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
Answer:
sweeps out equal areas in equal times.
Explanation:
As we know that there is no torque due to Sun on the planets revolving about the sun
so we will have

now we have

now we also know that

so rate of change in area is given as

so we will have


since angular momentum and mass is constant here so
all planets sweeps out equal areas in equal times.
Answer:19ohms
Explanation:
equivalent resistance=5+2+12
equivalent resistance=19ohms
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.