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3 years ago

11

Why can't we see the sun's corona under normal condition?

1 answer:

Wittaler [7]3 years ago

5 0

The naked eye cannot see the corona of the sun because the photosphere' is light is much to bright. When we look at the sun, we are virtually not even looking at it. We only see the photosphere.

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7Which of the following terms describes how glaciers move?

tatuchka [14]

Answer:

D is the answer I think (0 w 0 )

Explanation:

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3 years ago

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Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28

Ludmilka [50]

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{28.1\text{ years}}$

$k=2.47\times 10^{-2}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $2.47\times 10^{-2}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant = 1.00 g

a - x = amount left after decay process = 0.600 g

Now put all the given values in above equation, we get

$t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}$

$t=20.7\text{ years}$

Therefore, the time passed in years is 20.7 years.

8 0

4 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov

son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The eyepiece focal length is $f_e = 0.027 \ m$

Explanation:

From the question we are told that

The focal length is $f_o = 5.5 \ mm = -0.0055 \ m$

This negative sign shows the the microscope is diverging light

The angular magnification is $m = 300$

The distance between the objective and the eyepieces lenses is $Z = 19 \ cm = 0.19 \ m$

Generally the magnification is mathematically represented as

$m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]$

Where $f_e$ is the eyepiece focal length of the microscope

Now making $f_e$ the subject of the formula

$f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }$

substituting values

$f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }$

$f_e = 0.027 \ m$

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3 years ago

Can somebody google or pls if u know the correct answer help!!!

love history [14]

Answer:

Nutrients

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Stable body temperature

Atmospheric pressure

Explanation:

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