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Fofino [41]
3 years ago
12

A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv

e and the eyepiece are separated by 19.6 cm. The final image is at infinity. a) What is the angular magnification? (b) How far from the objective should the object be placed.
Physics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

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B: leaving metal outside in the rain until rust forms an it's surface

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An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav
fiasKO [112]

Answer:

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

      p₀ = m v₀

Final after the crash before starting the movement

     p_{f} = (m + M) v

     p₀ = p_{f}

     m v₀ = (m + M) v

     v = v₀ m / (m + M)

     v = 2.0 0.015 / (0.015 +0.080)

     v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

    Em₀ = K = ½ (m + M) v²

Higher

    E_{mf} = U = (m + M) g y

   Em₀ = E_{mf}

   ½ (m + M) v² = (m + M) g y

   y = ½ v² / g

   y = ½ 0.316² / 9.8

   y = 0.00509 m

Let's look for the angle the height from the pivot point is

    L = 0.40 / 2 = 0.20 cm

The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º

4 0
3 years ago
the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0
3 years ago
What important event takes place during the interphase of the cell cycle?
yan [13]
The dna is multiplied obviously
5 0
3 years ago
Read 2 more answers
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s
muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = \frac{1}{1609} mile

50.0 miles/hour = \frac{50 * 1609.3}{60 * 60} m/s

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from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0
3 years ago
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