The answer & explanation for this question is given in the attachment below.
Assume that television broadcasts are nonrival and nonexcludable (some TV stations, such as those on cable TV, are excludable, b
ut many are not) and that there are no live TV recording devices, such as VCRs or DVRs like TiVo. How can advertising solve the free‑rider problem?
2 answers:
Answer: Advertising acts in a method similar to a fee. People who watch TV broadcasts must watch ADs. TV stations turn this into money by selling airtime to advertisers.
Explanation:
A non-rival good is a good whose consumption by one person does not reduce the remaining quantity available. An example is a street light.
For non-excludable goods, it is impossible to prevent everyone from enjoying the benefits of the good. An example is a lighthouse. This is where the free rider problem comes in.
A free rider is someone enjoying the benefits of a good without paying for it. When a good is both non-rival and non-excludable, it is convenient for consumers to enjoy the benefit without paying for it.
If TV broadcasts are both non-rival and non-excludable, everybody can choose to become a free rider. Advertising can solve this problem by converting free riders to potential buyers of goods or services advertised during broadcasts. This way, stations can generate revenue by selling airtime.
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The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.