Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:
where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:
where in this problem:
is the initial pressure of the gas
is the initial absolute temperature of the gas
is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
It begins with a kick-off
