All categories

1 year ago

When you changed from low to high power, how did the change affect the working distance of the lens?

1 answer:

4 0

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about working distance

brainly.com/question/13551539

#SPJ4

You might be interested in

Explain the relationship between resistance and the velocity of shortening.

butalik [34]

The velocity of shortening refers to the speed of the contraction from the muscle shortening while lifting a load. The relationship between the resistance and velocity of shortening is inverse. The greater the resistance, the shorter the velocity of shortening and the smaller the resistance, the larger the velocity of shortening.

Hopefully this help :)

7 0

3 years ago

Which type radiation can be observed well from Earth's surface?

umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0

3 years ago

Read 2 more answers

Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t

Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0

2 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

2 years ago

Discuss the differences between Seismic Reflection and Seismic Refraction

Maru [420]

Explanation:

Seismic refraction is the bending of the seismic waves as they pass geological layers of the earth due to different densities. This is especially true for Primary waves because they can pass through all the layers of the earth both liquid and solid.

Seismic reflectivity is the bouncing back of seismic waves at a boundary of geological layers due to different densities or subsurface formation. This is especially true for secondary waves that are unable to pass through liquid layers of the earth like the outer core and mantle. When they reach the boundary of these layers they bounce back towards the earth's surface.

These properties of P and S waves are used to ‘auscultate’ the epicenter of an earthquake by triangulation.

Learn More:

For more on Seismic Reflection and Seismic Refraction check out;

brainly.com/question/13502364

#LearnWithBrainly

6 0

2 years ago