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:
One
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___
<h3>what is supported mounted?</h3>
- A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.
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Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
Answer:
μ = 0.136
Explanation:
given,
velocity of the car = 20 m/s
radius of the track = 300 m
mass of the car = 2000 kg
centrifugal force


F c = 2666. 67 N
F f= μ N
F f = μ m g
2666.67 = μ × 2000 × 9.8
μ = 0.136
so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136
Answer:
The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D
Explanation:
An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.