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Anastaziya [24]
1 year ago
14

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i

s 11,000 N. The speed v_1 of the elevator at the beginning of the 8 m descent is most nearly (A)4 m/s (B) 10 m/s (C) 13 m/s (D) 16 m/s E) 21 m/s
Physics
1 answer:
WARRIOR [948]1 year ago
6 0

The speed v_1 of the elevator at the beginning of the 8 m descent is most nearly 4 m/s.

The rate of change in the position of an object in any course. Speed is measured because of the ratio of distance to the time wherein the distance became included. Speed is a scalar amount because it has the handiest route and no significance.

Velocity can be the notion of the rate at which an object covers distance. A quick-moving object has a high speed and covers a notably big distance in a given amount of time, whilst a sluggish-shifting item covers a relatively small amount of distance in the same amount of time.

The primary unit or the S.I. unit of velocity is m/s or ms⁻¹.

Learn more about Speed here:-brainly.com/question/13943409

#SPJ4

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The air in a car tire la compressed when the car rolls over a rock. If the air
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Answer:

the signs of heat and work are; -Q and -W

Explanation:

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W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

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Is there information in the previous question which relates to this one?
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One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

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Using the same cost and time estimates, consider the time-effectiveness of each engineer.
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Answer:

Camilla

Explanation:

I got it right on edge. :)

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