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3 years ago

6

You are ordering steel cable for a 250 foot long zip-line you are building in your back yard. The cable can be ordered in diamet

ers with 1/16 inch increments. (1/16, 1/8, 3/16, 1/4, 5/16, 3/8, etc.) The manufacturer states that the cable has a Yield Stress of 108,000 psi and a Modulus of Elasticity of 10,186,000 psi. You have determined that the maximum load the cable should be able to support without exceeding the Yield Stress is 10,000 lbs.
a. What is the minimum diameter cable should you order?

2 answers:

AlekseyPX3 years ago

5 0

The answer would be 1/4 of an inch good luck!!

Margarita [4]3 years ago

3 0

Answer:

If i am correct It should be 1/4 of an inch

Explanation:

Sorry but i can't quite explain

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Which material has the highest cp value?

Nataly [62]

Answer: B. Water

Explanation:

6 0

2 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

3 0

2 years ago

Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

2 years ago

A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0

Aneli [31]

Answer:

Q=339.5W

T2=805.3K

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.

3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature

4 0

3 years ago

Air fl ows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 1258C, and the velocity is

abruzzese [7]

The correct temperature is 125°C

Answer:

A) M_a1 = 0.5

B) T2 = 232.17 K

C) V2 = 611 m/s

D) m' = 187 kg/s

Explanation:

We are given;

Pressure; P1 = 250 kPa

Temperature; T1 = 125°C = 398 K

Speed; v1 = 200 m/s

Area; A2 = 0.25 m²

M_a2 = 2

A) Formula for M_a1 is given by;

M_a1 = v/a1

Where;

v is speed

a1 = √kRT

k is specific heat capacity ratio of air = 1.4

R is a gas constant with a value of R = 287 J/kg·K

T is temperature

Thus;

M_a1 = 200/√(1.4 × 287 × 398)

M_a1 = 200/399.895

M_a1 = 0.5

B) To find T2, let's first find the Stagnation pressure T0

Thus;

T0/T1 = 1 + ((k - 1)/2) × (M_a1)²

T0 = T1(1 + ((k - 1)/2) × (M_a1)²)

T0 = 398(1 + ((1.4 - 1)/2) × (0.5)²)

T0 = 398(1 + (0.2 × 0.5²))

T0 = 398 × 1.05

T0 = 417.9 K

Now,similarly;

T0/T2 = 1 + ((k - 1)/2) × (M_a2)²

T2 = T0/[(1 + ((k - 1)/2) × (M_a2)²)]

T2 = 417.9/(1 + (0.2 × 2²))

T2 = 417.9/1.8

T2 = 232.17 K

C) V2 is gotten from the formula;

T0 = T2 + (V2)²/(2C_p)

Cp of air = 1005 J/Kg.K

Thus;

V2 = √(2C_p)[T0 - T2]

V2 = √((2 × 1005) × (417.9 - 232.17))

V2 =√373317.3

V2 = 611 m/s

D) mass flow is given by the formula;

m' = ρA2•V2

Where;

ρ is Density of air with an average value of 1.225 kg/m³

m' = 1.225 × 0.25 × 611

m' = 187 kg/s

6 0

2 years ago