What is the gravitational field value (g) on planet Saturn with a mass of 5.69 X 1026

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

Plz answer will give Brainliest answer to whoever does.

Jobisdone [24]

Answer:

A foot kicking ball.

An iPhone being charged

Explanation:

A foot kicking a ball is example of transportation of Kinetic energy

$\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2$

An iPhone being charged means the electronic energy is converted into machinery energy

6 0

3 years ago

in an isolated system, two cars, each with a mass of 1,500 kg, collide. car 1 is initially at rest while car 2 is moving at 5 m/

kirza4 [7]

Answer:2.5 m/s

Explanation:

6 0

3 years ago

In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m

Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}$

F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a = 8.20 x 10⁻⁸

$a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}$

a = 9 x 10²² m/s²

c) speed of the electron

$a =\dfrac{v^2}{r}$

$9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}$

v² = 4.77 x 10¹²

v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}$

T = 1.53 x 10⁻¹⁶ s

8 0

3 years ago

Read 2 more answers

What is the average velocity of the object between 14 and 22 seconds?

Slav-nsk [51]

-1.1m/s

Explanation:

Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) - U.

4 0

3 years ago