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
if drppoed then locally path vertical down ... towards centre of earth
knowledge of first aid ... eg St John Ambulance, Red Cross etc. I think that everyone in a school should be taught First Aid.
Answer:
0.08 N/C
Explanation:
Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,
E = Kq/r².............................. Equation 1
Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.
making q the subject of the equation,
q = Er²/k............................... Equation 2
Given: E = 2 N/C, r = 4 m,
Substitute into equation 2
q = 2(4)²/k
q = 32/k C.
When r is increased to 20 m,
E = k(32/k)/20²
E = 32/400
E = 0.08 N/C.
Hence the electric Field = 0.08 N/C
Answer:
0.32 V
Explanation:
N = 10, A = 0.23 m^2, B = 0.47 T, t = 0.34 s
The average induced emf is given by
e = - N dФ / dt
Where, dФ be the change in magnetic flux in time dt.
dФ / dt = d / dt (B A) = A dB/dt
So,
e = - 10 x 0.23 x 0.047 / 0.34 = - 0.32 V
The negative sign shows the direction of induced emf.
