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sergiy2304 [10]
2 years ago
7

Two marbles, one twice as massive as the other, are dropped from the same height. When they strike the ground, how does the kine

tic energy of the more massive marble compare to that of the other marble
Physics
1 answer:
Vsevolod [243]2 years ago
7 0

Answer: The more massive one will have larger kinetic energy.

Explanation:

We know that when we drop an object, the acceleration of the object will be equal to the gravitational acceleration.

a(t) = -9.8m/s^2

And to get the velocity, we need to integrate over time, to get:

v(t) = (-9.8m/s^)*t + v0

Where v0 is the initial speed of the object.

You can see that the mass of the object does not affect the velocity of it.

Then when we drop two marbles of different masses from the same height, we know that the final velocity of them will be equal.

Now, we also know that the kinetic energy can be written as:

K = (m/2)*v^2

where m is the mass, and v is the velocity.

Then the kinetic energy of the marble with less mass can be written as:

k = (m/2)*v^2

And the kinetic energy for the more massive one is:

K = (M/2)*v^2

And we know that both of them have the same velocity, and M is larger than m, then we can conclude that the marble with larger mass will have larger kinetic energy.

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Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

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7. When initially-unpolarized light passes through three polarizing filters, each oriented at 45 degree angles from the precedin
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Answer:

Explanation:

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After second filter , the intensity  will be  I₀ / 2 x cos²45 =  I₀ / 4

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1 / 8 the of initial light passes through the last filter .

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Answer:

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