How to calculate area with force and pressure

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba divers we

grin007 [14]

Answer:

Check the explanation

Explanation:

A) There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface).

To find out how they relate, draw a picture with the path the light follows in the plastic and the normal to both surfaces.

Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normal are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law, θa = asin[(nw/na)*sin(θw)]

B) D = l/tan(θw)

C) D = l/θw

D) d/D = na/nw

8 0

2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a

Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0

2 years ago

Read 2 more answers

A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the

Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

Concepts and Principles

The intensity at a distance r from a point source that emits waves of power P is given as:

I=P/4π*r^2 (1)

Given Data

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.

Required Data

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.

solution:

(a)

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:

I_1/I_2=(r_2/r_1)^2 (2)

Solve for I_2 :

I_2=I_1(r_2/r_1)^2

I_2=0.0625 I_1

(b)

Solve Equation (2) for r_2:

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

=3.16 m

3 0

2 years ago