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3 years ago

9

At what distance from the centre of the earth its acceleration due to gravity becomes one third only? [Mass and radius of earth

= 6x10 24 kg & 6.4x10 6 m resp.]

2 answers:

Anarel [89]3 years ago

5 0

235 is the answer sir

Kipish [7]3 years ago

3 0

Answer:

235

Explanation:

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Which statement places events during the Space Race in the correct order from the earliest to the most recent?

leva [86]

Answer:

3,1,4,2

Explanation:

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3 years ago

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9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J

givi [52]

Answer:

The amount of energy that would be released is equal to 4182 Joules.

Explanation:

Total amount of coke = 2 kg = 2000 g

1 calorie per gram is equal to 4.184 Joules of energy

4.184 J/gC*2000g = 8368 J

1 food calorie is roughly equal to 4186 J

8368 - 4186

Therefore, the amount of energy that would be released is equal to 4182 Joules.

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3 years ago

An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

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3 years ago

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Which event happens in a solar cell when the photoelectric effect occurs?

KiRa [710]

Answer:

current forms when electromagnetic waves strike a semiconductor, removing some of its electrons.

Explanation:

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2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago