Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
Answer:
a) 2.5m/s
b) 0.91m/s
c) 0m/s
Explanation:
Average velocity can be said to be the ratio of the displacement with respect to time.
Average speed on the other hand is the ratio of distance in relation to time
Thus, to get the average velocity for the first half of the swim
V(average) = displacement of first trip/time taken on the trip
V(average) = 50/20
V(average) = 2.5m/s
Average velocity for the second half of the swim will be calculated in like manner, thus,
V(average) = 50/55
V(average) = 0.91m/s
Average velocity for the round trip will then be
V(average) = 0/75, [50+25]
V(average) = 0m/s
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here 
is the velocity after the collision.
Newton's laws A force cannot act alone is the THIRD LAW!
