Explanation:
A) The distance between the two successive compressions (or rarefactions) is actually called the wavelength of the longitudinal waves.
B) Wavelengths of longitudinal and transverse waves are comparable in the fact that in a transverse wave, the particles move perpendicular to the direction the wave travels whereas in a longitudinal wave the particles are displaced along the direction to the direction the wave travels
The kinetic energy of any moving object is
(1/2) (mass) (speed²) .
For the object you described, that's
(1/2) (100 kg) (12.5 m/s)²
= (50 kg) (156.25 m²/s²)
= 7,812.5 joules
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Your attachment is way out of focus, and impossible to read.
This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:
F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²]
Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Answer:
Explanation:
Given
Speed while running towards east is 
Distance traveled in east direction 
For Another interval you run with velocity


Total displacement

Time for first interval


Time for second interval

total time 

average velocity 

Therefore average velocity is less than