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ki77a [65]
2 years ago
7

An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11

cm from it, there is a converging lens of the same focal length. The distance between the lenses is 5.0 cm. Find the location and size of the final image.
Physics
1 answer:
Arada [10]2 years ago
8 0

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

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Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

v_i = 5.10 m/s

displacement of the package as it is dropped on ground

d = -105 m

acceleration is due to gravity

a = -9.8 m/s^2

now by kinematics

y = v* t + \frac{1}{2}at^2

-105 = 5.1 * t - \frac{1}{2}*9.8*t^2

4.9 t^2 - 5.1 t - 105 = 0

by solving above equation we have

t = 5.2 s

so it will take 5.2 s to reach the ground

7 0
3 years ago
In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to
Vlad1618 [11]

Answer:

I hope this helps and I'm not to late

A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.

Explanation:

you also can use paraphrase to help you reword bye bye!!

7 0
2 years ago
One gallon of paint (volume = 3.79 10-3 m3) covers an area of 17.8 m2. What is the thickness of the fresh paint on the wall?
AURORKA [14]

Answer:

0.2129 mm

Explanation:

We have given volume of the paint = 1 gallon=3.79\times 10^{-3}m^3

Area that covers the paint =17.8 m^2

We have to find the thickness of the fresh paint

So thickness=\frac{volume}{area}=\frac{3.79\times 10^{-3}}{17.8}=0.2129\times 10^{-3}m=0.2129mm

So the thickness of fresh paint on the wall is 0.2129 mm

6 0
3 years ago
Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus
ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N  (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

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8 0
1 year ago
You can increase the vapor pressure of a liquid by:
irga5000 [103]

Answer: Option (c) is the correct answer.

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

When we increase the temperature of a liquid substance then there will occur an increase kinetic energy of the molecules. As a result, they will move readily from one place to another.

Hence, liquid state of a substance will change into vapor state of the substance. This means that an increase in temperature will lead to an increase in vapor pressure of the substance.

Thus, we can conclude that you can increase the vapor pressure of a liquid by increasing temperature.

4 0
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