Answer:
option b and E are true
Explanation:
A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:
option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.
<u>option b is true.</u> A rigid bar remains unchanged regardless of the load that it carries.
option c is incorrect, a rigid bar does not deform with loads on it
option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.
<u>option e is true.</u> A rigid bar would not experience any deformation whatsoever.
Answer:
Following are the solution to the given question:
Explanation:
The material is necessary to cook since frying is a speedy process for evaporation.
Drug A is now in the compressed fluid area, and the material would not boil if the pressure is chilled. Because the ship is solid, its substance A claim is false.
Unless the volume comprises of a drop in the heat, the B substance reaches a vapor pressure area and a wet region. That's the area that melting may occur. His claim for material B could therefore be true.
B
Now this is a guess sorry but i am positive on this guess
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
Making a the subject formula;
Substituting into the above formula;
a = 0.07m or 70mm.
Also, shearing stress;
Making b the subject formula;
Substituting into the above equation;
b = 0.205m or 205mm