Answer:. Option c
Explanation: the speed of an electromagnetic wave is simply the vector product of the magnetic field and the electric field.
The direction of the velocity is the direction of the electromagnetic wave.
The wave is already moving towards the negative y axis (-j) and the magnetic field is already pointing towards the positive x axis (i)
From cross product of unit vectors
i × j = k
i × k = - j
With the second identity, we can see that the electric field will be pointing towards the positive of the x axis (k).
Option c is validated
5 n to the right I think Hope this helps
Answer:
Explanation:
a) using the energy conservation equation
mgh = 0.5mv^2 + 0.5Iω^2
I(moment of inertia) (basket ball) = (2/3)mr^2
mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)
gh = 1/2v^2 + 1/3v^2
gh = v^2( 5/6)
v = 
putting the values we get
solving for h( height)
h = 3.704 m apprx
b) velocity of solid cylinder
mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)
g*h = 1/2v^2 + 1/4v^2
g*h = 3/4v^2
putting the value of h and g we get
v= = 6.957 m/s apprx
The crest of the wave has traveled 415 cm in 11.7s. then the crest should have travelled 415/11.7 cm in one second i.e. 35.47cm. Then the crest should have travelled 26*35.47 cm in 26 seconds i.e. 922.2cm which has 46 vibrations in it.
so for length of each vibration (Wavelength) is 20 cm
Answer:
The capacitor having less distance of separation has a stronger electric field.
Explanation:
The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,
C1= Aε/d and C2=Aε/2d
The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,
Q1=VC1
Q1=VAε/d
Q2=VC2
Q2=VAε/2d
Therefore, the surface charge density σ1 and σ2 for the capacitors is,
σ1=Q1/A
σ1=VAε/(d*A)
σ1=Vε/d
Similarly,
σ2=Q2/A
σ2=Vε/2d
The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.
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