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Yuki888 [10]
4 years ago
6

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

e distance the worker’s eyes must travel is 20 in. The perpendicular distance from her eyes to the line of travel is 24 in. No refocus is required. What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element?
Physics
1 answer:
iogann1982 [59]4 years ago
7 0

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

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Answer:

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Mass of the star = 5.09 × 10³⁰ kg

Explanation:

Given;

Diameter = 1.8 × 10⁷ m

Therefore,

Radius = \frac{\textup{Diameter}}{\textup{2}}  = \frac{\textup{1.8}\times10^7}{\textup{2}}

or

Radius of the planet = 0.9 × 10⁷ m

Rotation period = 22.3 hours

Radius of star = 2.2 × 10¹¹ m

Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s

free-fall acceleration = 12.2 m/s²

Now,

we have the relation

g = \frac{\textup{GM}}{\textup{R}^2}

g is the free fall acceleration

G is the gravitational force constant

M is the mass of the planet

on substituting the respective values, we get

12.2 = \frac{6.67\times10^{-11}\times M}{(0.9\times10^7)^2}

or

M = 1.48 × 10²⁵ Kg

From the Kepler's law we have

T² = \frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3

on substituting the respective values, we get

35164800² = \frac{\textup{4}\pi^2}{6.67\times10^{-11}\timesM_{star}}(2.2\times10^{11})^3

or

M_{star} = 5.09 × 10³⁰ kg

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8. In the billiard ball room at the basement of Memorial Union, two identical balls are traveling toward each other along a stra
mel-nik [20]

Answer:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

Explanation:

In an elastic collision, both the total momentum and the total kinetic energy of the system are conserved.

The conservation of the momentum can be written as:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

m_1 is the mass of ball 1

m_2 is the mass of ball 2

u_1 is the initial  velocity of ball 1

u_2 is the initial velocity of ball 2

v_1 is the final velocity of ball 1

v_2 is the final velocity of ball 2

The conservation of kinetic energy can be written as

\frac{1}{2}m_1 u_2^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2

Working together the two equations, it is possible to find two expressions for the final velocities in terms of the initial velocities:

v_1=\frac{m_1 -m_2}{m_1 +m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\v_2=\frac{2m_1}{m_1+m_2}u_1 -\frac{m_1 -m_2}{m_1 +m_2}u_2

In this problem we have:

m_1 = m_2 = m since the mass of the two balls is identical

u_1=+4.67 m/s is the initial velocity of ball 1

u_2=-7.89 m/s is the initial velocity of ball 2

Substituting into the equations, we find the final velocities:

v_1=\frac{2m}{m+m}u_2=u_2 =-7.89 m/s\\v_2=\frac{2m}{m+m}u_1=u_1=+4.67 m/s

Therefore:

- Ball 1 will move backward at 7.89 m/s

- Ball 2 will move forward at 4.67 m/s

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