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Elena L [17]
3 years ago
12

Our rule-of-thumb for presenting final results is to round to three significant digits or four if the first digit is a one. By t

his rule, write the result for
σ = Me/I , where M= 1835 lbf in and I/c = 0.875 in^3
Engineering
1 answer:
olga_2 [115]3 years ago
4 0

Answer:

To four significant digits = 2097 psi

Explanation:

<u>Applying the rule of thumb </u>

σ = Mc/I  ---- ( 1 )

M = 1835 Ibf in ,  I/c = 0.875 in^3

∴ c/l = 1 / 0.875 = 1.1429

back to equation 1

σ = 1835 * 1.1429 = 2097.2215 psi

To four significant digits = 2097 psi

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A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0
3 years ago
A bar having a length of 5 in. and cross-sectional area of 0.7 i n . 2 is subjected to an axial force of 8000 lb. If the bar str
barxatty [35]

Answer:

E=1.969 × 10¹¹ Pa

Explanation:

The formula to apply is;

E=F*L/A*ΔL

where

E=Young modulus of elasticity

F=Force in newtons

L=Original length in meters,m

A=area in square meters m²

ΔL= Change in length in meters,m

Given

F= 8000 lb = 8000*4.448 =35584 N

L= 5 in = 0.127 m

A= 0.7 in² =0.0004516 m²

ΔL = 0.002 in = 5.08e-5 m

Applying the formula

E=(35584 * 0.127)/(0.0004516*5.08e-5 )

E=1.969 × 10¹¹ Pa

8 0
3 years ago
One horsepower is the ability to lift 550 pounds a height of one foot in one second or one horsepower is equal to 550ft-lb/s. Ho
madam [21]

Answer:

32 horsepower will be equal to 17600 ft-lb/sec

Explanation:

We have given 1 horsepower = 550 ft-lb/sec

We have to convert 32 horsepower into ft-lb/sec

As we know that 1 horsepower is equal to 550 ft-lb/sec

So for converting horsepower to ft-lb/sec we have multiply with 550

We have given 32 horse power

So 32\ horsepower =32\times 550=17600ft-lb/sec

So 32 horsepower will be equal to 17600 ft-lb/sec

5 0
4 years ago
S1.1 The Acre-Foot. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as
Mars2501 [29]

Answer:

1072 acre foot

1331424000 kg

Explanation:

1 feet has 12 inches, so 2 in is 0.167 feet.

1 km^2 has 1 million m^2.

1 acre is 4074 m^2.

So, 1 km is 247 acres.

Then 26 km^2 is 6422 acres.

So, the volume of water is

6422 * 0.167 = 1072 acre-foot

Since one cubic meter of water has 1000 kg

One inch is 25.4 mm = 0.0254 m

One feet is 12 * 0.0254 = 0.3048 m

An acre-feet has a volume of

4074*0.3048 = 1242 m^3

And that is a mass of water of

1242 * 1000 = 1242000 kg/acre-feet

Therefore the mass of rainwater in the town is of

1072 * 1242000 = 1331424000 kg = 1331424 tons

4 0
3 years ago
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