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Sergeeva-Olga [200]

4 years ago

5

If the force between two fixed Charged spheres is F and you triple the charge on one sphere and also triple the distance between

the spheres, what fraction/multiple of original force of attraction is now between the two spheres?

1 answer:

mezya [45]4 years ago

4 0

Since the charge has a direct proportionality, the increase will be a factor of three. But, since you triple the distance, which has an inversely squared proportion, it will be 3/9 F or 1/3 F.

*This is all based on the fact that this is true:
$F_s = \frac{Kq_1q_2}{r^2}$

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Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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A circuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(20 t) volts. Wr

Zinaida [17]

Answer:

q = 24 cos (20t) (1- e (-t / 1.2))

Explanation:

To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero

Vg + Vr + Vc = 0

Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage

Vg = 120 cos 20t = V

Vr = i R

Vc = q / C

We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two

i = dq / dt

Let's replace in the initial equation

V + R dq / dt + q / C = 0

Reorder the terms

Rdq / dt + q / C - V = 0

dq / dt + q / rC - V / R = 0

dq / dt = V / r -q / RC

dq / (V / R -q / RC) = dt

we integrate

∫I dq / (V / R - 1 / RC q) = ∫ dt

We change variables

u = (V / R - 1 / RC q)

du = -1 / RC dq

∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u

-RC ln u = -RC ln (V / R - 1 / RC q)

We evaluate between the limits of integration, the lower t = 0 q (0) = 0

-RC [ln (V / R - 1 / RC q) - ln (V / R)] = t

[ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC

Ln (V / R - 1 / RC q) / (V / R)] = -t / RC

Ln (1 - 1 / VC q)) = (-t / RC)

Ln (VC- q) = ln (VC) (-t / RC)

VC-q = VC e (-t / RC)

q = VC - Vc e (-t / RC)

q = VC (1- e (-t / RC)

We substitute the values they give us

q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))

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7 0

4 years ago

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

So

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

So

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

5 0

3 years ago