All categories

1 year ago

I have no clue what to do please help.

Physics

1 answer:

Yakvenalex [24]1 year ago

3 0

Answer:

Step one : read the directions

Step two: complete the assignment

Explanation:

You might be interested in

Two particles with charges are initially very far apart (effectively an infinite distance apart). They are then fixed at positio

ddd [48]

Answer:

potential energy increases.

Explanation:

The potential energy between the two charged particles is given by

U = k Q q / r

If they are very far apart then r tends to infinity and the potential energy is zero.

If they come closer then the potential energy between the two charged particles increases.

Thus, the potential energy increases.

3 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

vovikov84 [41]

Gravitational force is given by, $F= G\frac{mM}{R^{2}}$

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, $F_{1}= G\frac{m_{1}M}{R^{2}}$

Gravitational force of the star on planet 2, $F_{2}= G\frac{3m_{1}M}{(3R)^{2}}$

Ratio, $\frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}$

$\frac{F_{1}}{F_{2}}= \frac{3}{1}$

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0

2 years ago

Read 2 more answers

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

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

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

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

2 years ago

Label and describe what is happening in this picture

SOVA2 [1]

Something is reproducing.

7 0

2 years ago

What happens during the process of deposition

frez [133]

Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .

8 0

2 years ago