A storage tank contains a liquid at depth y where y=0 when the tank is half full. liquid is withdrawn at a constant flow rate q
1 answer:
Answer:
y(t) =
Explanation:
Given data :
storage tank contains a liquid at depth ; Y
when tank is half full ; y = 0
flow rate = q
sinusoidal rate = 3Q
Determine how the equation can be written for this system
solution is attached below
note ; from the question it can be seen that the surface area ( A ) is constant
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a) 52.5 m
b) 16.0 m/s
Explanation:
a)
The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:
where
v = 0 is the final velocity (the car comes to a stop)
u = 28.7 m/s is the initial velocity of the car
a is the acceleration
s is the stopping distance
For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:
where:
is the coefficient of friction
is the acceleration due to gravity
Substituting and solving for s, we find:
b)
In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is
Here we want the stopping distance of the car to remain the same as part a), so
We can use again the same suvat equation:
And since the final velocity is zero
u = 0
We can find the initial velocity of the car: