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Lostsunrise [7]
3 years ago
7

Which of these statements about light microscopes and electron microscopes is NOT true? A.) You can use light microscopes to obs

erve both living and nonliving cells; you can use electron microscopes to observe only nonliving cells. B.) Light microscopes use visible light rays; electron microscopes use beams of electrons. C.) Preserving and staining samples for use with light microscopes is necessary;preserving and staining samples for use with electron microscopes is optional.
Physics
2 answers:
andrew-mc [135]3 years ago
6 0
I think the correct answer from the choices listed above is option C. Preserving and staining samples for use with light microscopes is necessary;preserving and staining samples for use with electron microscopes is optional is not true about <span>light microscopes and electron microscopes</span>
tangare [24]3 years ago
4 0

Correct answer choice is :


C) Preserving and staining samples for use with light microscopes is necessary;preserving and staining samples for use with electron microscopes is optional


Explanation:


A light microscope is an apparatus that uses evident light and magnifying lenses to observe small objects not visible to the naked eye, or in less detail than the naked eye allows. Magnification, however, is not the most significant point in microscopy. An electron microscope is a microscope that uses a beam of stimulated particles as a source of information. As the wavelength of an atom can be up to 100,000 times less than that of visible light photons, electron microscopes have a greater resolving energy than light microscopes and can expose the formation of miniature objects.


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Which of the following statements about Boyle's law are correct? The temperature and pressure must remain constant for the law t
kvasek [131]
Answer: The temperature and the number of molecules must reamain constant for the law to apply, and as the pressure increases, the volumen decreases proportionally.

Boyle's law states that if the temperature, T, of a given mass of gas, remains constant, the Volume, V, of the gas is in inverse relation to the pressure, p; i.e.

pV = constant (for a given mass of gas, at constant T)

Then, if p increases, V decreases proportionally to keep the relation pV = constant.
8 0
3 years ago
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If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its in
Flauer [41]
Increase in temperature of water = 0.53 °C
Explanation:
Change in mechanical energy = Potential energy
Potential energy = mgh
Mass, m = Mass of 1 L water = 1 kg
Acceleration due to gravity, g = 9.81 m/s²
Height, h = 225 m
Potential energy = 1 x 9.81 x 225 = 2207.25 J
Because of this 2207.25 J water gets heated.
Heat energy, E = mcΔT
Mass, m = Mass of 1 L water = 1 kg
Specific heat of water, c = 4200 J/kg/C
Energy, E = 2207.25 J
Change in temperature, ΔT = ?
Substituting
2207.25 = 1 x 4200 x ΔT
ΔT = 0.53 °C
Increase in temperature of water = 0.53 °C
3 0
3 years ago
A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, awa
igomit [66]

Answer:

At a distance of 1376.49 candle emits 0.2 watt power

Explanation:

Distance between Sun and earth 6.6\times 10^{16}m

Sun emits a power of P=3.8\times 10^{26}watt

Power emitted by candle = 0.20 watt

We know that brightness is given by

B=\frac{P}{4\pi d^2}

So \frac{3.8\times 10^{26}}{4\pi (6\times 10^{16})^2}=\frac{0.20}{4\pi d^2}

3.8\times 10^{26}d^2=7.2\times 10^{32}

d^2=1.89\times 10^6

d=1376.49m

So at a distance of 1376.49 candle emits 0.2 watt power

3 0
3 years ago
An aurora can be found in the _____.
Tomtit [17]
I believe the correct answer is d. thermosphere. 
7 0
3 years ago
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An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. Ther
astraxan [27]

Answer:

q = 1.815 \times 10^{-8} C

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Explanation:

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According to the Gauss's theorem in electrostatics

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Charge, q = surface charge density x area

q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}

q = 1.815 \times 10^{-8} C

5 0
3 years ago
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