The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per
1 answer:
Answer:
0.2631 N/C
Explanation:
Given that:
The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m
The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m
The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons
Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)
= 0.0006
The number of electron flow per second is calculated as follows:
The magnitude of the electric field is:
E =
E =
E =
E = 0.2631 N/C
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Answer:
0.34
Explanation:
2.5 Mg = 2500 kg
The change in speed from 100 km/h to 40 km/h is
The deceleration caused by friction force is the change in speed per unit of time
Using Newton 2nd law we can calculate the friction force that caused this deceleration:
F = ma = 2500 * 3.33 = 8333 N
Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient
Explanation:
The total resistance used to calculate the total resistance of the two parallel resistor is given by :
Taking reciprocal of the above equation.
We can also write the above equation as :
On simplification of above equation,
or
Hence, proved.
This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
(-) ---------------->(+)
d
so let distance between the dipoles be d
∴ P = d
Let = 1 nC
so
P = d
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let = 2 nC
so
P = d
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let = 3 nC
so
P = d
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m