Answer:
Explanation:
given wave = 8.5 x 10⁻³ cos ( 172 x - 2730t )
here wave no k = 172
angular frequency ω = 2730
velocity of wave on the string
a )
v = ω / k
= 2730 / 172
= 15.87 m/s
time taken to travel full length
= 1.2 / 15.87
= 75.6 x 10⁻³ s
b )
For velocity of wave on the wire the formula is
T is tension and m is mass per unit length of wire
m = .0121 / ( 9.8 x 1.2 )
= 1.03 x 10⁻³ kg / m
15.87 =
W = .259 N
c ) wavelength λ = 2π / k
= 2 x 3.14 / 172
= .0182 m
no of wave length
n = 1.2 / .0182
= 66 approx .
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
Newton's second law of motion can be used to calculate weight.
Answer:
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E₀ and B₀ are the amplitudes of the electric and magnetic fields.
Explanation:
Generally, the variables that specify the nature of the wave are:
(1) The amplitude of the electric field, designated as E₀, and
(2) The amplitude of the electric field, designated as B₀.