PLEASE HELLP ILL GIVE CROWN AND HEART
2 answers:
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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
Answer:
Explanation:
From the question we are told that:
Height
Time
Generally the Newton's equation for Initial velocity upward is mathematically given by
Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28
Generally the Newton's equation for time to reach initial velocity is mathematically given by
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut + at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) + x a x 5.2
53 = 2.6a
a = = 20.4m/s²