The answer is to this question D
The elevation in reservoir at the rate of flow using is 03m/s is 114m.
The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.
The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.
calculation,
Let L = 100 m pipe
L1 = 150 m pipe
H f = friction losses
Using Reynolds number, relative roughness, friction co- effiicients and friction losses
Substitute the value in equation
Z = 110= 0.48= 3.54
Z = 114m
Therefore water surface elevation at reservoir is 114 meter.
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Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer:
75 N
Explanation:
In this problem, the position of the crate at time t is given by
The velocity of the crate vs time is given by the derivative of the position, so it is:
Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:
[m/s^2]
According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:
where
m = 5.00 kg is the mass of the crate
At t = 4.10 s, the acceleration of the crate is
And therefore, the force on the crate is:
Answer:
a) Vf = 27.13 m/s
b) It would have been the same
Explanation:
On the y-axis:
Solving for t:
t1 = 0.67s t2= -2.4s
Discarding the negative value and using the positive one to calculate the velocity:
So, the module of the velocity will be:
If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.
