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slega [8]
1 year ago
9

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

Physics
1 answer:
Basile [38]1 year ago
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

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Summarize what you read on the three websites you visited by identifying the advantages and disadvantages of using plastics.
nexus9112 [7]
There are both advantages and disadvantages in using plastics:
The advantages are:
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3 0
3 years ago
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The photograph shows a kettle made of the element copper. Which statement describes a chemical property of copper?
Aleks04 [339]
A copper forms a patina when it interacts with air
7 0
2 years ago
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Draw a winding path that covers a distance of 70 miles and finishes with a displacement 20 miles soulthwest of the starting poin
EleoNora [17]

Distance is indeed a scalar amount that also refers to "<em><u>how the ground an object has encased</u></em>", and the Displacement is a vector thing that leads "<em><u>to the extent to which an object is located</u></em>", and the further calculation can be defined as follows:

Given:

distance= 70 miles

displacement = 20 miles

  • Displacement formula: \bold{Velocity = \frac{displacement}{time}}
  • Distance formula: \bold{Distance = speed \times time}

Please find the graph in the attached file.

Learn more:

brainly.com/question/9290794

7 0
2 years ago
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
A soccer player kicks a 0.44 kg ball with a force of 57.6 N, what is the ball’s acceleration
noname [10]

THE BALL'S ACCELERATION IS 130.90\frac{m}{s}

Explanation:

According to Newton's second law "the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)", meaning that when the soccer player kicks a ball, a force is acting on the ball, therefore increasing it's acceleration

SO FOR CALCULATING THIS WE WILL USE NEWTON LAW,

   FORCE = MASS × ACCELERATION

WE ARE GIVEN

FORCE = 57.6N

MASS= .44KG

SO HERE WE APPLYING FORMULA ,WE WILL GET

ACCELERATION = \frac{FORCE}{MASS}

ACCELERATION = \frac{57.6N}{.44KG}

ACCELERATION =130.90\frac{m}{s}

8 0
3 years ago
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