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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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5 0

2 years ago

A snowball with a mass of 85 g hits the top hat of a 1.5 m tall snowman and sticks to it. the hat and the snowball, with a combi

LenaWriter [7]

Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,

Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.

v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s

Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2

Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds

Then,
x = 8*0.553 = 4.424 m

7 0

3 years ago

An electrician finds that a 1 m length of a certain type of wire has a resistance of 0.24 Ω . What is the total resistance of th

zlopas [31]

The resistance of a given conductor depends on its electrical resistivity ( $\rho$ ), its length(L) and its cross-sectional area (A), as follows:

$R=\frac{\rho L}{A}$

In this case, we have $L'=138L$ , $\rho'=\rho$ and $A'=A$ . So, the total resistance of the wire with length of 138m is:

$R'=\frac{\rho' L'}{A'}\\R'=\frac{\rho 138L}{A}\\R'=138\frac{\rho L}{A}\\R'=138R\\R'=138(0.24\Omega)\\R'=33.12\Omega$

5 0

3 years ago

6. Show that the weight of an object on the moon is 1/6 its weight on earth.

mojhsa [17]

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

5 0

2 years ago

What is the passenger's apparent weight at t=1.0s?

Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight = m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight = m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0

3 years ago

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