Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
I think that this is false but I am not sure
Answer:
ε = 6.617 V
Explanation:
We are given;
Number of turns; N = 40 turns
Diameter;D = 18cm = 0.18m
magnetic field; B = 0.65 T
Time;t = 0.1 s
The formula for the induced electric field(E.M.F) is given by;
ε = |-NAB/t|
A is area
ε is induced electric field
While N,B and t remain as earlier described.
Area = π(d²/4) = π(0.18²/4) = 0.02545
Thus;
ε = |-40 × 0.02545 × 0.65/0.1|
ε = 6.617 V
(we ignore the negative sign because we have to take the absolute value)
Answer:
The answer cannot be determined.
Explanation:
The energy of the diver when he hits the pool will be equal to its potential energy 
, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.
The change in temperature 
then will be
Where m is the mass of water in the pool, c is the specific heat capacity of water, and 
is the added heat which in this case is the energy of the diver.
Since we do not know the mass of the water in the pool, we cannot make this calculation.
