Electromagnets in radios can be used to A) play a CD. B) create sound. C) change the station. D) make a remote control.

Which statement best describes the skier?

Olegator [25]

its c!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

8 0

3 years ago

Read 2 more answers

Guys please helpp!!!!1

Setler79 [48]

Answer:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy ( $E$ ) is the sum of gravitational potential energy ( $U$ ) and translational kinetic energy ( $K$ ), all measured in joules. In addition, gravitational potential energy is directly proportional to height ( $h$ ) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

3 0

2 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

A uniform magnetic field B=5.210 T is perpendicular to the plane of the paper (into the page). The current-carrying wire of valu

ehidna [41]

Answer:5.21 N

Explanation:

Given

B=5.210 T

I=2 A

L=0.5 m

Given Wire is perpendicular to Magnetic field

$\theta =90^{\circ}$

$F=IL\times B$

$F=BIL sin\theta$

$F=5.210\cdot 2\cdot 0.5 sin(90)$

F=5.210 N

as 1 Tesla =1 N/A/m

4 0

3 years ago

An office worker has his lunch on the 53rd floor of an office building. In which case is the gravitational potential energy of t

Ann [662]

Options

a. The worker carried the lunch up to the 53rd floor in the elevator.

b. The worker got a ride in a helicopter to the top of the 68-floor building and then carried the lunch down in the elevator to the 53rd floor?

c. The Worker carried it up the stairs to the 53rd floor.

d. Gravitational potential energy does not depend on the path taken.

Answer:

The gravitational Energy would remain the same in all cases.

Explanation:

Gravitational Energy is a conservative energy and its potential energy comes from gravitational actions.

An instance is when someone lifts a weight from the floor to a certain height and then dropped the weight back to the floor; the work done by the field will not be altered at all; it'll be the same throughout.

If otherwise, it'll be against thermodynamics 2nd law of physics because the object would have remaining energy after returning to the starting point,

For this reason, the gravitational potential energy only depends from the height, and is independent from the path taken to reach there.

4 0

3 years ago