Problem:

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

2 years ago

What is the cancer treatments

ad-work [718]

One is chemo. Chemo is a special magnetic field like to treat cancer

3 0

2 years ago

Read 2 more answers

If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0 ∘C to 85.0 ∘C ?

andrey2020 [161]

Answer:

Explanation:

$C_{water}$ = 4190 J/kg.K

$C_{Al}$ = 910 J/Kg. K

$m_{Al}$ = 1.50 kg

$m_{water}$ = 1.80 kg

$Q_{added} = Q_{Al} + Q_{water}$

$=m_{Al} C_{Al}$ ΔT + $m_{water} C_{water}$ ΔT

= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)

= 578,955 J

= 579 kJ

3 0

3 years ago

in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th

Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

$R=\dfrac{u^2\sin2\theta}{g}$

Plugging all the values we get :

$R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m$

So, the maximum range of the ball is 73.25 m

8 0

2 years ago

While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

mote1985 [20]

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

(a) The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

(b) The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

(c) If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

5 0

3 years ago