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svetoff [14.1K]
3 years ago
10

A converging lens focuses a set of light rays entering the lens on one side parallel to its axis to a single focal point on the

opposite side of the lens. if the principal rays emitted from a single point on an object all intersect at a single point after passing through a converging lens, then this intersection point helps in determining the location and size of a real image of the object. the most useful principal rays for converging lenses are the following:
Physics
1 answer:
Vladimir79 [104]3 years ago
7 0

1)The P ray is directed parallel to the axis of the lens. It emerges from the lens directed toward the focal point on the side of the lens opposite the object.


2) The M ray is directed toward the midpoint of the lens. It emerges from the lens unchanged in direction.


3) The F ray is directed toward the focal point on the same side of the lens as the object. It emerges from the lens directed parallel to the lens axis.


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Torque can be calculated by multiplying the force
lana66690 [7]
<span>Torque can be calculated by multiplying the force and the "perpendicular distance"

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3 years ago
To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 ✕ 10−10 m,
Elodia [21]
1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
\Delta V =- \int\limits^{d}_{\infty} {E} \, dr
Where E is the electric field generated by the proton, which is
E=k_e  \frac{q}{r^2} 
where k_e=8.99\cdot10^9~Nm^2C^{-2} is the Coulomb constant and q=1.6\cdot10^{-19}~C is the proton charge.
Replacing the electric field formula inside the integral, we obtain
\Delta V =- \int\limits^{d}_{\infty} {k_e  \frac{q}{r^2} } \, dr = k_e  \frac{q}{d}= 27~V

2) Then, we can calculate the work done by the electric field to move the electron (charge q_e=-1.6\cdot10^{-19}C) through this \Delta V. The work is given by
W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J

5 0
3 years ago
Read 2 more answers
The frequency is slowly increased. Once it passes the critical value fb, the student hears the ball bounce. There is now enough
vitfil [10]

Answer: g = acceleration = A*w^2 = A*(2*pi*fb)^2.

Explanation:

The ball bounces when the acceleration of the ball exceeds that of gravity. If A and fb are measured at that point, g = acceleration = A*w^2 = A*(2*pi*fb)^2.

3 0
3 years ago
A student is investigating thermal energy transfer. The student touches a piece of metal to a hot object, causing the metal to h
aliina [53]
It should be C. Conduction
3 0
3 years ago
A flowerpot falls from a window sill 36.5 m
hram777 [196]

When a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

From Newton's third equation of motion,

v^2 = u^2 + 2gh

where,

h is the height of the object or body from ground

u is the initial velocity of the body or object

v is the final velocity of the body or object

g is the acceleration due to gravity

Now, as we know that

Flowerpot is at rest. So, u = 0

g = 9.81m/s^2

h = 36.5m

By substituting all the values, we get

v^2 = 2 × 9.81 × 36.5

= 716.13

v = 26.7m/s

Thus, we concluded that when a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

learn more about Newton's equation of law of motion:

brainly.com/question/8898885

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7 0
1 year ago
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