Answer:
0.035 N
Explanation:
Parameters given:
Charge q1 = -3.31x10^(-7) C
Charge q2 = -5.7x10^(-7) C.
Distance between them, R = 22 cm = 0.22 m
Electrostatic force between to particles is given as:
F = (k* q1 * q2) / R²
F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²
F = 0.035 N
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

- From this, the value of weight will be,

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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Answer:
1.62 atm
Explanation:
We can solve the problem by using the ideal gas equation:

where:
p = ? is the pressure of the gas in the tire
V = 8.5 L is the volume of the tire
n = 0.55 mol is the number of moles of the gas
R = 0.0821 atm L / K mol is the gas constant
T = 305 K is the temperature of the gas
By re-arranging the equation and substituting the numbers in, we find:

833.33 sec
5000m/6ms
Divide 5000 by 6 and you get your answer !!
Octopus
1. Octopus/Squid. These similar creatures are both cephalopods and have three hearts in total, one systematic to go along with two “gill hearts” that force blood to the gills.
Hope This Helps! Have a Nice Day!!