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ehidna [41]
3 years ago
7

Describe Kinetic Energy.​

Physics
1 answer:
mina [271]3 years ago
4 0
In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

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The speed of a car is 40m/s after 10 sec suddenly there was a crowd and the driver reduces its speed to 20m/s. What is the decel
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Explanation:

the number 0.4 in P4 from east

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The half life of Po-218 is three minutes. How much of a 2.0 gram sample remains after 15 minutes? Suppose you wanted to buy some
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After 15 minutes :

M = M₀ (1/2)^(t/T)

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Find an expression for the frequency of oscillations perpendicular to the rubber bands. assume the amplitude is sufficiently sma
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3 years ago
An object is released from rest near a planet’s surface. A graph of the acceleration as a function of time for the object is sho
Natasha_Volkova [10]

Answer:

the body has descended a height (4a)

Explanation:

This exercise should use the acceleration given in the graph, but unfortunately the graph is not loaded, but we can build it, using the law of universal gravitation and the fact that you indicate that the movement is near the surface of the planet

           F = m a

f

orce is gravitational force

           G M m / r² = m a

           a = G M / r²

where G is the universal constant of gravitation, M the mass planet and r the distance from the center of the planet of radius R to the body, if we measure the height of the body from the surface of the planet (y), we can write

           r = R + y

for which

           a = G M/R²     (1+ y/R)⁻²

if we use y«R we can expand the function in series

           a = (G M /R²)   (1 -2 y/R - 2 (-2-1) /2!   y² / R² +…)

as the height is small we can neglect the quadratic term and in many cases even the linear term, for this exercise we will remain only constant therefore the acceleration is constant

           a = G M / R²

from this moment we can use the relations of motion with constant acceleration for the exercise

a) they ask us for the position for t = 2s

            y = y₀ + v₀ t - a t²

as the body is released v₀ = 0

            y-y₀ = - a t²

            y-y₀ = - a 2²

             y-y₀ = - a 4

therefore  

therefore the body has descended a height (4a) where a is the acceleration of the planet's gravity

5 0
3 years ago
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