All categories

2 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]2 years ago

5 0

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).

You might be interested in

PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!

SVETLANKA909090 [29]

Answer:

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 $m/s^2$

Explanation:

1) Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2) Average speed = Total distance covered / Time taken for that distance to cover.

Total distance covered = 2+0.5+2.5 = 5 km

Time taken = 3 hours

Average speed = 5/3 = 1.66 km /hour

3) Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

Change in velocity = 14 - 6 = 8 m/s

Time = 4 seconds

So acceleration = 8 / 4 = 2 $m/s^2$

6 0

3 years ago

When was hawaii volcanoes national park established

Lerok [7]

1906 i Believe..........

6 0

3 years ago

An apple suddenly drops a distance of 3.2 m from a tree. If the acceleration due to gravity is 9.8 m/s2, how long does it take t

mart [117]

Answer:

0.33 seconds

Explanation:

7 0

3 years ago

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

P = 0.816 Watt

7 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago