Ask question

Ask question

All categories

3 years ago

12

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n

ewtons?

1 answer:

hichkok12 [17]3 years ago

5 0

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q = r*sq rt F[4pieo]

q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

891 excess electrons must be present on each sphere

You might be interested in

A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t

k0ka [10]

Answer:

20 ft

Explanation:

5 0

3 years ago

Read 2 more answers

a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work

ioda

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation for v, we find the final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s$

So, the final speed is 12.5 m/s.

3 0

3 years ago

A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?

WITCHER [35]

Answer:

6.25 ms²

Explanation:

..................

3 0

2 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

3 years ago

Find the temperature

NNADVOKAT [17]

Answer:

-------1

Explanation:

beacuse that is what i know

8 0

2 years ago