Answer:
Solution
λ=v/n
Here, v=344 m s−1
n=22 MHz =22×106 Hz
λ=344/22×106=15.64×10−6m=15.64μm.
Answer:
0.050 m
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.
In this problem, we have:
(current in the wire)
(strength of magnetic field)
Solving for r, we find the distance from the wire:
Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".
Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.
Answer:
Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell
Use VFR1 = VFR2 to discover the velocity at in the hose VFR =
A * V
D hose =10 * D nozzle, R hose = 5 * D nozzle
Area of a circle = πR^2
Area h=3.14*25*D^2 = 75.5D^2
(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2
Use VFR = VFR v2 = 0.4m/s
0.4*.785D^2 = 75.5*D^2* v1 D^2
= .314 =75.5*V1
v1 = 0.004m/s
Now we have the velocity, we can use Bernoulli's equation.
P1+ρgh1+ρV1^2 /2 = constant
There is no atmospheric pressure before so the P1= the gauge
pressure at the pump, let’s call the height of the hose 0m and the height of
the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at
the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3
Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2
Pg + .008= 100000+9800+80
Pg+.008= 109880
Pg=109880.008 PA
