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2 years ago

Which of the following power tools has a revolving vertical shaft and a cutter? *

Engineering

2 answers:

Natalka [10]2 years ago

5 0

Saber saw I believe

Morgarella [4.7K]2 years ago

4 0

Saber saw I’m pretty sure

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If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respect

Natasha2012 [34]

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance $x!$

= cos 75⁰ = $\frac{x^!}{10ft}$

$x! = 10 * cos 75^{o}$ = 2.59 ft

calculate distance x

= cos 75⁰ = $\frac{x}{30ft}$

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * $x!$ + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

= 4014.5 + 1950.35 + 2030.535

= 7995.385 ft. Ib ≈ 7.9954 kip.ft

6 0

3 years ago

What is the command this line of code is telling the robot?

podryga [215]

Answer:

drive the car

Explanation:

4 0

2 years ago

Read 2 more answers

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

3 years ago

An electric kettle is required to heat 0.64 kg of water from 15.4°C to 98.2°C in six

skelet666 [1.2K]

Answer:

Almost done

Explanation:

I am just finishing up my work

7 0

2 years ago

The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a rate of 15

KATRIN_1 [288]

Answer:

attached below

Explanation:

7 0

2 years ago