Answer:
since each loop is ewuvivalent to one half wave lenght . the length of the string is equal to two halves of a wavelength . put in the form of an equation in the same reasoning also
<span>v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s </span>, hope this helps:)
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
Answer:
A) 0.50 mV
Explanation:
In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:
where
is the strength of the magnetic field
v = 13 m/s is the speed of the bird
L = 1.2 m is the wingspan of the bird
is the angle between the direction of motion and the direction of the magnetic field
Substituting numbers into the formula, we find
