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Elza [17]
4 years ago
9

What power objective lens has the greatest Field of view

Physics
1 answer:
Serhud [2]4 years ago
3 0
The eyepiece, also called the ocular lens, is a low power lens. The objective lenses of compound microscopes are parfocal. You do not need to refocus (except for fine adjustment) when switching to a higher power if the object is in focus on a lower power. The field of view is widest on the lowest power objective.
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Which of the following regions of the electromagnetic spectrum has the highest energy
garri49 [273]

Answer:

B. Radiowaves

Explanation:

Radiowaves are on the higher end of the spectrum, then infrared, visible light, and UV light (which is the lowest energy)

6 0
3 years ago
A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a bui
Nimfa-mama [501]

Answer:

t = o.6 s

Explanation:

Let ball thrown from below be A and ball dropped from above be B.

A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know

d + h = 15 m  ---------------(1)

Now apply s = ut + \frac{1}{2}at²

To A upwards,

d = 25t - \frac{1}{2}gt²  -----------------(2)

To B downwards,

h = 0 + \frac{1}{2}gt² ----------------(3)

(1) = (2) + (3) ⇒ 15 = 25t

                         t = 0.6 s

5 0
3 years ago
Tracy's foot is pushing on her cars gas pedal, is it pushing or pulling back?
UkoKoshka [18]

Answer:

pushing

Explanation:

7 0
2 years ago
Read 2 more answers
Careers in this career cluster also include related professional and technical support activities such as production planning an
qwelly [4]
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8 0
4 years ago
A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume
Andreyy89

Answer:

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

4 0
4 years ago
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