She is
kilometers away from her starting point
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
<span>Is the following sentence true or false? Newton's first law does object's mass concentration and its axis of rotation increases, its rotational inertia The bicycle wheels at rest have no angular momentum, and the bicycle will fall over easily.</span><span>
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Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
Total work done is 0.13 Joules
<h3>What is work done ?</h3>
The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.
According to the given information
We need to find the work done
work done = force × distance
We are given,
force = 26 N
Distance = 0.0005 meter
hence ,
Work done = 26 × 0.005
= 0.13 Joules
Total work done is 0.13 Joules
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