an electron is held at a distance from a stationary, positively charged object . If the electron is released , what energy change will take place?

Answers

Answer 1

Answer:

Answer:

Electrostatic potential energy will convert into kinetic energy

Explanation:

When electron is released from rest at some distance from positive charge then due to electrostatic attraction force these two charges will attract each other.

Now in the form of energy we can say that

electrostatic potential energy = $(kq_1q_2)/(r)$

now as the two charges comes closer to each other then the electrostatic potential energy will decrease and hence this energy will convert into kinetic energy of two charges

so we can say

$(kq_1q_2)/(r_2) - (kq_1q_2)/(r_1) = (1)/(2)mv^2$

Answer 2

Answer: Electric potential energy will be transformed into kinetic energy.

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a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

Answers

In this question we have given

velocity of missile=1350m/s

angle at which missile is moving=25degree

distance between missile and targets=23500m

angle between target and missile=55degree

time=10.2s

To find the final velocity of missile we will first find the acceleration required

Let x be the horizontal component of distance

x - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity

horizontally: x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

Therefore

V = √(V'x² + V'y²) = 3504 m/s

therefore magnitude of final velocity of missile=3504m/s

A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be operated from 200V mains giving the same power?

Answers

The resistance of the lamp is apparently 50V/2A = 25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.

A motor with a power of 7.0 × 104 watts lifts a loaded elevator to a height of 20.00 meters for 40 seconds. What is the force exerted by the motor on the elevator?

Answers

Answer:

$1.4\cdot 10^5 N$

Explanation:

First of all, we can find the work done as follows:

$W=Pt$

where P is the power and t is the time taken. In this problem, $P=7.0\cdot 10^4 W$ and $t=40 s$ , so the work done is

$W=Pt=(7.0\cdot 10^4 W)(40 s)=2.8\cdot 10^6 J$

The work done is equal to the force applied F times the displacement d:

$W=Fd$

since we know d=20.0 meters, we can re-arrange the equation to find the force exerted:

$F=(W)/(d)=(2.8\cdot 10^6 J)/(20.0 m)=1.4\cdot 10^5 N$

Answer:

C) 1.4 × 105 newtons

Explanation:

Correct answer on plato/edmentum

IN a physics lab, a student discovers that the magnitude of the magnetic field at a certain distance from a long wire is 4.0μT. If the wire carries a current of 5.0 A, what is the distance of the magnetic field from the wire?

Answers

Answer:

0.25 m

Explanation:

The intensity of the magnetic field around a current-carrying wire is given by:

$B=(\mu_0 I)/(2 \pi r)$

where

$\mu_0 = 4\pi \cdot 10^(-7)Tm/A$ is the permeabilty of free space

I is the current

r is the distance from the wire

In this problem, we know:

$B=4 \mu T=4 \cdot 10^(-6) T$ is the magnetic field

$I=5.0 A$ is the current in the wire

Re-arranging the equation, we can find the distance of the field from the wire:

$r=(\mu_0 I)/(2 \pi B)=((4\pi \cdot 10^(-7))(5.0 A))/(2\pi(4\cdot 10^(-6) T))=0.25 m$

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit the ground first? (Assume that there is no air resistance.) lead ball glass ball both at the same time either one, but not both at the same time

Answers

If he dropped them both at the same time, then as close as anyone could tell, they both hit the ground at the same time.

Describe the Milky Way. What does it contain?

Answers

It consists of our Solar system. Also, it contains many stars and alot of dust and gases. Our milky way is a galaxy full of amazing stuff, many stars blow up and cause auroras in the sky. If you go to watch the aurora borealis(northern lights) , its amazing and beautiful. The sky turns green and blue and the stars appear.

Answer:

The milky way consists of our solar system

Explanation:

edge 2022

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