Which group is the most common sedimentary rock
1 answer:
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Answer:
7 m/s
Explanation:
To solve this problem you must use the conservation of energy.
That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.
The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:
K1=U2
Solve for v
Input known values and you get 7 m/s.
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