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Crank
3 years ago
10

If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can

see
Physics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Explanation:

The length of the telescope is

         L = f_ocular + f_objetive

the magnification of the telescope is

         m = - f_objective / f_ocular

These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

3 0
3 years ago
The type of lens that spreads out parallel light is a
ohaa [14]

the answer is concave lens

7 0
3 years ago
Witch sentence states Newton third law
Dominik [7]

Answer:

For every action, there is an equal and opposite reaction.

Explanation:

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7 0
3 years ago
Balance the following equations: CUCO3+H2SO4- CUSO4+H2O+CO2​
Setler79 [48]

Answer:

It is already balanced

Explanation:

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6 0
3 years ago
A centrifuge is a device used to separate materials by their masses. A sample in a centrifuge is rotated at high speeds along a
german

Answer:

 F = 1.047 10⁻² N

Explanation:

Let's use kinematics to find the angular acceleration

             w = w₀ + α t

as for rest w₀ = 0

             w = α t

             α = w / t

let's reduce the magnitudes to the SI system

              w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s

              m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg

              r = 10.0 cm (1 m / 100 cm) = 0.100 m

let's calculate

              α = 104.72 / 1

              α = 104.72 rad / s²

angular and linear variables are related

               a = α  r

               a = 104.72 0.100

               a = 10.47 m / s²

finally we substitute in Newton's second law

               F = 1 10⁻³ 10.47

               F = 1.047 10⁻² N

8 0
3 years ago
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