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Oxana [17]
3 years ago
6

Part A What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torqu

e with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torque with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. 2000x + 1400 N-m 2000x N-m –2000x +1800 N-m –2000x – 1400 N-m 2000x – 1800 N-m –2000x – 200 N-m 2000x + 200 N-m –2000x N-m
Engineering
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

2000x + 1400 N-m

Explanation:

The torque at a given point is determined as the perpendicular force multiply by the distance of the force from the given point. The conventional method is to take clockwise direction as positive and anticlockwise direction as negative. Therefore:

Torque at the segment AB = 2000*x + 400 = (2000x + 400) N-m

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A 2-bit positive-edge triggered register has data inputs d1, d0, clock input clk, and outputs q1, q0. Data inputs d1d0 are 01 an
ale4655 [162]

Answer:

  q1q1 ⇒ 01

Explanation:

The outputs of a positive edge triggered register will match the inputs after a rising clock edge.

  q1q1 ⇒ 01 . . . . matching d1d0 = 01

7 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
the tire restraining device or barrier shall be removed immediately from service for any of these defects except
lora16 [44]

Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air.

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5 0
2 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
3 years ago
If 5000 N of thrust is acting to the left, and 4300 N of drag is acting to the right, what is the magnitude and direction of the
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Answer:

700 N acting to the left.

8 0
3 years ago
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