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
Yes thank u teehee
.................... x
Answer:
Hi
Final temperature = 250.11 °C
Final volume = 0,1 m3.
Process work = 0
Explanation:
The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.
This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:
v=0.05 m3/kg
v1=0.057061 m3/kg
T1=242.56°C
v2=0.049779 m3/kg
T2=250.35°C
T=
The process work is zero because there is no change in volume during heating:
W=PxΔv=Px0=0
where
W=process work
P=pressure
Δv=change of volume, is zero because the piston was blocked so the volume remains constant.
The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.
At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.
They would both be measuring what they see correctly.
The photosynthetic wave interaction between visible light and a photosensitive part of a plant is very important t how plants use light to grow.
The frequency range and intensity levels of this light, I don't know.
Maybe the above ???
