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:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
Answer: C. the motion of a spacecraft under gravitational influence.
Explanation:
A is Metallurgy, B is Biology, C is astro-physics, I am not sure what D is, but it's safe to say it's not physics, E, micro-biology, and the study of radiation. C is the only one involving physics.
Answer:
The boat's acceleration are: a= 0.3 m/s² in west direction.
Explanation:
Fm= 4100 N
Fwi= 800 N
Fwa= 1200 N
Fm - Fwi - Fwa= m *a
clearing a:
a= 0.3 m/s²
Answer:
Any variable would be affected if the product is varied.
Explanation:
If you have any questions feel free to ask in the comments - Mark
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