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2 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Physics

1 answer:

Brut [27]2 years ago

8 0

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

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2 years ago

You need a 450 microgram sample of gold, but you only have a mass balance that measures in decigrams. Convert the amount of gold

sveta [45]

The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

<h3>How to convert micrograms to decigrams?</h3>

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1 micrograms = 1 × 10-⁵ decigrams

This means 450 micrograms is equivalent to 450 × 1 × 10-⁵ = 4.5 × 10-³ decigrams

Therefore, the amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

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7 0

1 year ago

Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).

JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

$M_1$ = Mass of Earth

$M_2$ = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

$F=\frac{GM_1m}{r^2}$

The force of gravity on an object on the other planet is

$F=\frac{GM_2m}{(2r)^2}$

As the forces are equal

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So, the other planet would have 4 times the mass of Earth

6 0

3 years ago

An AC generator has 80 rectangular loops on its armature. Each loop is 12 cm long and 8 cm wide. The armature rotates at 1200 rp

Sauron [17]

Answer: 28.96 V

Explanation:

Given

No of loops on the armature, N = 80

Length of the loop, l = 12 cm = 0.12 m

Width of the loop, b = 8 cm = 0.08 m

Speed of the armature, 1200 rpm

Magnetic field of the loop, B = 0.30 T

To solve this, we use the formula

V(max) = NBAω

Where,

A = area of loop

A = l*b = 0.12 * 0.08

A = 0.0096 m²

ω = 1200 rpm = 1200 * 2π/60 rad/s

ω = (1200 * 2 * 3.142) / 60

ω = 7540.8 / 60

ω = 125.68 rad/s

Substituting the values into the formula

V(max) = NBAω

V(max) = 80 * 0.30 * 0.0096 * 125.68

V(max) = 80 * 0.362

V(max) = 28.96 V

Therefore, the maximum output voltage of the generator would be 28.96 V

5 0

2 years ago