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>
120n
since the speed is doubled, her force is doubled
Velocity. Since velocity consists of a speed and a direction, acceleration is a change in speed, or direction, or both.
Answer:
Pilots depend on the precision of INSTRUMENT LANDING SYSTEMS to guide them when they have limited visibility.
Explanation:
A pilot is a professionally trained individual that has the ability to control an aircraft following directions from a flight control.
Instrument landing system is a guidance system that makes available an instrument based procedures for guiding the pilot of an aircraft to approach and land safely. This system guides the pilot during unfavorable conditions such as low visibility through the use of radio signals.
There are two types of guidance provided by this system:
-> Lateral guidance subsystem and
-> Vertical guidance subsystem.
Lateral guidance subsystem prevents the aircraft approaching a runway to shift laterally from the path recommended.
Vertical guidance subsystem prevents the aircraft approaching a runway to shift laterally from the recommended path.
