Answer:
The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.
Explanation:
Taking moment along point A
Also normal stress is given as 100 psi
Now
The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.
Answer:
The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.
The wall is an insulator.
Explanation:
When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.
Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.
A school bus is a compound machine is because it has more than 2 simple machines operating together
Answer:
a)n= 3.125 x electrons.
b)J= 1.515 x A/m²
c) =1.114 x m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
=
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area