A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
1 answer:
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
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Answer:
By using the acceleration formula,
Answer:
The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Explanation:
Hi there!
The total momentum is calculated as the sum of the momenta of the pieces.
The momentum of each piece is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
The 300 g-piece flies along the y-axis. Its momentum vector will be:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
The total momentum is the sum of each momentum:
Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Total momentum = (16.4 kg m/s, 13.5 kg m/s)
The magnitude of the total momentum is calculated as follows:
The direction of the momentum vector is calculated using trigonometry:
cos θ = px/p
Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)
Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.