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Gnom [1K]
3 years ago
13

When using the lens equation, a negative value as the solution for di indicates that the image is

Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

The Anatomy of a Lens

Refraction by Lenses

Image Formation Revisited

Converging Lenses - Ray Diagrams

Converging Lenses - Object-Image Relations

Diverging Lenses - Ray Diagrams

Diverging Lenses - Object-Image Relations

The Mathematics of Lenses

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)

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An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni
Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, v=(2\times 10^6i+3\times 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

Charge of electron, q_e=-1.6\times 10^{-19}\ C

(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v\times B)

F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]

F_e=-1.6\times 10^{-19}\times (-390000)(k)

F_e=6.24\times 10^{-14}k\ N

(b) The charge of electron, q_p=1.6\times 10^{-19}\ C

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
A man of mass 50kg ascends a flight of stairs 5m high in 5 seconds. If acceleration due to gravity is 10ms-2, the power expended
Setler79 [48]

The power expended is 500 W

Explanation:

First of all, we start by calculating the work done by the man in order to ascend: this is equal to the gravitational potential energy gained by the man, which is

W=mg\Delta h

where

m = 50 kg is the mass of the man

g=10 m/s^2 is the acceleration of gravity

\Delta h = 5 m is the change in height

Substituting,

W=(50)(10)(5)=2500 J

Now we can calculate the power expended, which is given by

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 5 s is the time elapsed

Substituting, we find

P=\frac{2500}{5}=500 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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Sunny_sXe [5.5K]
100 J

Explanation:
multiply the force by the distance
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