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3 years ago

An advantage of light microscopes compared to electron microscopes is that light microscopes _____.

2 answers:

8 0

C is the correct answer because electron microscopes can’t be used to view living specimens because the methods used to prepare them for viewing kills cells.

ivolga24 [154]3 years ago

5 0

<span>C is the correct answer. Electron microscopes require a vacuum to work, so living cells cannot be seen because they cannot respire. Light microscopes use a ray of visible light instead of a beam of electrons to magnify something so it can be seen by the naked eye. There are two different types of electron microscope: transmission (TEM) and scanning (SEM).</span>

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An object with a mass of 90 grams, moving at a constant velocity of 6 meters per second, has __________.

son4ous [18]

<span>A. a momentum of 540 grams per meters per second The momentum equation is p = mv. P is the variable for momentum, m is the variable for mass and v is the variable for velocity. To find the momentum, you must multiple the mass of the object by the velocity it is traveling by. The unit for momentum is the mass per meter per second.</span>

5 0

3 years ago

Read 2 more answers

I will mark as the brainliest answer<br><br>plz 8,9,10

blagie [28]

Answer:

8. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 unit.

10. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units.

Explanation:

8. i) acceleration = velocity / time

ii) In this figure velocity = time

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. i) acceleration = velocity / time

ii) In this figure 4 = m + 5, therefore m = -1

therefore velocity = (-0.5 $\times$ time) + 5

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 units.

10.) velocity is constant at 2

therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units

5 0

3 years ago

How can an electron in an atom lose energy to go from a higher energy level to a lower energy level? select one:

Romashka-Z-Leto [24]

I belive that the answer is b.

8 0

3 years ago

PLEASE HELP AS SOON AS POSSIBLE ILL MARK BRALIEST TO WHOEVER IS CORRECT

Reptile [31]

Answer:

There are two significant figures in 2.200 x 10^7

3 0

3 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago