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Aleksandr-060686 [28]
3 years ago
13

CI. Design challenge: Your goal is to build a compound microscope than can at least double the size of the object. We will const

rain ourselves to values that will work later in a simulation, which has the unrealistic- for microscopes - focal lengths in meters. (Imagine instead that those values are in cm) You need to choose two converging lenses. You have available lenses with focal lengths of 2, 4, 8 and 12 m. First choose an objective lens focal length, and a distance to place the object from the lens. Then choose an eyepiece lens focal length and a distance to place the lens from the objective lens. C2. Draw your microscope design including a ray diagram. Your diagram should include: 1. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the objective. 2. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the eyepiece. C3. Calculate the magnification of your microscope

Physics
1 answer:
lilavasa [31]3 years ago
6 0

Answer:

2. 17.7cm

3. -7 that is magnification

Explanation:

See attached handwritten document

You might be interested in
In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc
Harman [31]

Answer:

A=\frac{A_0}{\sqrt 3}

Explanation:

Initial amplitude=A_0

We are given that

Amplitude after 2 s=A=\frac{1}{3}A_0

We have to find the amplitude after 1 s.

We know that amplitude at any time t

A=A_0e^{-\alpha t}

Using the formula

\frac{A_0}{3}=A_0e^{-2\alpha}

\frac{1}{3}=e^{-2\alpha}

3=e^{2\alpha}

ln 3=2\alpha

\alpha =\frac{ln 3}{2}=ln\sqrt 3

e^{\alpha}=\sqrt 3}

When t=1 s

A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}

8 0
4 years ago
A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what
kondor19780726 [428]
The kinetic energy of the tomato is : 

K.E =  1/2 mv^2

K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

Hope this helps
7 0
3 years ago
If you dropped a ball off a building, how far would it fall in 3 seconds?
tangare [24]
If you drop a ball off a building it increases its speed by 10 m/s so in 3 seconds it’s speed would be 30 m/s and it’s distance would be 45m
4 0
4 years ago
Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal spee
e-lub [12.9K]

Answer:

   r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

        Em₀ = U_g + U_e + K = -G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2}  m v^2)

the two in the kinetic energy is because they are two particles

final point. When it is detained

        Em_f = U_g + U_e = -G \frac{m_1m_2}{r_1} + k \frac{q_1q_2}{r_1}

the energy is conserved

        Em₀ = em_f

the charges and masses of the two particles are equal

         -G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}        

         

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm

4 0
3 years ago
1. Determine the energy released per kilogram of fuel used.
yarga [219]

200 MeV of energy  
E1/E2=7.61=8

U is equal to 1 kilogram or 1000 g.
There are 6.02310 23 atoms in one mole, or 235 g, of uranium. Therefore, 6.02310 23 atoms are present in 1000 g of 92/235 U.
It is understood that one atom releases 200 MeV of energy during its fission.

As a result, the energy released from the fission of one kilogram of 92/235 is given by E 2 = 6.02310 23 1000200/235 =5.10610 26 MeV E1/E2=7.61=8

In light of this, the energy released during the fusion of one kilogram of hydrogen is roughly eight times greater than the energy generated during the fission of one kilogram of uranium.

To learn more about Fission please visit -
brainly.com/question/27923750
#SPJ1

6 0
2 years ago
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