Answer:
72.54 degree west of south
Explanation:
flow = 3.9 m/s north
speed = 11 m/s
to find out
point due west from the current position
solution
we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west
so it become like triangle with 3.3 point down and the hypotenuse is 11
so by triangle
hypotenuse ×cos(angle) = adjacent side
11 ×cos(angle) = 3.3
cos(angle) = 0.3
angle = 72.54 degree west of south
Answer:
electric potential, V = -q(a²- b²)/8π∈₀r³
Explanation:
Question (in proper order)
Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings
<em>consider the attached diagram below</em>
the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below
Va = q/4π∈₀ [1/(a² + b²)¹/²]

Also
the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Sum of the potential at point p is
V = Va + Vb
that is


![V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%28a%5E%7B2%7D%20%2B%20r%5E%7B2%7D%20%29%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%28b%5E%7B2%7D%20%2B%20r%5E%7B2%7D%20%29%5E%7B1%2F2%7D%20%7D%5D)
the expression below can be written as the equivalent

likewise,

hence,
![V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%7Br%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%7Br%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%5D)
1/r is common to both equation
hence, we have it out and joined to the 4π∈₀ denominator that is outside
![V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%5D)
by reciprocal rule
1/a² = a⁻²
![V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B-1%2F2%7D%20-%20%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B-1%2F2%7D%5D)
by binomial expansion of fractional powers
where 
if we expand the expression we have the equivalent as shown

also,

the above equation becomes
![V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%28%281-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%29%20-%20%281-%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%29%5D)
![V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B1-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20-%201%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D)
![V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D%5C%5C%5C%5CV%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D)


Answer

OR

Answer:
the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 
Explanation:
given information:
power, P = 25 W
wavelength. λ - 580 nm = 5.80 x
m
time, t = 1 s
to calculate the number of photon(N), we use the following equation
N = λPt/hc
where
λ = wavelength (m)
P = power (W)
t = time interval (s)
h = Planck's constant (6.23 x
Js)
c = light's velocity (3 x
)
So,
N = λPt/hc
= (5.80 x
)(25)(1)/(6.23 x
)(3 x
)
= 7 x 
The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,
1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.
It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.
Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.
A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.
So, in analyzing the four choices given, we look for low P and high T.
A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.
B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.
We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.