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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.

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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:
Plastic is cheap to manufacture because it is easy to mold. It is light but durable and unbreakable. Plastic does not rust and it is water resistant.
Plastic is reusable over and over again and when it is worn, the cost of replacement is cheaper than glass or rubber. It is odorless and chemical resistant.
The disadvantages of plastic are:
Plastic can cause cancer. Recycling plastic can cause toxic chemicals to be released during the process. It has low heat resistance and melts when placed near fires. Plastic containers for food absorb the smell of the food even when washed thoroughly. Small parts in plastic toys can cause choking in small children, the same with plastic bags when accidentally wrapped around children's heads. Plastics can clog up sewer systems when people through them away indiscriminately.

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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

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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 "how the ground an object has encased", and the Displacement is a vector thing that leads "to the extent to which an object is located", 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:

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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