Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
How do fission nuclear reactions differ from fusion nuclear reactions?
A. Fission reactions involve the conversion of matter into energy, but fusion reactions do not.
B. Fusion reactions involve the conversion of matter into energy, but fission reactions do not.
C. Fission reactions are used to generate electricity for consumers, but fusion reactions are not.
D. Fusion reactions are used to generate electricity for consumers, but fission reactions are not.
Answer:
C
Explanation:
Both fission and fusion are nuclear reactions that produce energy, but their applications differs.
Fission is the splitting of a large (heavy, unstable) nucleus into smaller ones, and fusion is the process where nuclei of small atoms are combine together to form the nuclei of larger atoms releasing vast amounts of energy.
The correct answer is c. Fission reactions are used to generate electricity for consumers, but fusion reactions are not.
The physics of fusion is the process that makes the sun shine, and that makes the hydrogen bomb explode.
Answer:
F = 1958.4 N
Explanation:
By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side
so we have
so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m
So here the net pressure on the smaller area is given as
excess pressure exerted on the smaller area is given as
now the force required on the other side is given as
Answer:
Explanation:
It is given that,
Mass of the golf ball, m = 46 g = 0.046 kg
Terminal speed of the ball, v = 44 m/s
The drag force, 
Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.
Hence, this is the required solution.
