At t = 0 a truck starts from rest at x = 0 and speeds up in the positive x-direction on..., the time the truck passes the car and the coordinate the truck passes the car is mathematically given as
<h3>What time does the truck pass the car and coordinate does the truck pass the car?</h3>
Generally, the equation for Distance traveled is mathematically given as
By Truck
By Car
In conclusion, when the conditions the truck passes the car
Therefore
Therefore
differentiation of T
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Answer:
The frequency will decrease by a factor of square root of 2 (<em>ω = √(2 (g / L))</em>.
Explanation:
A simple pendulum consists of a mass m hanging from a string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's second law for rotational systems, the equation of motion for the pendulum may be obtained
τ = I α ⇒ - mg sin(θ) L = mL² (d²θ/dt²)
where
- τ is torque
- I is the moment of inertia
- α is the angular frequency
- g is the acceleration due to gravity
- L is the length of the string
- m is the mass of the ball
The above expression can be rearranged as
(d²θ/dt²) + g / L (sin(θ)) = 0
If the amplitude of angular displacement is small enough that the small angle approximation () holds true, then the equation of motion reduces to the equation of simple harmonic motion
(d²θ/dt²) + g / L (θ) = 0
The simple harmonic solution is
θ(t) = θ₀ cos(ωt + Ф)
where
- ω is the frequency of the pendulum
- Ф is the phase angle
The frequency is expressed as
ω = √(g / L)
If the pendulum is pulled from equilibrium by 2 times theta, The simple harmonic solution will be
θ(t) = θ₀ cos(2 ωt + Ф)
and therefore,<em> the frequency will be</em>
<em>ω = √(2 (g / L))</em>
The important point here is that volumetric flow rate in the pump and the pipe is the same.
Q = AV, where Q = Volumetric flow rate, A = Cross sectional area, V = velocity
Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V
Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s
Therefore, the flow of water in the pipe is 5000 cm/s.