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hodyreva [135]
3 years ago
6

Any 3 differences between telescope and microscope

Physics
1 answer:
Keith_Richards [23]3 years ago
4 0
An instrument used to observe or imagine very small object using an optical mangifier
mirco cell.
Telescope is a magnifer of distance object
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Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it
MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface  R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .  

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r =  r-h / r

2h = r-h

3h = r

h = r / 3

5 0
3 years ago
Write any two features of capital​
amid [387]
Washington DC and new Mexico
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2 years ago
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If the same quantity of heat is added to both a 2-liter and a 4-liter container of water, the temperature change for water in th
VikaD [51]

Answer:

brainly.com/question/10966794

Explanation:

3 0
2 years ago
Un automóvil se mueve 80 m en 40 s con una velocidad constante. ¿Cuál es la velocidad del automóvil? Ayuda plissssssssssssssssss
Alexxandr [17]

Answer:

Velocidad = 2 m/s

Explanation:

Dados los siguientes datos;

Distancia = 80 m

Tiempo = 40 s

Para encontrar la velocidad del automóvil;

La velocidad se puede definir como la tasa de cambio en el desplazamiento (distancia) con el tiempo.

La velocidad es una cantidad vectorial y, como tal, tiene magnitud y dirección.

Matemáticamente, la velocidad viene dada por la ecuación;

Velocidad = \frac {distancia} {tiempo}

Sustituyendo en la fórmula, tenemos;

Velocidad = 80/40

Velocidad = 2 m/s

5 0
3 years ago
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe
never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

  • ∑Fᵧ = maᵧ  
  • T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

  • T₁ = m₂aᵧ + m₂g
  • T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

  • ∑Fₓ = maₓ
  • F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

  • (F_n · μ_k) - m₁g sinΘ = m₁aₓ
  • (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
  • [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
  • [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
  • [2.539595871] - [-58.0962595] = 6aₓ
  • 60.63585537 = 6aₓ
  • aₓ = 10.1059759 m/s²

Now let's go back to this equation:

  • T₁ = m₂(a + g)  

We have 3 known variables and we can solve for the tension force.

  • T = 2(10.1059759 + 9.8)
  • T = 2(19.9059759)
  • T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0
2 years ago
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