The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;
The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;
Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
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Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X = t + (0.5) t²
100 = (4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y = t + (0.5) t²
0 = (4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt(()² + ()²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg
1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
and from this, we find
For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s
