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timurjin [86]
3 years ago
14

As the temperature of a sample of gas decreases, the kinetic energy of the particles _____.

Physics
1 answer:
topjm [15]3 years ago
5 0
Hello friend!

We know that kinetic energy is the energy possessed due to the motion of the object.
And we know if the object is in a fast motion then the temperature would be high, whereas if the object is slow in motion then it will have lower temperature. 
So we know that the kinetic energy is indirectly related to temperature.
From our knowledge we can conclude that HIGHER THE TEMPERATURE, HIGHER THE KINETIC ENERGY and LOWER THE TEMPERATURE, LOWER THE KINETIC ENERGY.

Hence, the answer to your question here is, DECREASES.

Hope it helps!!
All the best!!
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An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
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Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

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3 years ago
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marysya [2.9K]

Answer:

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3 years ago
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A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra
givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a

divide both sides by 40

a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2

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Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
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Let say for every 5 s of time interval the speed will remain constant

so it is given as

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so here we will have

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now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

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