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
Observe that the given vector field is a gradient field:
Let 
, so that
Integrating the first equation with respect to 
, we get
Differentiating this with respect to 
gives
Now differentiating 
with respect to 
gives
Putting everything together, we find a scalar potential function whose gradient is 
,
It follows that the curl of is 0 (i.e. the zero vector).
Answer:
Frequency of incident light = 995×10 12
Explanation:
hope this helped you :)
During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s
