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A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

6 0

3 years ago

Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerz

Maksim231197 [3]

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

Learn more here: brainly.com/question/24629099

3 0

3 years ago

What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT

sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0

3 years ago

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

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

3 years ago