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3 years ago

18

Engineering

1 answer:

svp [43]3 years ago

5 0

Answer:

серйозних порушень точності,

∵∴⊕∴∵

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The _______ is a tendency to assume that people with one positive attribute (e.g., who are physically attractive) also have othe

morpeh [17]

Answer:

Halo effect

Explanation:

In psychology, halo effect is defined as the tendency to assume that people with one positive attribute (e.g. physical attractiveness) also have other positive traits. Mere Exposure is the tendency to feel more positively toward a stimulus as a result of repeated exposure to it.

7 0

3 years ago

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0

3 years ago

Reverse masking forms a soft edge on the panel.

valina [46]

True I think I’m not sure?

7 0

2 years ago

If the reading of mercury manometer was 728 mmHg, what is the reading for another liquid such as water in mH20 units?

vodka [1.7K]

Answer:

mH275 units

Explanation:

that was true

4 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$