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nasty-shy [4]
3 years ago
10

When it is 2 PM Mountain Standard Time (MST) on February 3 in North Platte, NE(L = 101◦ W, φ = 41.1◦ N), what is the solar time?

b When it is 2 PM MST in Boise, ID(L = 116◦ W, φ = 43.6◦ N), on February 3, what is the solar time? c What Eastern Daylight Time corresponds to solar noon on July 31 for Portland, ME(L = 70.5◦ W, φ = 43.7◦ N)? d What Central Daylight Time corresponds to 10:00 AM on July 31 for Iron Mountain, MI(L = 90◦ W, φ = 45.8◦ N)?
Engineering
1 answer:
barxatty [35]3 years ago
3 0

Solar time for

North Platte is 13.02

Boise is 13.02

Eastern daylight time is 12.48

Central daylight is 12.06

<u>Explanation:</u>

  • 6 hours "central standard time" is the standard time zone. The saving time of daylight is + 1 hour. current time zone is the 5 hours central daylight time.
  • 7 hours is the mountain standard time. The saving time of daylight is + 1 hour. current time zone is the 6 hours central daylight time.
  • 5 hours is the eastern daylight time. The time shifts to +1 hour forwarded to eastern daylight zone and GMT is behind 4 hours.
  • 5 hours is the central daylight time. The GMT is behind 5 hours during summer season

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When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

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An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine t
lesya692 [45]

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the maximum bending stress in the strap is 3.02 ksi

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Given the data in the question;

steel strap thickness = 0.125 in

width = 2 in

circular arc radius = 600 in

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Now, using simple theory of bending

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substitute equation 1 into 2

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2 years ago
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