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Law Incorporation [45]
3 years ago
12

You place an object 20 cm from a lens and find an image on the opposite side 30 cm from the lens. You calculate a focal length o

f 12 cm. You then move the object 30 cm from the same lens and once again measure the image distance. All your measurements and calculations are correct. What should you find when you calculate this focal length? a) It increased. b) It decreased. c) It stayed the same. d) There is not enough information to determine this focal length.
Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

The correct option is 'c': It stays same.

Explanation:

The focal length of any lens is obtained by using lens maker's formula as follows

\frac{1}{f}=(\frac{n_{2}}{n_{2}}-1)(\frac{1}{R_{1}}-\frac{1}{R_{2}})

where

n_{2} is the refractive index of the material of lens

n_{1} is the refractive index of the medium surrounding the lens

R_{1},R_{2} are the radii of the 2 surfaces of the lens.

As we see that in the formula the focal length of the lens only depends on the refractive indexes and the radius of the lens thus we conclude that the focal length of the lens is independent on the position of object or image.  

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A conducting rod (length = 2.0 m) spins at a constant rate of 2.0 revolutions per second about an axis that is perpendicular to
vladimir2022 [97]

Answer:

0.050V

Explanation:

To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.

\int dV = \int_{0}^{l/2} Bv (dl)

Where,

v= tangencial velocity

B = Magnetic Field

We know for definition that,

v= l\omega

Where,

L = length

\omega = Angular velocity

We can replace this values in our first equation then,

\int dV = \int_{0}^{l/2} B (l\omega) (dl)

Integrating we have,

V = \frac{1}{8} Bl^2 \omega

Replacing the values,

V= \frac{1}{8} (8*10^{-3})(12.56)(4)

V = 0.050V

Therefore the potential difference between the  center of the rod and the other rod is 0.050V

7 0
3 years ago
Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning.
iragen [17]

Answer:

y = 27 cm

Explanation:

given,

After 11 hours of burning height of candle is 27.9 cm

After 30 hours of burning height of candle is 26 cm

height after 20 hour = ?

the candle height decreases linearly

using linear equation

y = m x+ c

27.9 = m (11 ) + C.......(1)

26 = m (30) + C..............(2)

on solving equation (1) and (2)

1.9 = -19 m

m = -0.1

from equation 2

26 = -3 + C

C = 29

y = m x + 29

at 20 hour

y = -0.1× 20 + 29

y = 27

height of candle after 20 hours is 27 cm.

3 0
3 years ago
Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, tra
castortr0y [4]

Answer:

A. Its translational kinetic energy is larger than its rotational kinetic energy.

Explanation:

Given that

Radius = R

Mass = M

We know that mass moment of inertia for the solid sphere

I=\dfrac{2}{5}MR^2

Lets take angular speed =ω

Linear speed =V

Condition for pure rolling , V= ω R

Rotation energy ,RE

RE=\dfrac{1}{2}I\omega^2

RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2

RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2

RE=\dfrac{1}{5}\times MR^2\times \omega^2

RE=\dfrac{1}{5}\times MV^2

RE= 0.2  MV²

The transnational kinetic energy TE

TE=\dfrac{1}{2}MV^2

TE= 0.5 MV²

From above we can say that transnational energy is more than rotational energy.

Therefore the answer is A.

3 0
3 years ago
I appreciate any help I get on this, thank you in advanced :)
Reika [66]
The picture isn’t clear so I can’t read the dimensions of the box but I can try my best to guide u through the question.

For part a u need to find the volume of the box as that will equal the volume of sand that can be filled inside.
For this u multiply the height, width and length of the box.

For part b the mass of sand alone will be
=Mass of box + sand - Mass of empty box
=216 - 40
=176 grams

For part c the density of sand can be calculated by the formula
Density= Mass/Volume
So the mass (176g) / volume from part a

For part d u need to know that something will float if it has a lower density than what it is floating in. If the final density of sand that was found in part c is less than the density of gold (19.3 g/cm^3) it will float. Otherwise it will sink.

Hope this helped!
3 0
3 years ago
Date: 12-3-21<br> 4.<br> The momentum of a 5-kilogram object moving at<br> 6 meters per second is
ASHA 777 [7]

Answer

30 kg . m/sec

Explanation:

mark me as brainliest!

5 0
2 years ago
Read 2 more answers
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