Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
First establish the summation of the forces acting int the
ladder
Forces in the x direction Fx = 0 = force of friction (Ff) –
normal force in the wall(n2)
Forces in the y direction Fy =0 = normal force in floor (n1)
– (12*9.81) –( 60*9.81)
So n1 = 706.32 N
Since Ff = un1 = 0.28*706.32 = 197,77 N = n2
Torque balance along the bottom of the ladder = 0 = n2(4 m) –
(12*9.81*2.5 m) – (60*9.81 *x m)
X = 0.844 m
5/ 3 = h/ 0.844
H = 1.4 m can the 60 kg person climb berfore the ladder will
slip
Your answer is correct. No problem and Have a nice day
