The answer to your question is true.
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
Answer:
Van der Waal's equation
Explanation:
The Van der Waal's equation is use to calculate the properties of a gas under nonideal or real gases conditions.
.
Here P, V ,T ,n and R have usual meaning as in the ideal gas equation
that is PV=nRT
with the difference of constant a and b. a and b are constants representing magnitude of intermolecular attraction and excluded volume respectively respectively.
Yes.
The acceleration vector WILL point south when the car is slowing down while traveling north
Answer:
Taking gravity to be 9.8m/s2, The velocity is 24.5m/s2.
Taking gravity to be 10m/s2, The velocity is 25m/s2.
Explanation:
According the first formula of motion under the influence of gravity for upward motion, v=u-gt, where v=final velocity, u=initial velocity, and t= time taken.
Here the time taken for the ball to reach the maximum point is half of 5, which is 2.5 seconds.
And v is 0, since at the maximum point gravity slows down the velocity to 0.
Finding the initial velocity,
v=u-gt
0=u-10(2.5)
u=10(2.5)
u=25m/s