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Zanzabum
3 years ago
6

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by

a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)
Physics
1 answer:
Marizza181 [45]3 years ago
6 0

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

f_o = Focal length of objective = 0.9 cm

f_e = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52

The angular magnification of the compound microscope is -252.52

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Answer:

B. Value you have recorded during your experiment.

Explanation:

In the observation section, those data is present which is collected during an experiment. Measurement of each key should be reported carefully. Data is usually present in the form of table, graph and figures.

This data can be represented in the tabular form because it is the best method to represent any data.

Therefore, in the observation section of lab report observer usually find the values which are recorded during an experiment.

4 0
4 years ago
What is the net Force needed to get a 16 kg box moving 4 m/s^2?
igomit [66]

Answer:

The net force should be of a magnitude of 64 N

Explanation:

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F_{net} = m\,*\,a

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3 years ago
An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect
klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is  B= 0.0048 T

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

                             \Delta V q = \frac{1}{2}mv^2

      Making v the subject

                             v = \sqrt{[\frac{2 \Delta V * q }{m}] }

 Where m is the mass of electron

              v is the velocity of electron

              q charge on electron

               \Delta V is the potential difference  

Substituting values

         v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }f

            = 4.5556 *10^ {7} m/s

For the electron to move in a circular path the magnetic force[F = B q v] must be equal to the centripetal force[\frac{mv^2}{r}] and this is mathematically represented as

                  Bqv = \frac{mv^2}{r}

making B the subject

                B = \frac{mv}{rq}

r is the radius with a value = 5.4cm = = \frac{5.4}{100} = 5.4*10^{-2} m

Substituting values

                B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}

                     = 0.0048 T

                 

7 0
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Net force of 200 newtons is applied to a wagon for 3 seconds
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1. Which is an example of chemical potential
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Answer:

D

Explanation:

3 0
3 years ago
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