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3 years ago

1. What's the maximum overall length of the part? 2. What material is used to fabricate the part

Engineering

1 answer:

kiruha [24]3 years ago

8 0

Answer:

1. The maximum length overall length of the part is 4 inches

2. The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel

Explanation:

The information given in the drawing are;

The angle of projection of the drawing = Third angle projection

The thickness of the material = 0.25" plate

The material with which the part is fabricated = 0.25 inch thick HRS

HRS is the acronym for Hot Rolled Steel

Therefore;

The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel

The scale of the drawing = 1:1 = which is 1 is to 1 drawing

The maximum length overall length of the part, l = 4 units

By visually testing the 1:1 scale drawing measurements of 4 units is comparable 4 inches measurement with an online ruler

Therefore;

The maximum length overall length of the part, l = 4 inches

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4 years ago

The 2-kg package leaves the conveyor belt at with a speed of vA = 1 m/s and slides down the smooth ramp. Determine the required

Kisachek [45]

Answer:

Incomplete question: Find the normal reaction the curved portion of the ramp exerts on the package at B if pB = 2 m, the height is 4 m

Answer: The required speed is 8.9107 m/s

The normal reaction is 99.0006 N

Explanation:

Given data:

m = 2 kg

vA = speed = 1 m/s

h = 4 m

g = gravity = 9.8 m/s²

Questions: Determine the required speed of the conveyor, vc = ?

Find the normal reaction, Nc = ?

The required speed applying the conservation of energy:

$\frac{1}{2} mv_{A}^{2} +mgh=\frac{1}{2} mv_{c} ^{2}$

Solving for vc:

$v_{c} =\sqrt{v_{A}^{2}+2gh } =\sqrt{1^{2}+(2*9.8*4) } =8.9107m/s$

The normal reaction applying the equilibrium of forces:

$N_{c} -mg=m(\frac{v_{c}^{2} }{p_{B} } )$

$N_{c} =2*(\frac{8.9107^{2} }{2} )+(2*9.8)=99.0006N$

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3 years ago

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

Virty [35]

Answer:

a) The additional time required for the truck to stop is 8.5 seconds

b) The additional distance traveled by the truck is 230.05 ft

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

$v^2 - u^2 = 2*a*s$

$54.12^2 - 108.24^2 = 2*a*690$

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

$v^2 - u^2 = 2*a*s$

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

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