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telo118 [61]
3 years ago
13

A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat

e of purchase, at an annual interest rate of 4%. Immediately after the second payment, the terms are changed such that the company can pay off the loan immediately in a lump sum during the following year. What is the final single payment?
Engineering
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

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The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

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3 years ago
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Answer:

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Explanation:

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When mining diamonds with a stone pick what will be the outcome
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Which of the following is the LEAST-Likely cause of tire wear? A) Underinflation B) Braking C) Acceleration D) Tire rotation
mihalych1998 [28]

Answer:

Tire rotation is the least likely cause of tire wear. So, the option D is correct.

Explanation:

Step1

Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.

Step2

On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.

Step3

Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.

Step4

Tire rotation has least amount of wear and tear due to no rubbing action.  It has less amount surface contact with the surface in rotation.  

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2 years ago
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