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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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The average diameter of one tennis ball in a package of three is 6.8 cm. Which of the following is the combined volume of all th
Crank

We want to find the combined volume of 3 tennis balls. We will get that the combined volume is 493.7 cm^3

First, remember that for a sphere of diameter D, the volume is:

V = \frac{4}{3}*3.14*(\frac{D}{2})^3

Where 3.14 is pi.

Here we know that the average diameter of a tennis ball is 6.8cm, then we can replace that in the above equation to find the volume (in average) of a single tennis ball:

V = \frac{4}{3}*3.14*(\frac{6.8cm}{2})^3 = 164.5 cm^3

Now, in 3 balls of tennis, the combined volume will be 3 times the above one, this is:

3*V = 3*164.5cm^3 = 493.7 cm^3

If you want to learn more about volumes, you can read:

brainly.com/question/10171109

4 0
2 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

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\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
Can I PLEASE get some help?
Rus_ich [418]
1.) C
2.) B
3.) D
4.) B
Good luck with your work!
3 0
3 years ago
According to this equation, F=ma, how much force is needed to accelerate an 82-kg runner at 7.5m/s2?
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You multiply the mass by the acceleration 82*7.5=615; that's what I would do
5 0
3 years ago
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Elis [28]

Answer:

Generally, 40 to 50 degrees

Explanation:

About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.

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