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 liquid is withdrawn at a constant flow rate q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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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
Explanation:
Given that,
Heat required, Q = 1200 J
Mass of the object, m = 20 kg
The increase in temperature,
We need to find the specific heat of the object. The heat required to raise the temperature is given by :
So, the specific heat of the object is .