For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity.
Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g))
The equation reduces to 3.85 m = 6 * L so L = 0.642 m
chrsclrk · 7 years ago
Answer:
a. 0.143 mm b. 77.6 rad/m c. 483.18 rad/s d. +1
Explanation:
a. ym
Since the amplitude is 0.143 mm, ym = amplitude = 0.143 mm
b. k
We know k = wave number = 2π/λ where λ = wavelength.
Also, λ = v/f where v = speed of wave in string = √(T/μ) where T = tension in string = 19.3 N and μ = mass per unit length = 5.12 g/cm = 5.12 ÷ 1000 kg/(1 ÷ 100 m) = 0.512 kg/m and f = frequency = 76.9 Hz.
So, λ = v/f = √(T/μ)/f
substituting the values of the variables into the equation, we have
λ = √(T/μ)/f
= √(19.6 N/0.512 kg/m)/76.9 Hz
= √(38.28 Nkg/m)/76.9 Hz
= 6.187 m/s ÷ 76.9 Hz
= 0.081 m
= 81 mm
So, k = 2π/λ
= 2π/0.081 m
= 77.6 rad/m
c. ω
ω = angular frequency = 2πf where f = frequency of wave = 76.9 Hz
So, ω = 2πf
= 2π × 76.9 Hz
= 483.18 rad/s
d. The correct choice of sign in front of ω?
Since the wave is travelling in the negative x - direction, the sign in front of ω is positive. That is +1.
B. The solar energy is responsible for the currents being made in the atmosphese and the hydrosphere. It creates a convection current in the planet that maintains the flow of the air and the water in it. That's why you experience changes in temperature on the air during daytime and night, as well as varying currents on the seaside depending on the time of the day/night.
Answer:
Explanation:
Given that,
Radius of the circular loop, r = 10 cm = 0.1 m
Current flowing in the loop, I = 3.6 A
Uniform magnetic field, B = 12 T
To find,
The magnetic dipole moment of the loop.
Solution,
Let M is the magnitude of magnetic dipole moment of the loop. We know that the product of current flowing and the area of cross section. Its formula is given by :
A is the area of circular wire
Therefore, the magnetic dipole moment of the loop is 
. Hence, this is the required solution.
