What is the new current?

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

One advantage to using this form of circuit is that

777dan777 [17]

It has to be the last one because whenever lights are turned on it decreases because all lights are on at the same time. It's good to just have one light on. It doesn't use as much electricity.

8 0

3 years ago

Two +1 C charges are separated by 30000 m, what is the magnitude of the force?

Kipish [7]

Answer:

The magnitude of the force is 10 N

Explanation:

Coulomb's Law

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

$\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}$

F = 10 N

The magnitude of the force is 10 N

7 0

3 years ago

Ice floats on water but iron nail sinks in it why give reason?

levacccp [35]

Explanation:

If you like my answer than please mark me brainliest thanks

4 0

3 years ago

Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, t

makvit [3.9K]

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

6 0

3 years ago