Assume earth is approximately a uniform, solid sphere that has a mass of 5.98×1024 kg and a radius of 6.38×106 m. calculate the
1 answer:
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Answer:
a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m - m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N
Answer:
Young's modulus for the rope material is 20.8 MPa.
Explanation:
The Young's modulus is given by:
Where:
F: is the force applied on the wire
L₀: is the initial length of the wire = 3.1 m
A: is the cross-section area of the wire
ΔL: is the change in the length = 0.17 m
The cross-section area of the wire is given by the area of a circle:
Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:
Where:
: is the tension of the wire
: is the weigh of the object = mg
m: is the mass of the object = 1700 kg
g: is the acceleration due to gravity = 9.81 m/s²
Hence, the Young's modulus is:
Therefore, Young's modulus for the rope material is 20.8 MPa.
I hope it helps you!