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vlabodo [156]
3 years ago
13

How does the use of a scanning electron microscope differ from that of a transmission electron microscope?. the options are....

. A: A scanning electron microscope can produce only a two-dimensional black and white image, while a transmission electron microscope can produce a three-dimensional color image.. . B: A scanning electron microscope can be used to study live specimens, while a transmission electron microscope would harm the specimens that are being viewed.. . C: A scanning electron microscope can be used to study the details of a specimen’s surface, while a transmission electron mi
Physics
2 answers:
Ratling [72]3 years ago
7 0
"A scanning electron microscope can be used to study the details of a specimen’s surface, while a transmission electron microscope can be used to study the internal structures of a cell" is the one that explains how <span>the use of a scanning electron microscope differ from that of a transmission electron microscope. The correct option among all the options that are given in the question is option "C". However you did not give the complete sentence present in option "C". </span>
elena-14-01-66 [18.8K]3 years ago
3 0

Well, we know A is wrong because scanning electron microscope can produce a three-dimensional image. And B is wrong because both microscopes cant be used to study live specimens because of how strong they are so D is also wrong.


So your answer would be (C) or "a scanning electron microscope can be used to study the details of a specimen’s surface, while a transmission electron microscope can be used to study the internal structures of a cell."

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A simple ideal Brayton cycle uses argon as the working fluid. At the beginning of the compression, P1 = 15 psia and T1 = 70°F, t
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B I thing hope this helps

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→15 points← Waiting for his new game to come in the mail, Billy races up the stairs in 3 seconds from the basement whenever he h
stich3 [128]

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500 Newton’s 3 meters high

Explanation:

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Taylor Swift weighing 794 N gets on an elevator. The elevator uses 313 W of power to lift the person 22.0 m. How much time did t
mezya [45]

Answer:

55.80s

Explanation:

Power is calculated using the expression

Power = Work done/Time

Workdone= Force ×distance

Workdone = 794×22

Work done = 17468Joules

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Time = Workdone/Power

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3 years ago
The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How
sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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