Answer:
0.5 m/s north
Explanation:
Take east to be +x, west to be -x, north to be +y, and south to be -y.
His displacement in the x direction is:
x = 20 m − 20 m = 0 m
His displacement in the y direction is:
y = 10 m
His total displacement is therefore 10 m north.
His velocity is equal to displacement divided by time.
v = 10 m north / 20 s
v = 0.5 m/s north
The correct response that will be used to describe this particular element would be the third option, since all of the other options are incorrect and apply to different elements in their groups. The element is a metal and will react with a non metal.
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
![E_2 = \frac{E_1q_1}{q_2}](https://tex.z-dn.net/?f=E_2%20%3D%20%5Cfrac%7BE_1q_1%7D%7Bq_2%7D)
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
![E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C](https://tex.z-dn.net/?f=E_2%20%3D%20%5Cfrac%7B4%2A10%5E6%2A13%7D%7B23%7D%20%3D%202.26%20%2A10%5E6%20%5C%20N%2FC)
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
The answer is 15 kilometers in 20 minutes.
Answer:
(a) The resistance R of the inductor is 2480.62 Ω
(b) The inductance L of the inductor is 1.67 H
Explanation:
Given;
emf of the battery, V = 16.0 V
current at 0.940 ms = 4.86 mA
after a long time, the current becomes 6.45 mA = maximum current
Part (a) The resistance R of the inductor
![R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7BV%7D%7BI_%7Bmax%7D%7D%20%3D%20%5Cfrac%7B16%7D%7B6.45%2A10%5E%7B-3%7D%7D%20%3D%202480.62%20%5C%20ohms)
Part (b) the inductance L of the inductor
![\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}](https://tex.z-dn.net/?f=%5Cfrac%7BRt%7D%7BL%7D%20%3D%20-ln%281-%5Cfrac%7BI%7DI_%7Bmax%7D%7D%29%5C%5C%5C%5CL%20%3D%20%5Cfrac%7BRt%7D%7B-ln%281-%5Cfrac%7BI%7DI_%7Bmax%7D%29%7D%7D)
where;
L is the inductance
R is the resistance of the inductor
t is time
![L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7BRt%7D%7B-ln%281-%5Cfrac%7BI%7DI_%7Bmax%7D%29%7D%7D%20%3D%20%5Cfrac%7B2480.62%2A0.94%2A10%5E%7B-3%7D%7D%7B-ln%281-%5Cfrac%7B4.86%7D%7B6.45%7D%29%7D%20%5C%5C%5C%5CL%20%3D%5Cfrac%7B2.3318%7D%7B1.4004%7D%20%3D%201.67%20%5C%20H)
Therefore, the inductance is 1.67 H