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Korolek [52]
3 years ago
5

User Location: Inside Viewport Viewport Location: Fresh Viewport in downtown San Fransisco, California Query: [Burger King] Resu

lt: Burger King (located just outside the Fresh Viewport) What steps should you take to rate this result?
Engineering
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

Steps should you take to rate this result are

  • Check the pin location.
  • Check the official website for the address .
  • Check if there are closer results that we are not returning.
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Mile markers appear as green signs.
aksik [14]

Answer:

Mile markers appear as small, rectangular green signs with white numbers. Mile markers are especially useful in determining your precise location on desolate stretches of road. For example, by paying attention to mile markers, you will be able to inform a tow truck driver that you are stalled on the fifty-eighth mile of a particular roadway.

Explanation:

why Mile markers appear as green signs.

4 0
3 years ago
Please help i will give brainilest
Eduardwww [97]

Answer:

I hot u

Explanation:

it makes things function better and more fluently than a manual version of the certain object

8 0
3 years ago
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What is the main reason for using a complex data table?
wolverine [178]

Answer:

D. To interpret the possible meaning of the data

7 0
3 years ago
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Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e
marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

\omega =\sqrt{\frac{k}{m}}

where

'm' is the mass of the load

Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

8 0
3 years ago
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Which option identifies the type of power system Tommy will design in the following scenario?
Sedaia [141]

Answer:

diagram of an electrical curcuit

an sketch of an HVAC system

Also 3D image of a hydrualic piston

se

7 0
3 years ago
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