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3 years ago

What is the normal balance side of an asset?

1 answer:

8 0

The normal balance for asset and expense accounts is the debit side, while for income, equity, and liability accounts it is the credit side. An account's assigned normal balance is on the side where increases go because the increases in any account are usually greater than the decreases

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Two technicians are discussing scan-tool use. Technician A says that a special diagnostic interface device is needed to connect

Anna71 [15]

Answer:

both technicians are correct, I haven't taken the test

7 0

3 years ago

Read 2 more answers

The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress

sasho [114]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0

3 years ago

scrapers are used to haul dirt from a borrow pit to the cap of a landfill. the estimated cycle time for the scrapers is 9.5 minu

ladessa [460]

Answer is: 12.8 because when you multiply it by the 2nd power of 8 then the scraper equals to be 12.8 in height and that’s how much each scraper requires to operate :)

7 0

3 years ago

An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh

Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in

B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

(ii) Average shear stress at the web = 18.289 * 10^5 psi

(iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 * 10^6 * 2.375}{53.54}$ = 0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 *10^6*(8.5-2.375)}{53.54}$ = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = $\frac{wL}{2}$ = $\frac{1000*6*12}{2}$ = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - $\frac{0.5}{2}$ ) * $\frac{(2.375 - (\frac{0.5}{2} ))}{2}$

= 5.878 in^3

t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 ) psi

= 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = $\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}$

= 1.143 *10^5

4 0

4 years ago

A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 23-lb bar of length L

Sergeeva-Olga [200]

Answer: hello attached below is the missing image the slender weight is different from what is in the question here so I worked with 23-Ib as requested in the question

answer

≈ 12.17 Rad/sec

Explanation:

weight of bullet ( Wb ) = 0.08 Ib

horizontal velocity = 1800 ft/s

Slender(Wr) = 23-Ib bar with

length ( L ) = 30

h = 12 inches

Vro = 0

<u>Calculate the angular velocity of the bar immediately after the bullet becomes embedded </u>

attached below is a detailed solution

6.708 = ( 0.05011 + 0.5011 ) w'

w' = 6.708 / 0.55121 ≈ 12.17 Rad/sec

6 0

3 years ago