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3 years ago

How would you describe the two projectors used to screen a three-dimensional movie?

Physics

2 answers:

Anit [1.1K]3 years ago

8 0

the answer to this question is b

True [87]3 years ago

4 0

the answer to this question is B

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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

3 years ago

A tiger started running when it saw a deer running at uniform velocity 2m/s at 15m far from it ..if the tiger ran at acceleratio

Masteriza [31]

You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
Time t is when the tiger catches the deer
t^2-2t-15=0
(t-5)(t+3)=0 factored
t=5s is the answer you use (t=-3 is a meaningless solution)

7 0

3 years ago

What is a mixture of rock and mineral fragments, organic matter, air, and water called?

Gekata [30.6K]

The answer should be soil

7 0

3 years ago

Your roommate is working on his bicycle and has the bike upside down. He spins the 60 cm

tigry1 [53]

Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.

The pebble in the tread goes by 3 times every second.

This is the same as 3 times per second.

Recall the unit of frequency is Hertz or per second, s⁻¹

So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz

For angular motion:

Angular speed, ω = 2πf

   = 2*π*3

   = 6π   rad/s

Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s

Linear acceleration, a = v² / r

   a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²

Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²

Angular speed = 6π rad/s ≈ 18.840 rad/s

The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s

The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²

The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²

5 0

3 years ago

A 3.39-kg block is sliding down a 7.92-kg frictionless incline. Both are initially moving to the right at 10.3 m/s. q7 (a) If th

Agata [3.3K]

Answer:

Final Speed of Incline = 5.549 m/s

Explanation:

To answer this question, we must remember the fact that any momentum gained by the block is equal to the momentum lost by the incline.

So lets start by find the initial and final momentum of the box in the horizontal direction:

Initial Momentum = mass * initial velocity

Initial Momentum = 3.39 * 10.3 = 34.917 kg*m/s

Final Momentum = mass * final velocity

Final Momentum = 3.39 * 21.4 = 72.546 kg*m/s

Change in momentum: 72.546 - 34.917 = 37.629 kg*m/s

This is equal to the momentum lost by the incline.

Initial momentum of incline = 7.92 * 10.3 = 81.576 kg*m/s

Final momentum of incline = 81.576 - 37.629 = 43.947 kg*m/s

Plugging this into the momentum equation to find the speed:

Final Speed * Mass = Final Momentum

Final Speed = 43.947 / 7.92

Final Speed = 5.549 m/s

3 0

3 years ago