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Natasha_Volkova [10]
3 years ago
10

What is the correct measure of an astronomical unit?

Physics
1 answer:
ipn [44]3 years ago
4 0
<span>In our system an astronomical unit is the average distance from the earth to the sun which is about 93 million miles. (92,957,000 miles) This is known as 1 AU.</span>
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We are usually not aware of the electric force acting between two everyday objects because
Semmy [17]
Most everyday objects have as many plus charges as they have minus charges
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3 years ago
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Convert the following as instructed:
Elis [28]
3.4m
6700g
172800s
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7 0
2 years ago
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of
Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0
3 years ago
The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati
Shalnov [3]

Answer is

9.773m/s^2

-----------------------------------------------------------------------------

Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

3 0
2 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
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