Which of these would best describe the outcome of increasing the kinetic energy of the molecules in the above picture? A) The te
2 answers:
Answer: Option (A) is the correct answer.
Explanation:
Relation between kinetic energy and temperature is as follows.
As, kinetic energy is directly proportional to temperature. So, when there will be increase in temperature then there will also occur increase in kinetic energy of the particles of a substance.
And, when gas particles move slowly then it means kinetic energy of gas particles is very low. It also implies that temperature is low.
Thus, we can conclude that temperature will increase best describes the outcome of increasing the kinetic energy of the molecules in the above picture.
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Answer:
a) 6 times farther. b) 6 times longer.
Explanation:
Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:
vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s
In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:
vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s
Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:
If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:
On earth, this time could be calculated in the same way:
As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:
Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m
On earth, the distance travelled had been as follows:
Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m
⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00
b) As we have just found, the time of flight on the moon and on the earth are as follows:
tmoon = 11. 9 s
tearth = 1.98 s
⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = radians
So, 2.73 revolutions =
Therefore, the angular velocity of the tack is,
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for . This gives,
Therefore, the radial acceleration of the tack is 110.9 m/s².