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2 years ago

Which is the correct way of dual dimensioning using the position method

Engineering

1 answer:

erastova [34]2 years ago

3 0

Answer:

by placing one of the two measurements below the other or by separating them with a stroke

Explanation:

I had a test on this and it was correct

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A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

F = 90 x 9.99 N

F= 899.59 N

3 0

3 years ago

The primary of an ideal transformer has 400 turns and its secondary has 200 turns. Neglecting electrical losses, if the power in

Vanyuwa [196]

Answer:

A.C. Voltage (transformers only work with A.C.) from input (primary) to output (secondary) is purely a function of the turns ratio.

Explanation:

6 0

1 year 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

Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un a

Nitella [24]

Answer:

can you translate

Explanation:

what Is that?

4 0

2 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

2 years ago