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3 years ago

9

Which of the following diagrams involves a virtual image ?

1 answer:

sergiy2304 [10]3 years ago

7 0

Answer:

The third diagram

Explanation:

<u>A virtual image</u> is an image that can not be formed on a screen.
<u>A convex lens</u> can form both virtual and real image depending on the position of the object from the lens.
A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.

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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

<h3>b. </h3>

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

For a particular pipe in a pipe-organ, it has been determined that the frequencies 296 Hz and 370 Hz are two adjacent natural fr

user100 [1]

Answer:

fundamental frequency of pipe will be equal to 74 Hz

Explanation:

We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz

Speed of the sound in air is 343 m/sec

We have to find the fundamental frequency for the pipe

Fundamental frequency will be equal to difference of the two adjacent frequency

So fundamental frequency = 370 - 296 = 74 Hz

So fundamental frequency of pipe will be equal to 74 Hz

8 0

3 years ago

A person's prescription for bifocals is -0.25 diopter for distant vision and +2.50 diopters for near vision, the near vision bei

Darina [25.2K]

Answer:

net power is + 2.25 D

Explanation:

Given data

distance vision = -0.25 D

near vision = + 2.50 D

to find out

net power

solution

we have given a person lens power for near is - 0.25 diopter and lens power for near power is +2.50 diopter so

net power is sum of both the power vision

so

net power = distance + near power

put both value we get net power

net power = ( -0.25 D) + ( + 2.50 D)

net power = + 2.25 D

so net power is + 2.25 D

6 0

3 years ago

A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t

grandymaker [24]

Answer:

13.23J

Explanation:

PE = m*g*h

PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)

3 0

3 years ago

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In a given chemical reaction the energy of the products is less than the energy of the reactants. Which statement is true for th

Westkost [7]

Answer:

B. Energy is released in the reaction.

7 0

3 years ago