The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
Answer:
Explanation:
Let the volume of the unknown bulb = X L
The volume of the system , after opening valve = (X + 0.72 L )
Use Boyles law gas equation,
P1V1 = P2V2 ( at temperature is constant )
Given:
P1 = 1.2 atm
P2 = 683 torr
Converting mmHg to atm,
1 atm = 760 mmHg(torr)
683 mmHg = 683/760
= 0.8987 atm
1.2X = 0.8987*(X + 0.720)
1.2X = 0.8987X + 0.6471
0.3013X = 0.6471
X = 2.15 L
Newton's 3rd law: The ball exerts a force on the bat that is equal and opposite to the force exerted by the bat on the ball.
Winds are named based on which compass direction the wind is blowing. For example some common ones are NE or N or SE or SW. NE stands for Northeast, N for North, SE for South East and SW for Southwest.
The kinetic energy gained by the air molecules is 0.0437 J
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Given:
Mass of a coffee filter, m = 1.5 g
Height from which it is dropped, h = 3 m
Speed at ground, v = 0.7 m/s
Initially, the coffee filter has potential energy. It is given by :

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m
P = 0.0441 J
Finally, it will have kinetic energy. It is given by :

×
× 10⁻³ × (0.7)²
E = 0.000343 J
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
E = 0.000343 - 0.0441
= 0.0437 J
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J
Learn more about kinetic energy here:
brainly.com/question/8101588
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