We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
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<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Answer:
Explanation:
Diameter of pool = 12 m
radius of pool, r = 6 m
Total height raised, h = 3 + 2.5 = 5.5 m
density of water, d = 1000 kg/m³
Mass of water, m = Volume of water x density
m = πr²h x d
m = 3.14 x 6 x 6 x 5.5 x 1000
m = 113040 kg
Work = m x g x h
W = 113040 x 9.8 x 5.5
W = 6092856 J
Answer: a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Explanation:
Acceleration is the rate of change in the velocity per time
a = change in velocity/time
a = ∆v/t
average acceleration a = (v2 -v1)/t. ....1
Given;
Final velocity v2 = 1.63m/s
Initial velocity v1 = -1.15ms
time taken t = 2.11s
Substituting into eqn 1
a = [1.63 - (-1.15)]/2.11
a = (1.63+1.15)/2.11
a = 2.78/2.11
a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Answer:
True
Explanation: If this is a true or false question it is *T*
