Answer:
492.6 kN
Explanation:
Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus
P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m
P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².
Since P = F/A were F = force and A = area,
F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²
F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN
A reasonable estimate is 1 x 10^3 feet. By elimination, the first and second can't be since it will yield a negative answer and height can't be negative. Next, the fourth option will yield almost 1800 miles tall which is an impossible height. So the third is the reasonable estimate. Hope it helps.
Given:
Acceleration is uniform and acceleration (a) = 10 m/s^2
Now it has been mentioned for the first 2 secs the acceleration is 10m/s^2 .
Hence velocity= acceleration x time
Velocity= 10 x 2 = 20 m/s
Consider s as the distance traveled in the 3rd second.
Now we know s= ut+1/2(at^2)
Where s is the distance measured in m.
u is the initial velocity measured in m/sec
t is the time taken for the object to travel the above distance. This is equal to one second as we need to calculate the distance traveled between 3rd and 2nd second.
t = (3-2)= 1 sec
Substituting the given values in the above formula we get
s = 20 x1 + 1/2 (10 x 1 x 1)
s = 25 m
Thus the distance traveled by the object in the 3rd second is 25 m
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