An object moves a distance of 10 meters in 5 seconds. What is the average speed of the object?
1 answer:
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Answer:
a) Check Explanation
b) Check Explanation
c) The rate at which water is being pumped into the tank = 2.631 cm³/min
Explanation:
Let the rate of flow of water into the tank be k cm³/min
a) The image of the conical tank is presented in the attached image
Note, the radius and height of a cone are related through the similar triangles principle.
As shown in the attached image, it is evident that
r/h = 3/10
r = 3h/10 = 0.3 h
b) The quantities given in the problem.
- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3
- total height of the tank, H = 10 cm
- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm
- rate at which water is leaking from the tank = 1.5 cm³/min
- water is being pumped into the tank at constant rate of k cm³/min
- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min
c) volume of the tank at any time = πr²h/3
Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)
dV/dt = k - 1.5
V = πr²h/3 and r = 0.3 h, r² = 0.09 h²
V = 0.03πh³
dV/dt = (dV/dh) × (dh/dt)
dV/dh = 0.09π h²
dV/dt = 0.09π h² (dh/dt)
dV/dt = k - 1.5
0.09π h² (dh/dt) = k - 1.5
But at h = 2 cm, (dh/dt) = 1.0 cm/min
0.09π h² (dh/dt) = k - 1.5
0.09π 2² (1) = k - 1.5
k - 1.5 = 1.131
k = 1.5 + 1.131 = 2.631 cm³/min
Sound level at distance of 15 m is given as 20 dB
so intensity at this distance is given as
now if we move closer to some some distance the sound level is now 50 dB
now the intensity is given as
now we know that
so now the distance from friend must be 47 cm