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
The answer is to the ground.
Gravity refers to the force that holds together the universe. On Earth, the gravity attempts to change the velocity of all the objects on the Earth's surface toward the ground at a rate of 9.8 meters per second squared according to Galileo.
1.commensalism
2. pred-prey
3. parasite-host
4.commensalism
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
Force = (mass) · (acceleration)
= (1,000 kg) · (9.8 m/s²)
= 9,800 newtons
Why are you still having a problem with F = M · a ?
