10. John does 176 J of work lifting himself a distance of 0.40 m. How

A QL = -26 μC charge is placed on the x-axis at x = - 0.2 m. A QR = 26 μC charge is placed at x = +0.2 m. (for all answers below

Kaylis [27]

You are given a QL = -26 μC charge that is placed on the x-axis at x = - 0.2 m and a QR = 26 μC charge that is placed at x = +0.2 m. The answers are:

The x-component of the electric field at x = 0 m and y = 0.2 m is 3.
The y-component of the electric field at x = 0 m and y = 0.2 m is 2.

4 0

3 years ago

A car with a mass of 1,324 kilograms, traveling at a speed of 20 meters/second, crashes into a wall and stops. What is the kinet

Valentin [98]

I think the correct answer from the choices listed above is option A. The kinetic energy after the perfectly inelastic collision would be zero Joules. A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. Hope this answers the question.

7 0

3 years ago

Read 2 more answers

An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute the values then we get

$-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}$

$\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}$

$v=-\frac{20}{3}=-6.7 cm$

Image distance of apple=-6.7 cm

Magnification=m= $\frac{v}{u}=\frac{-\frac{20}{3}}{-20}$

Magnification of apple= $\frac{1}{3}=0.33$

Hence, the magnification of apple=0.33

5 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

Here, M is the magnification of an astronomical telescope, $f_e$ is the focal length of the eyepiece lens and $f_o$ is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put $f_o=74 cm$ and $f_e=2.4 cm$ in the above expression.

$M=-\frac{74}{2.4}$

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0

3 years ago

A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p

Andrei [34K]

Answer:

13.4 x 10 raise to power -19 C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is E = 214 N/M

, The decrease in electrical potential energy is d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C

5 0

3 years ago