Answer:
3.34×10^-6m
Explanation:
The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain
can be expressed as
shear stress/(shear strain)
= (F/A)/(Lo/ . Δx)
Stress=Force/Area
The sheear stress can be expressed below as
F Lo /(A *Δx)
Where A=area of the disk= πd^2/4
F=shearing force force= 600N
Δx= distance
S= shear modulus= 1 x 109 N/m2
Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m
If we make Δx subject of the formula we have
Δx= FLo/(SA)
If we substitute the Area A we have
Δx= FLo/[S(πd^2/4]
Δx=4FLo/(πd^2 *S)
If we input the values we have
(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2
= 3.35×10^-6m
Therefore, its shear deformation is 3.35×10^-6m
A=area of the disk= πd^2/4
= [3.142×(4×10^-2)^2]/4
Answer:
light is an example of a wave that is not mechanical .
it is different as it does not need material medium for its propagation
Answer:
- a.
- b.
- c.
Explanation:
The spacetime interval 
is given by
please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:
.
Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.
<h3>a.</h3>
so
<h3>b.</h3>
so
<h3>c.</h3>
so
