Answer:
The electric force will be 0 N
Explanation:
From the question we are told that
The magnitude of the charge is 
Generally from Coulombs law the electric force between two charges is mathematically represented as
Here r is the distance of separation between that two charges.
Now from the question we are told that the charge is far away from any other charge hence we can say that the distance between the charge and any other charge is 
So
=> 
Hence the electric force will be 0 N
Basically, the temperature is a result of the average kinetic energy of all the atoms comprising the solid/liquid/gas. In solid, these atoms can just vibrate in place, leaving them to only be able to conduct and radiate heat. However, as you probably know liquids and solids take the shape of their container because the bonds between atoms are loose enough to allow them to freely move around. Due to each individual atom having its own energy, and these atoms being free to move about the liquid/gas they collide with other atoms in the substance. These collisions result in a transfer of energy. Finally, lower energy atoms "sink" and higher energy atoms "rise" thus creating a "convection current".
Answer:
Drag
Explanation:
It is caused by air resistance and acts in the oooosite direction to the motion.
Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.
(a) Does an electric field exert a force on a stationary charged object? YES ( F = Eq)
(b) Does a magnetic field do so?- NO ( F= qvB)
(c) Does an electric field exert a force on a moving charged object? YES
(d) Does a magnetic field do so? YES ( F = qvB)
(e) Does an electric field exert a force on a straight current-carrying wire? ( NO)
(f) Does a magnetic field do so? Yes
(g) Does an electric field exert a force on a beam of moving electrons? Yes
(h) Does a magnetic field do so? YeS
To know more about magnetic field visit : brainly.com/question/10353944
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