Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
where, S = length of the rod
r = radius
Putting the given values into the above formula as follows.
= 
= 
= 
Now, we will calculate the charge density as follows.
= 
= 
Now, at the center of arc we will calculate the electric field as follows.
E = 
= 
= 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
I think this is the solution:
1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4
<span>D. price ceiling
</span><span>This is a government regulation that establishes a maximum price for a particular good.</span><span>
</span>
Answer:
The correct answer is 231 Mpa i.e option a.
Explanation:
using the equation of torsion we Have
where,
= shear stress at a distance 'r' from the center
T = is the applied torque
= polar moment of inertia of the section
r = radial distance from the center
Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.
We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals
Applying values we get
I think it’s D) all of the above
