Ask question

Ask question

All categories

3 years ago

8

what is the force between two small charged spheres that have charges 2x10(power -7)c and 3x10(power -7)c and are placed 30 cm a

part in air?

1 answer:

Andre45 [30]3 years ago

6 0

F=1/(4*pi*e0)*q1*q2/r^2
e0 ≈ 8,854187817·10^−12
F=1/(4*pi* 8,854*10^(-12))*2*10^(-7)*2*10^(-7)/(0,3)^2 H = <span> 0,003995 </span>H ~ 0,004 H

You might be interested in

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago

What situation would give you a mechanical advantage? *

const2013 [10]

D. Using a fixed-pulley system

3 0

3 years ago

A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0

3 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$

Replacing,

$a =\frac{42N}{83000kg}$

$a =5.06*10^{-4}m/s^2$

The total speed change

$\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity }$ we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

$a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}$

We have that,

$t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }$

$t = 1403.16s$

Therefore the Rocket should be fired around to 1403.16s

7 0

3 years ago

Uses of concave lens

natita [175]

Answer:

Concave Lens Uses. Telescope and Binoculars Spectacles Lasers Cameras FlashlightsPeepholes. ...

Used in telescope and binoculars. ...

Concave lens used in glasses. ...

Uses of concave lens in lasers. ...

Use of concave lens in cameras. ...

Used in flashlights. ...

Concave lens used in peepholes.

3 0

2 years ago