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3 years ago

The uniform slender rod has a mass m.

Engineering

1 answer:

Nikolay [14]3 years ago

6 0

Answer: 3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

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irina1246 [14]

Answer:

It B

Explanation:

4 0

2 years ago

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

2 years ago

What is a material safety data sheet? A. a categorized list of hazardous substances in an industry B. a document with safety inf

horsena [70]

Answer: A.

Explanation:

safety data sheet are documents that list information relating to occupational safety and health for the use of various substances and products.

3 0

3 years ago

(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a

Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 × $10^{-5}$ %

Explanation:

No. of phosphorus atoms = 5 × $10^{17} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100$

PCT = $10^{-3} = 0.001$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × $10^{15} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100$

PCT = 0.4 × $10^{-5}$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

7 0

3 years ago

A spherical tank is being designed to hold 10 moles of carbon dioxide gas at an absolute pressure of 5 bar and a temperature of

lesya692 [45]

Answer:

r=0.228m

Explanation:

The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases

PV=nRT

where

P=pressure =5bar=500.000Pa

V=volume

n=moles=10

R = universal constant for ideal gases = 8.31J / (K.mol)

T=temperature=80F=299.8K

solvig For V

V=(nRT)/P

$V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3$

we know that the volume of a sphere is

$V=\frac{4\pi r^3}{3} \\$

solving for r

$r=\sqrt[3]{ \frac{3 V}{4\pi } }$

solving

$r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m$

4 0

3 years ago