What distance is moved if we have a 8N force and the work done is 90J

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image (image distance can be negative or positive depending on the nature of the image).

According to the lens equation

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

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$

Hence the focal length of the convex lens is 20cm

7 0

3 years ago

Unlike other states of matter, what expand to fill their containers

padilas [110]

Answer:

gas

Explanation:

8 0

3 years ago

Read 2 more answers

A current-carrying wire 1.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10 A and there is

LenaWriter [7]

Answer:

0.2

Explanation:

F= BIL sin©

3= B×10×1.5 sin90

B=3/15

B= 0.2

8 0

2 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

Where:

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago

How is the setting in chapter one significant to a possible theme in chapter one of mice and men?

ra1l [238]

In Steinbecks of nice and men, the setting at the start of chapter one is “ a few miles south of Soledad” Lenny and George have been walking for what seems like four miles. George is aggravated having been dropped if by a truck driver who said they didn’t have far to go

4 0

3 years ago