A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
2 answers:
Explanation:
The given data is as follows.
Initial velocity; u = 0, Final velocity; v = 19.6 m/s
time; t = 2 seconds
As the relation between initial velocity, final velocity and acceleration is as follows.
v = u + at
Hence, putting the given values into the above formula as follows.
v = u + at
19.6 m/s = 0 +
a = 9.8
Thus, we can conclude that acceleration of the dropped ball is 9.8 .
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Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m