I believe that it is the first one just a guess tho. So don't trust me, just in case
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)
Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm
Answer:

Explanation:
<u>Charge of an Electron</u>
Since Robert Millikan determined the charge of a single electron is

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 
Let's test the possible charges listed in the question:
. We have just found it's a possible charge of a particle
. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets
this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge
cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6
Finally, the charge
is four times the charge of the electron, so it is a possible value for the charge of an oil droplet
Summarizing, the following are the possible values for the charge of an oil droplet:

Answer:
The current in the primary is 0.026 A
Explanation:
Using the formula
I1 = (V1/V2)*I2
we have
I1 = (6.4/120)*0.500
I1 = 0.026 A