Answer:
PLEASE MARK AS BRAINLIEST!!
Explanation:
ANSWER IS IN THE IMG BELOW
Answer: The time required for the impluse passing through each other is approximately 0.18seconds
Explanation:
Given:
Length,L = 50m
M/L = 0.020kg/m
FA = 5.7×10^2N
FB = 2.5×10^2N
The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.
Ca(t) + CB(t) = 50
Where CA and CB are the velocities of the wire A and B
t = 50/ (CA + CB)
But C = Sqrt(FL/M)
Substituting gives:
t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))
t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))
t = 50 / (168.62 + 111.83)
t = 50/280.15
t = 0.18 seconds
Answer:
Part a)
Average EMF for half cycle is

Part b)
For one complete cycle we will have

Part c)
Maximum induced EMF will be at
t = 0.025 s and 0.075 s
minimum induced EMF is at
t = 0.05s and 0.1 s
Explanation:
As we know that magnetic field is oscillating in direction as well as magnitude
so induced EMF is given as

Part a)
For average value of EMF from positive maximum to negative maximum which is equal to half cycle
so we have




Part b)
For one complete cycle we will have


Part c)
Maximum induced EMF will be at

here we know

t = 0.025 s and 0.075 s
minimum induced EMF is at

so it is
t = 0.05s and 0.1 s
How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>
Answer: The Sound of Silence.