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3 years ago

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Two 2.0 cm * 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. a. What

are the charge on each electrode and the potential difference between them? The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. b. What are the charge on each electrode and the potential difference between them?

1 answer:

NARA [144]3 years ago

5 0

Answer: a) 31.86 *10^-12 C=31.86 pC; b) 18 V

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

C=Q/V where C is equal to C=εo*A/d where A and D are the area and the separation between the plates.

also we have

Q=C*V=ε(o*A/d)*V=(8.85*10^-12*0.02*0.02/1*10^-3)*9=31.86*10^-12 C=31.86pC

Then if the plates apart to a new spacing of 2.0 mm the new capacitance is equal

Cnew=εo*A/2*d so Cnew =Cinitial/2

then Cnew =Q/Vnew (Q is constant after disconnection to the battery)

Finally Vnew= Q/(Cinitial/2)= 2*(Q/Cinitial)= 2*Vinitial= 2*9=18V

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let's do it for this case:
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Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
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3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
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$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

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A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

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Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

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$P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J$

$P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)$

$K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\ v_{max} = 24.2 \ m/s$

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2 years ago

The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o

sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

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Potential energy of the spring, $P_f=20\ J$

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

$P_i=K_f+P_f$

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20$

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$50-20=\dfrac{1}{2}mv^2$

$30=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{60}{100}}$

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

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3 years ago