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2 years ago

There are two questions about SolidWorks.

Engineering

1 answer:

Nataliya [291]2 years ago

4 0

Answer:

1. It is a good practice to fully define a sketch to avoid having erroneous dimensions on the faces of a solid, this avoids that when it is required to make an assembly with the drawn part appear assembly errors.

2. The 2D sketch should always be done on a plane, so solidworks would ask you to select a plane on which you want to make the sketch, on the other hand, if it is a 3D sketch, solidworks allows you to do it without the need for Select any plane.

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An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n= 1200

Sample mean : $\overline{x}=2.45$

Standard deviation : $\sigma=0.07$

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, $z=\dfrac{x-\mu}{\sigma}$ , the z-value corresponds to 2.39 will be :-

$z=\dfrac{2.39-2.45}{0.07}\approx-0.86$

z-value corresponds to 2.60 will be :-

$z=\dfrac{2.60-2.45}{0.07}\approx2.14$

Using the standard normal table for z, we have

P-value = $P(-0.86$

$=P(z$

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0

3 years ago

Which of the following is not a relationship set between elements in a sketch

Svet_ta [14]

Answer:

The entity relationship (ER) data model has existed for over 35 years. It is well suited to data modelling for use with databases because it is fairly abstract and is easy to discuss and explain. ER models are readily translated to relations. ER models, also called an ER schema, are represented by ER diagrams.

6 0

2 years ago

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0

2 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

Levart [38]

Answer:

230.51 m

Explanation:

Pb = 695 mmHg

Pt = 675 mmHg

Pb - Pt = 20 mmHg

Calculate dP:

dP = p * g * H = (13600)*(9.81)*(20/1000) = 2668.320 Pa

Calculate Height of building as dP is same for any medium of liquid

dP = p*g*H = 2668.320

H = 2668.32 / (1.18 * 9.81) = 230.51 m

8 0

3 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago