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butalik [34]
2 years ago
11

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. What is the potential differenc

e between the plates? Express your answer with the appropriate units. Part B If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? Express your answer with the appropriate units. Part C How much work is required to double the separation? Express your answer with the appropriate units (mJ)
Physics
1 answer:
horsena [70]2 years ago
5 0

Answer: a)235.9*10^-6 V; b) 471.8 *10^-6 V; c) 1.08*10^-13 J

Explanation: In order to explain this problem we have to use the expression for a capacitor od two parallel plates, which is given by:

C=Q/V where Q and V are the charge and voltage difference in teh capacitor, respectively.

Then we have;

V=Q/C=920*10^-12/(3.9*10^-6)= 235.9*10^-6 V

If we double the separation between the plates, the potencial difference will be:  ( Q is constant)

Vnew=Q/Cnew  where Cnew is given by;

Cnew=ε *A/(2*d) = 1/2*(ε *A/d)=1/2* Co ; then:

Vnew=2*Q/(Co)=2*Vo=2*4.24 *10^3 V=471.8 *10^-6 V

Finally, the work made oin the capacitor is the difference of the final and initial potential energy so:

Uo=Q^2/(2Co)=

Uf= Q^2/(2Cnew)               Cnew=Co/2

W=Uf-Uo= Q^2(1/Co)-Q^2(1/2*Co)=Q^2(1/2*Co)=1.08*10^-13 J

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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
2 years ago
A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

6 0
2 years ago
What was the vertical component of her acceleration during push-off? the positive direction is upward?
gladu [14]
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5 0
2 years ago
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