{2K}{m}) \frac{1}{2y}  

           F= \frac{K}{y}  

Let's calculate.

           F= \frac{13720}{5}  

           F = 2744N

In conclusion using conservation of energy and Newton's second law we can find the results about Jason's falling motion are;

  • The energy when reaching the water is K = 13720 J
  • The average force of the water to stop it is: F = 2744 N

Learn more about energy here:  brainly.com/question/14274074

6 0
2 years ago
What subatomic particle has a nuetral charge?​
kirill [66]
An atom is a particle with a neutral charge
6 0
4 years ago
There are three identical train locomotives on separate tracks. Each applies equal force to begin moving. The first train consis
Luda [366]

Answer:

From lowest to highest acceleration:

3rd train

2nd train

1st train

Explanation:

The acceleration of an object can be found by using Newton's second law:

a=\frac{F}{m}

where

a is the acceleration

F is the net force on the object

m is the mass of the object

We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.

So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.

3 0
4 years ago
Given the Earth's density as 5.5 g/cm3 and the Moon's density as 3.34 g/cm3, determine the Roche limit for the Moon orbiting the
Assoli18 [71]

Answer:

The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth

Explanation:

The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.

The relation to determine Roche limit is:

Roche\ limit=2.423\times R_{P}\times\sqrt[3]{\frac{D_{P} }{D_{M} } }     ....(1)

Here R_{P} is radius of planet and D_{P}\ and\ D_{M} are density of planet and moon respectively.

According to the problem,

Density of Earth,D_{P} = 5.5 g/cm³

Density of Moon,D_{M} = 3.34 g/cm³

Consider R_{E} be the radius of the Earth.

Substitute the suitable values in the equation (1).

Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }

Roche\ limit= 2.86R_{P}

8 0
3 years ago
A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At
Gekata [30.6K]

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

\frac{x'}{x+x'}=\frac{3}{9}

3x'=x+x'

x=2x'

and it is given \frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s

Differentiating

\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}

4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}

\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s

6 0
3 years ago
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