'a', 'b', and 'c' are all reasonable statements.
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.
</span>
<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:
At low pressure- 
At high pressure- 
Explanation:
Initial speed, 
Final speed, 
Net horizontal force due to rolling friction 
mg where m is mass, g is acceleration due to gravity, 
is coefficient of rolling friction
From kinematic relation, 
For each tire,
Making the subject
Under low pressure of 40 Psi, d=18 m
Therefore,
At a pressure of 105 Psi, d=93.7
Therefore,
Answer:
Final Velocity = 4.9 m/s
Explanation:
We are given;. Initial velocity; u = 2 m/s
Constant Acceleration; a = 0.1 m/s²
Distance; s = 100 m
To find the final velocity(v), we will use one of Newton's equations of motion;
v² = u² + 2as
Plugging in the relevant values to give;
v² = 2² + 2(0.1 × 100)
v² = 4 + 20
v² = 24
v = √24
v = 4.9 m/s
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Uses of laser
DNA sequencing instrument
Cutting and welding materials
Semiconducting chip manufacturing
Military devices
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