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Debora [2.8K]
3 years ago
8

A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at

this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________
Physics
1 answer:
Liono4ka [1.6K]3 years ago
4 0

Answer:

(a)  the runner's kinetic energy at the given instant is 308 J

(b)  the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2}  \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J

the change in the kinetic energy is calculated as;

\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4

Thus, the kinetic energy increased by a factor of 4.

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R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
Calculate the force generated by a car that hits the wall at an
Makovka662 [10]

This is a defective question. It was WRITTEN by someone who is unclear on the concepts.  DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard.  Several times.

Your mass is the same every time.  Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one.  Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time.  But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.  

6 0
3 years ago
As a transverse wave travels through a rope from left to right, the parts of the rope _______.
kupik [55]
<h3><u>Answer;</u></h3>

Are moving up and down.

As a transverse wave travels through a rope from left to right, the parts of the rope <u>are moving up and down</u>.

<h3><u>Explanation;</u></h3>
  • Transverse waves occur when a disturbance causes oscillations perpendicular to the propagation, that is the direction of energy transfer.
  • <em><u>Particles of the medium move perpendicular to the direction the transverse wave itself is moving.  For example, if the wave is moving to the right, the particles of the medium are moving up and down.</u></em>
  • <em><u>Therefore, as a transverse wave travels through a rope from left to right, the parts of the rope are moving up and down.</u></em>
8 0
4 years ago
A bus accelerates forward. If an apple were on the floor of the bus it would move forward.
Paladinen [302]

Answer:

False

Explanation:

Since it is on the bus, it would not move forward because the outside acceleration cannot be considered.

4 0
3 years ago
The height of the mercury column in a barometer is 756 mm Hg on
notsponge [240]

Answer:

    p = 1.0076 10⁵ Pa

Explanation:

Atmospheric pressure is given by the relation

         P = rho g h

In this case they indicate that the height of the column of mercury is h = 756 mm Hg

let's reduce the height to the SI system

          h = 756 mm (1m / 1000 mm)

          h = 0.756 m

let's calculate

        P = 13600 9.8 0.756

        p = 1.0076 10⁵ Pa

7 0
3 years ago
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