Ask question

Ask question

All categories

2 years ago

12

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

3.0 nc ?

1 answer:

Liula [17]2 years ago

4 0

Answer:

Electric field, $E=6.86\times 10^6\ N/C$

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, $q=-3\ nC=-3\times 10^{-9}\ C$

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

$ma=qE$

a = g

$E=\dfrac{mg}{q}$

$E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}$

$E=6860000\ N/C$

or

$E=6.86\times 10^6\ N/C$

Hence, the magnitude of electric field that balances its weight is $6.86\times 10^6\ N/C$ . Hence, this is the required solution.

You might be interested in

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

2 years ago

Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert

chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h

5 0

2 years ago

Read 2 more answers

What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and

polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

= 1.63 N ( 3 sig. fig.)

6 0

2 years ago

Marie is puzzled by her findings she has done several meticulous calculations and has gotten the numbers .37 rad, .89 rad and 1.

Irina-Kira [14]

Answer:

I think the answer is a

Explanation:

for it to be accurate has be to exactly 0.9 rad

it is not precise because the answer she is getting is different everytime and not even close. For instance,

It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...

6 0

2 years ago

A bus travels east for 3 km, then north for 4 km. What is its final displacement?

ikadub [295]

5 km northeast. Left and up would make northeast

8 0

2 years ago