The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec
<h3>How to find the displacement?</h3>
We are given the velocity equation as;
s' = 40 - 3t²
Thus, the speed equation will be gotten by integration of the velocity equation to get;
s = ∫40 - 3t²
s = 40t - ¹/₂t³
Thus, the displacement between times of t = 2 sec and t = 4 sec is;
∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]
∆S = 210 m
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Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top 
= 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π
/4) ×
= √(4Q/π)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².
<h3>What is normal stress?</h3>
If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.
σ = P/A
Where, σ = Normal stress
P = Pressure
A = Area
1 Kg = 9.81 N
800 kg = 7848 N
Since the rod is half bronze and half steel
800 kg = 7848/2
= 3924 N
Pₙ = Fₙ = 3924 N [n = Bronze]
Pₓ = 3924 N [x = steel]
Given,
σₙ = 90MPa
σₓ = 120MPa
Aₙ = ?
Aₓ = ?
Aₙ = Pₙ/σₙ
Aₙ = 3924/90
Aₙ = 43.6 mm²
Aₓ = Pₓ/σₓ
Aₓ = 3924/120
Aₓ = 32.7 mm²
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