<em>There are some placeholders in the expression, but they can be safely assumed</em>
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
<u>Sinusoidal Waves
</u>
An oscillating wave can be expressed as a sinusoidal function as follows

Where



The voltage of the question is the sinusoid expression

(a) By comparing with the general formula we have


(b) The period is the reciprocal of the frequency:


Converting to milliseconds

(c) The amplitude is

(d) Phase angle:

The coal is urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity
All that business about the crane and the rope and the falling
is only there to confuse us.
The piano ended up 5 meters above the ground.
Potential energy = (mass) (gravity) (height)
= (200 kg) (9.81 m/s²) (5 m)
= (200 · 9.81 · 5) (kg-m²/s²)
= 9,810 joules .
Answer:
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Explanation:
Given data
initial circumference = 165 cm
rate = 12.0 cm/s
magnitude = 0.500 T
tome = 9 sec
to find out
emf induced and direction
solution
we know emf in loop is - d∅/dt ........1
here ∅ = ( BAcosθ)
so we say angle is zero degree and magnetic filed is uniform here so that
emf = - d ( BAcos0) /dt
emf = - B dA /dt ..............2
so area will be
dA/dt = d(πr²) / dt
dA/dt = 2πr dr/dt
we know 2πr = c,
r = c/2π = 165 / 2π
r = 26.27 cm
c is circumference so from equation 2
emf = - B 2πr dr/dt ................3
and
here we find rate of change of radius that is
dr/dt = 12/2π = 1.91
cm/s
so when 9.0s have passed that radius of coil = 26.27 - 191 (9)
radius = 9.08
cm
so now from equation 3 we find emf
emf = - (0.500 ) 2π(9.08
) 1.91 
emf = - 0.005445
and magnitude of emf = 0.005445 V
so
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.