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1 year ago

What is the focal length (in meters) of a lens whose radius of curvature is 9. 2 m and has a refractive index 1. 2?

1 answer:

4 0

The focal length (in meters) of a lens whose radius of curvature is 9. 2 m and has a refractive index 1.2 will be 18.4 m

The focal length of a lens is determined when the lens is focused at infinity. Lens focal length tells us the angle of view—how much of the scene will be captured—and the magnification—how large individual elements will be.

Focal length of a lens is the distance between center of lens and focal point . Focal point is a point on principal axis , at which light rays parallel to principal axis meet after refraction through lens or seem to meet after refraction .

The radius of curvature is the radius of sphere formed by the convex or concave mirror. It is also equal to the distance between the pole and center of curvature. The sign convention for focal length and radius of curvature is the same.

focal length = 2 * radius of curvature

given

radius of curvature = 9.2 m

focal length = 2 * 9.2

= 18.4 m

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The

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Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force $F_{net} =\sum F = F_1 +F_2 +..... = ma$

therefore we have from above equation $F_{net} = F -f = ma$

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

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2 years ago

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Answer:

50% of it .

Explanation:

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2 years ago

How can you increase the gravitational potential energy (GPE) of an empty shoe box on the bottom shelf of a bookcase

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Explanation: I did the test

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2 years ago

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Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This

vova2212 [387]

Answer:

$1.554\times 10^{32}\ \text{kg}$

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

Radius of orbit is given by

$R=\dfrac{vT}{2\pi}$

We have the relation

$\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}$

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6 0

2 years ago

If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?

Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0

2 years ago