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kkurt [141]
3 years ago
10

Does anybody know how to take a screenshot on a HP pavilion computer?

Engineering
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

I do i do it everyday

Explanation:

Press windows and prt sc at the same time

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In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetn
nekit [7.7K]

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

7 0
3 years ago
The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
James River Jewelry is a small jewelry shop. While James River Jewelry does sell typical jewelry purchased form jewelry vendors,
Sveta_85 [38]

Answer:

This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.

I hope it will help you a lot.

Explanation:

7 0
3 years ago
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
In casting experiments performed using a certain alloy and type of sand mold, it took 170 sec for a cube-shaped casting to solid
poizon [28]

Answer:

Answer for the question is : Solidification time will be same i.e. 170. See attached file for explanation.

Explanation:

Download pdf
7 0
3 years ago
Read 2 more answers
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