3 years ago

Which law states that absolute zero cannot be reached?

Physics

1 answer:

seraphim [82]3 years ago

4 0

C) The Third Law of Thermodynamics states that absolute zero cannot be reached.

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An automobile tire is rated to last for 35,000 miles. to an order of magnitude, through how many revolutions will it turn?

Furkat [3]

<span>The tire will rotate about 10 million times. An automobile tire is slightly less than 2 and half feet in diameter. It's circumference is that times pi with is a bit over 3. So 2.5 * 3 = 7.5 ft as an estimate for how far the tire rolls per revolution. A mile is a bit over 5000 feet, so call it 700 revolutions per mile. For the 35000 miles, call it 7 times 5000 miles. Now 7 times 7 is a bit under 50, so call 7 * 700 = 5000. And 5000 times 5000 = 25000000. The nearest order of magnitude is 10 million. So as an order of magnitude estimate, a automobile tire will rotate about 10 million times during it's life.</span>

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3 years ago

A car travels a distance of 540km in 6hours.What speed did he travel at?

Sloan [31]

Answer: 90 km/hr

Explanation:

Speed= distance divided by time

540/6

= 90km/hr

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3 years ago

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How would you differentiate the theoretical and actual velocities of PE and KE in falling objects?

Solnce55 [7]

Answer: you've been trolled

Explanation:

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2 years ago

Earth orbits once around the sun every 365.25 days at an average radius of 1.5X10^11 m. Earth's mass is 6X10^24 kg. What distanc

Katyanochek1 [597]

(a) The distance the Earth travels every year is the perimeter of the orbit; since we have the average radius of the orbit, we can calculate the perimeter:
$d=2\pi r = 2 \pi (1.5 \cdot 10^{11}m)=9.4 \cdot 10^{11}m$

(b) The average centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the average speed of the Earth, that we can find by dividing the distance covered by the Earth in one year (=the perimeter of the orbit) by the the time corresponding to 1 year:
t=1 year=365.25 days=3.2 \cdot 10^7 s
So the velocity is:
$v= \frac{d}{t}= \frac{9.4 \cdot 10^{11}m}{3.2 \cdot 10^7 s} =2.94 \cdot 10^4 m/s$
And so the centripetal acceleration is:
$a_c = \frac{v^2}{r}= \frac{(2.94 \cdot 10^4 m/s)^2}{9.4 \cdot 10^{11}m} =9.2 \cdot 10^{-4} m/s^2$

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3 years ago