A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of
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Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
As we know that standard form of wave equation given as
A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s
The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.
<h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.
They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.
The given data in the problem is;
q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C
F is the repulsive force = 36 mN =36 ×10⁶ N
d is the distance = 0.70 m
The Coulomb force is found as;
Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.
To learn more about Coulomb's law, refer to the link;
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