First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,

. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:

Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:

Now we can use the following relationship to find the distance covered by the skier before stopping, S:

where

is the final speed of the skier and

is the initial speed. Substituting numbers, we find:
I need a picture plz I don’t know what to answer.
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:
1340.2MW
Explanation:
Hi!
To solve this problem follow the steps below!
1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid
W=αhQ
α=specific weight for water =9.81KN/m^3
h=height=220m
Q=flow=690m^3/s
W=(690)(220)(9.81)=1489158Kw=1489.16MW
2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator
Wr=(0.9)(1489.16MW)=1340.2MW
the maximum possible electric power output is 1340.2MW