Refer to the diagram shown.
F = 65 N, the force exerted by Dwight.
The horizontal component of the force is
65 cos(50°) = 41.8 N
The vertical component of the force is
65 sin(50°) = 49.8 N
Answer:
Horizontal: 41.8 N
Vertical: 49.8 N
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:
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Answer:
The value of tangential acceleration 
40 
The value of radial acceleration 
Explanation:
Angular acceleration = 50 
Radius of the disk = 0.8 m
Angular velocity = 10 
We know that tangential acceleration is given by the formula 
Where r = radius of the disk
= angular acceleration
⇒ 0.8 × 50
⇒ 40
This is the value of tangential acceleration.
Radial acceleration is given by
Where V = velocity of the disk = r 
⇒ V = 0.8 × 10
⇒ V = 8 
Radial acceleration
This is the value of radial acceleration.
