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-BARSIC- [3]
2 years ago
14

With light microscopy, if the objective lens (lens closest to the specimen) magnifies 40-fold, and the eyepiece lens magnifies 1

0-fold, the final magnification will be:_________
Physics
1 answer:
Fed [463]2 years ago
6 0

The final magnification will be 400-fold or 400 times the original size of the object.

For magnifying smaller objects, a compound microscope is used.

A compound microscope consists of an objective and an eyepiece, whose diagram is shown in the adjoining image.

The lens  near  the object is called an objective and the other one is the eyepiece.

Let the magnification of the objective be m1

Let the magnification of the eyepiece be m2

The final magnification by the microscope, M, will be

M = m1 x m2

Putting the values in the above equation

M = 40 x 10

M= 400

Thus, the final magnification will be 400-fold or 400 times the original size of the object.

To know more about "optical instruments", refer to the link given below:

brainly.com/question/13276240?referrer=searchResults\

#SPJ4

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Find the volume of the irregular object a marble
Lostsunrise [7]

By using Displacement method we can find volume of irregular object like marble.

<h3>What is displacement method?</h3>

In displacement method,  

First , we measuring the volume of water displaced by an object which tell us the volume of the object.

Secondly We can use the physical balance to determine its mass.

Lastly , calculate the density by dividing the mass by the volume.

Procedure to find the Volume of irregular object i.e. marble

step 1 - Fill the graduated cylinder about half full and measure the initial volume of water.

step 2 - Drop the marble in the graduated cylinder.

step 3 -Now measure the final level of water.

Step 4 - Subtract both the values.

Here , comes the Volume of Irregular Object.

For more volume related question visit here:

brainly.com/question/1578538

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7 0
2 years ago
PLEASE HELP!!!! OFFERING 99 POINTS!!
Nitella [24]

The Gravitational PE (U) depends on three things: the object’s mass (m), its height (h), and gravitational acceleration (g), which is 9.81 m/s^2 on Earth’s surface.

so U = mgh = 9.81mh on earth

mass of the car = 50.0 grams = 0.05kg

height, h:

Hill 1 = 90.0 cm = 0.9m,

Hill 2 = 65.0 cm = 0.65m,

Hill 3 = 20.0 cm = 0.2m

substitute into eqn U = mgh

U @ top of Hill 1 = 0.05*9.81*0.9 = 0.4415J

U @ top of Hill 2 = 0.05*9.81*0.65 = 0.3188J

U @ top of Hill 3 = 0.05*9.81*0.2 = 0.0981J

difference in Gravitational Potential Energy from the top of Hill 1 to the top of Hill 3 = 0.4415 - 0.0981

= 0.3434J where J is the unit for energy, Joules


7 0
3 years ago
Read 2 more answers
Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
Help me please. Please see attached for the questions with the graph
KatRina [158]

<u>Answer</u>

5) b-c

6)    a-b and

       e-f

7) f-g

9) a-b = 0 m/s

    c-d = 0.6667 m/s

    e-f = 0 m/s

    f-g = -3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

<u>Explanation</u>

Answer

5) b-c

In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.

6) a-b and e-f

At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.

7) f-g

At this section the slope is negative meaning the cart is moving back to where it came from.

9) a-b = 0 m/s

At a-b the cart is not moving. So the velocity is zero.

<u>     c-d = 0.66667 m/s</u>

Velocity = distance / time

               =(50-40)/(40-25)

                = 10/15

                 = 0.6667  m/s

   <u> e-f = 0 m/s</u>

At e-f the cart is not moving. So the velocity is zero.

 <u>   f-g = -3 m/s</u>

Velocity = distance / time

               = (60-30)/(65-75)

                = 30/-10

                = - 3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

     

3 0
3 years ago
The period of physical changes to sex characteristics that marks the beginning of adolescence in males and females is
tatyana61 [14]

The period of physical changes to sex characteristics that marks the beginning of adolescence in males and females is Puberty.

6 0
3 years ago
Read 2 more answers
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