9.
A 2.85-cm diameter coin is placed a distance of 31.4 cm from a diverging lens that has a focal length
of-11.6 cm. Determine the image distance and the diameter of the image.

Answers

Answer 1

Answer:

image distance = -8.47 cm

diameter of the image = 0.769 cm


Related Questions

1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
a) greater than b) smaller than c) equal to d ) double
2 ) when an object falls freely under the effect of gravity , the distance moved is
a ) directly proportional to time
b ) inversely proportional to time
c ) directly proportional to square of time
d ) inversely proportional to square of time.

Answers

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

What is one benefit to measuring your body’s flexibility?

A.
meeting the national requirement for flexibility
B.
determining your muscular strength
C.
tracking your flexibility improvements over time
D.
increasing the length of your life

Answers

Answer:

C

Explanation:

if you measure your body's flexibility then you can keep track of how flexible you have gotten over time

a car is travelling at 18m/s accelerates ti 30m/s in 3seconds. what's the acceleration of the car​

Answers

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have

Initial velocity, u = 18 m/sFinal velocity, v = 30 m/sTime taken, t = 3 seconds

We've been asked to calculate acceleration.

[tex]\qquad \implies\boxed{\red{\sf{ a = \dfrac{v-u}{t} }}}\\[/tex]

a denotes accelerationv denotes final velocityu denotes initial velocityt denotes time

[tex] \twoheadrightarrow \quad \sf {a = \dfrac{30-18}{3} \; ms^{-2} } \\ [/tex]

[tex] \twoheadrightarrow \quad \sf {a =\cancel{ \dfrac{12}{3} \; ms^{-2} }} \\ [/tex]

[tex]\twoheadrightarrow \quad \boxed{\red{\sf{ a = 4 \; ms^{-2} }}}\\[/tex]

Therefore, acceleration of the car is 4 m/.

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