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andrew-mc [135]
3 years ago
6

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

tion of 400x with an objective lens that has a focal length of 0.60 cm. The distance between the eyepiece and objective lenses is 16 cm. Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus).
Physics
1 answer:
grin007 [14]3 years ago
3 0

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

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You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rollin
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2 years ago
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8 0
3 years ago
In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio
Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

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6 0
2 years ago
What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?
lina2011 [118]
We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
\sum F = ma

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

mg \sin \theta - \mu_K N = ma (parallel direction) (1)
mg \cos \theta - N =0 (perpendicular direction) (2)
where
\theta=35^{\circ} is the angle of the inclined plane, N is the normal reaction of the plane, \mu_K N is the frictional force, with \mu_K=0.4 being the coefficient of friction.

From eq.(2), we find
N=mg \cos \theta
and if we substitute into eq.(1), we can find the acceleration of the block:
mg \sin \theta - \mu_k mg \cos \theta = ma
from which
a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2
7 0
3 years ago
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