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yanalaym [24]
3 years ago
10

Sometimes students are taught that "air expands as it is heated". And sometimes they are taught that "as air expands it cools".

Under what conditions could the first statement be true? Under what conditions could the second statement be true?
Physics
1 answer:
hram777 [196]3 years ago
7 0

Answer:

Both are true under specific circumstances. And are related to Boyle's law. volume and pressure in a gas are inversely proportional.

Explanation:

There is a tendency to entropy in our reality, that is, in particular true and visible with gases, they tend to occupy the whole space where they are confined, when we heat a volume of gas, then the movement of the particles and in consequence the pressure of the gas increases and to compensate this the volume tends to be increased too, according to Boyle's law. And the opposite happens when the volume is increased, then the pressure is  relieved and since the particles are further one from each other, then the temperature is lower, and therefore it cools down.

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If the car traveled 754 meters in a straight line between 0 and 36 seconds, what was the
ArbitrLikvidat [17]

Answer:

20.94 m/s

Explanation:

Recall that average velocity is defined as:

V = distance / time

Then, for our case:

V = 754 m / 36 sec = 20.94 m/s

4 0
3 years ago
How does an object's position and velocity change as the object accelerates?”
taurus [48]

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0
3 years ago
When analyzing a position time graph the direction of the slope positive or negative tells us:
Sphinxa [80]

Answer:

Explanation:

Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.

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