The average speed is determined by the following formula:
average speed = [sum of (speed * time for which that speed was traveled)] / total time
average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division
average speed = 50.65 km/hr
The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60
Distance = 116.5 kilometers
Answer: P = 36.75W
The additional power needed to account for the loss is 36.75W.
Explanation:
Given;
Mass of the runner m= 60 kg
Height of the centre of gravity h= 0.5m
Acceleration due to gravity g= 9.8m/s
The potential energy of the body for each step is;
P.E = mgh
P.E = 60 × 9.8 × 0.5
PE = 294J
Since the average loss per compression on the leg is 10%.
Energy loss = 10% (P.E)
E = 10% of 294J
E = 29.4J
To calculate the runner's additional power
given that time per stride is = 0.8s
Power P = Energy/time
P = E/t
P = 29.4J/0.8s
P = 36.75W
(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
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Answer: The answer is D: 300,000km/s
Explanation:
I am not sure but i think the answer is C
