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
<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
True if you look up the question Is velocity speed in a certain direction you would’ve gotten the answer but I’m pretty sure it’s true
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]
