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2 years ago

What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?

1 answer:

6 0

The potential at a distance r from a charge Q is given by
$V(r) = k_e \frac{Q}{r}$
where ke is the Coulomb's constant.

The charge in our problem is $Q=2.9 nC=2.9 \cdot 10^{-9} C$ ; for the point at $r=0.48 mm=0.48 \cdot 10^{-3} m$ , the potential is
$V_1 = k_e \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}= 5.43 \cdot 10^4 V$

For the point at infinity, we immediately see that the potential is zero, because $r= \infty$ and so $V_2 = 0$ .

Therefore, the potential difference between the two points is
$\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V$

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azamat

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2 years ago

A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?

Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

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a = ( 15000-30000)/60

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2 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

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3 years ago

Can a vector be shorter than one of its components

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It can never be shorter than a component - magnitude of avector is the square root of the sum of the components squared, and a square function never produces a negative number. However, it can be the same size as its component, if that component is the only one

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3 years ago

Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$