Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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Answer:
9517.2 lbm
Explanation:
Electricity consumption = 14000 kWh/year
Fuel consumption = 900 gal/year
Amount of CO₂ produced per gallon = 26.4 lbm/gal
Amount of CO₂ produced per kWh = 1.54 lbm/kWh
Amount of CO₂ produced in one year
Reduction would be
The reduction in the amount of CO₂ produced is 9517.2 lbm
Answer:
1) The mass of the student is approximately 73.39 kg
2) The net force on the student is approximately 947.523 N
3) The value the scale will read is approximately 96.59 kg
Explanation:
The given parameters are;
The weight of the student = 720 N
The speed at which the elevator is decreasing = 3.1 m/s²
1) The weight of the student = The mass of the student × The acceleration due to gravity
The acceleration due to gravity is a constant = 9.81 m/s²
Substituting the known values gives;
720 N = The mass of the student × 9.81 m/s²
∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg
2) The forces acting on the student are;
i) The force of gravity which is the weight of the student acting downwards
ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student
The net force, = The weight of the student + The inertia force of the slowing elevator
∴ The net force, = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N
3) The scale will read the mass of the student as follows;
Mass reading of student on the scale = Force on scale/9.81
∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg
The value the scale will read = 96.59 kg.