1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
SCORPION-xisa [38]
2 years ago
9

Compare and Contrast refracting and reflecting microscopes?

Physics
1 answer:
mario62 [17]2 years ago
4 0
Eyepiece is magnifies the image so you can see it clearly. Compare and contrast how images form in a refracting telescope reflection telescope and microscope. ...Refracting telescope is uses two lenses to locate somethings. Reflecting telescope uses a large concave mirror to gather light.
~Marley
You might be interested in
A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i
nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J

4 0
3 years ago
An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the
bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N   +  N--->N   +   N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda

The frequency is calculated as follows;

F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0
3 years ago
What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers
IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0
3 years ago
Which gas giant has a rotation axis so tilted that the planet rotated like a bowling ball as it orbits the sun?
Anestetic [448]
The answer to your question is OPTION B
3 0
3 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
Other questions:
  • The general trend for igneous crystal growth is that crystals are ____ when the molten rock is __
    14·1 answer
  • Gravity is affected by (2 points)
    5·1 answer
  • An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce
    9·1 answer
  • Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint,
    14·1 answer
  • Consider a particle launched at a horizontal velocity v0 from a height h above the ground. 1. Derive an expression for the time
    14·1 answer
  • An airplane starts at rest and accelerates down the runway for 25 s. At the end of the runway, its final velocity is 75 m/s east
    7·1 answer
  • The wave function for a particle must be normalizable because:________ a. the particle's angular momentum must be conserved. b.
    13·1 answer
  • True or False – All transformations fit the Law of Conservation of Energy.
    10·1 answer
  • What are some human causes for global warming?
    10·1 answer
  • What is a topgraphic map?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!