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valkas [14]
2 years ago
7

You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

ing New York, you take your bathroom scale into the elevator and stand on them. The scales read 140 lb as the door closes. The reading varies between 120 lb and 170 lb as the elevator travels 101 floors.i)What is the maximum acceleration upward?ii)What is the maximum magnitude of the acceleration downward?
Physics
1 answer:
love history [14]2 years ago
7 0

To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N

120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N

170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

F = 756 - 622.3

F = 133.7N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{133.7}{63.5}

a = 2.1m/s^2

PART B) For the maximum magnitude of the acceleration downward we have that:

F = 622.3 - 533

F = 89.3N

Through the Newtonian relationship of the Force we have to:

F= ma

a = \frac{F}{m}

a = \frac{89.3}{63.5}

a = 2.1m/s^2

a = 1.04 m/s^2

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Part a) When collision is perfectly inelastic

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Part b) When collision is perfectly elastic

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Explanation:

Part a)

As we know that collision is perfectly inelastic

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mv_m = (m + M)v

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v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

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now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

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2 years ago
The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
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The formula for potential energy is PE=mgh
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Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

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Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

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I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

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we are given in the problem the following dimensions or specifications 
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