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rosijanka [135]
3 years ago
7

When a object loses kinetic energy it gains , A - potential energy B- thermal energy C - total energy

Physics
2 answers:
Vadim26 [7]3 years ago
8 0

It would gain potential energy

artcher [175]3 years ago
4 0

The answer is A if you think about it something lose ptential energy it gain kinectic. So the answers A.

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___ + 3H2O + light —> C3H6O3 + 3O2. What amount and substance balance this reaction?
Alik [6]

Answer:3H2O + light-c3h603+302

Explanation:

5 0
3 years ago
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Suppose we could shrink the earth without changing its mass..?At what fraction of its current radius would the free-fall acceler
Drupady [299]

Answer:

at R/\sqrt{3}

Explanation:

The free-fall acceleration at the surface of Earth is given by

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

The formula can be rewritten as

R=\sqrt{\frac{GM}{g}} (1)

We want to shrink the Earth at a radius R' such that the acceleration of gravity becomes 3 times the present value, so

g' = 3g

Keeping the mass constant, M, and substituting into the equation, we have

3g=\frac{GM}{R'^2}

R'=\sqrt{\frac{GM}{3g}}=\frac{1}{\sqrt{3}}\sqrt{\frac{GM}{g}}=\frac{R}{\sqrt{3}}

5 0
3 years ago
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On the position vs time graph a curved line represents ____________.
Art [367]
The velocity is changing.
6 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
4 years ago
A runner is jogging in a straight line at a
murzikaleks [220]

Explanation:

The runner was 8.6km away from the finish line when the bird starts flying.

Therefore it takes the bird 8.6/14.4 = 0.60 hours for the bird to fly to the finish line.

In that 0.60 hours, the runner would have ran an extra 3.6km/h * 0.6h = 2.16km.

Now, the runner and the bird are flying towards each other. The distance between them is 8.6 - 2.16 = 6.44km and their combined speed is 18.0km.

Hence, they will meet in 6.44/18.0 = 0.36 hours.

Overall, the bird flew for 0.60 + 0.36 = 0.96 hours, and flew 14.4km/h * 0.96h = 13.8km.

4 0
3 years ago
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