On a camping trip on the Oregon coast, you decide to hike to the ocean, but you are not sure of the direction. The time is 4:00
2 answers:
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When object reached the terminal speed then its acceleration is zero
So as per Newton's II law we can say
now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed
so we can write
so here we are given that
so terminal speed will be nearly 2 m/s
Answer:
See explanation
Explanation:
Given:-
- The DC power supply, Vo = 600 V
- The resistor, R = 1845 MΩ
- The plate area, A = 58.3 cm^2
- Left plate , ground, V = 0
- The right plate, positive potential.
- The distance between the two plates, D = 0.3 m
- The mass of the charge, m = 0.4 g
- The charge, q = 3*10^-5 C
- The point C = ( 0.25 , 12 )
- The point A = ( 0.05 , 12 )
Find:-
What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
Solution:-
- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.
E = Vo / D
E = 600 / 0.3
E = 2,000 V / m
- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:
Fe = E*q
Fe = (2,000 V / m) * ( 3*10^-5 C)
Fe = 0.06 N
- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:
Fnet = m*a
Where, a : The acceleration of the object/particle.
- The only unbalanced force acting on the particle is (Fe):
Fe = m*a
a = Fe / m
a = 0.06 / 0.0004
a = 150 m/s^2
- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:
s = C - A
s = ( 0.25 , 12 ) - ( 0.05 , 12 )
s = 0.2 m
- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:
vf^2 = vi^2 + 2*a*s
vf^2 = 0 + 2*0.2*150
vf = √60
vf = 7.746 m/s
- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:
vf - vi = a*t
t = ( 7.746 - 0 ) / 150
t = 0.0516 s
- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:
- Where, The constant c : The capacitance of the capacitor.
- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:
- Again, apply the Newton's second law of motion and determine the acceleration (a):
Fe = m*a
a = Fe / m
- Where the acceleration is rate of change of velocity "dv/dt":
- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:
Where, eo = 8.854 * 10^-12 .... permittivity of free space.
- The differential equation turns out ot be:
- Separate the variables the integrate over the interval :
t : ( 0 , t )
v : ( 0 , vf )
Therefore,
- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:
- The final velocity of particle while charging the capacitor would be:
vf = 7.264 m/s ... slightly less for the fully charged capacitor