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When you push a child on a swing, you are doing work on the child because you are pushing against the force of gravity
hope this helps :)
You measure the open distance between the floor and the bottom surface of the gas pedal. Then you press the gas pedal down 1/2 of that distance.
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Answer:</h2>
<em>Hello, </em>
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QUESTION)</u></h3>
Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².
<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>
- Δt = (√2xgxh)/9,8
- Δt = (14√10)/9,8
- Δt ≈ 4,5 s
Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg