Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36 
The strain of the bar when subjected to 150 MPa is
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
or 1.8m
Answer:
W = 1.432 KJ
Explanation:
given,
mass = 22.2 Kg
angle of the rope = 27.5°
distance on the ground = 24 m
kinetic friction= μ = 0.32
acceleration due to gravity, g = 9.8 m/s²
Work done = ?
W = F d cosθ
a = 0 because it is moving with constant speed
equating all the forces acting in x direction
F cosθ = F friction = μN
equating all the forces acting in y direction
F sinθ + N -mg =0
now,
N = mg - F sinθ
putting value of N
F cosθ = μ mg -μ F sinθ
F (cosθ + μsinθ ) = μ mg
F =67.28 N
now,
W=F d cosθ
W =67.28 x 24 x cos(27.5)
W =1432.27 J
W = 1.432 KJ
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
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