Answer:
42.0×10² second²
Explanation:
Here, time is given in milisecond
(64800 ms)²
= 4199040000 ms²
The SI unit is seconds
1 second = 1000 milisecond



42.0×10² second²
1) 333.6 C
In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:

The electric field strength at the surface of a charged sphere is given by

where
is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
Here we have a cloud of radius

So we can re-arrange the previous equation in order to find the charge on the cloud:

2)
excess electrons
The total charge of the cloud must be (in magnitude)
Q = 333.3 C
We know that one electron carries a charge of

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

where
N is the number of excess electrons
Solving for N, we find:

Answer:

Explanation:
<u>Displacement</u>
It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.
The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

Finally, he drives 9.5 km 35° north of east.

The total displacement is


The direction can be calculated with


Pressure = Force/ Area = 3000/2 = 1500 pascal.
Answer:
i. rubbing the material against a magnet
ii. placing the material in a magnetic field of opposite polarity
iii. placing the material near a strong magnet
Explanation:
Ferromagnetic materials are majorly metals which can be easily attracted by a magnet. ferromagnetic materials are made up of domains, behaves as minute pieces of magnet. They can be rearranged to align when under the influence of an external magnetic field.
The alignment of the domains in a ferromagnetic material can be caused by either of the following: rubbing the material against a magnet, placing the material in a magnetic field of opposite polarity, placing the material near a strong magnet.