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Umnica [9.8K]
3 years ago
9

1. There are two categories (shapes) for the Virginia driver's license. The ________________________ shape license represents th

e driver who is under the age of 21, and the _____________________________ shaped license represents the driver who is over the age of 21.
Engineering
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

Vertical; horizontal.

Explanation:

The Virginia Department of Motor Vehicles started issuing sets of newly designed driver's licenses to drivers in 2009. Although, the cards that were issued to drivers prior to the introduction of the new cards remained valid until they were expired.

There are two categories (shapes) for the Virginia driver's license. The vertical shape license represents the driver who is under the age of 21, and the horizontal shaped license represents the driver who is over the age of 21.

Additionally, the Virginia's driver license (vertical in shape) issued to drivers who are under the age of 21 has a background image of a dogwood flower while the horizontal shaped license issued to drivers who are over the age of 21 has a background image of the state capitol.

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At the beginning of last year, tarind corporation budgeted $1,000,000 of fixed manufacturing overhead and chose a denominator le
maxonik [38]

The tari's total standard machine-hours allowed for last year's output 600,000 hours

Budgeted at start of year:  $100,000 fixed manufacturing overhead  for 500,000 machine hours

Standard = $100,000 / 500,000 hours = $0.2 fixed overhead / maching hour

At end of year, manufacturing overhead volume was $20,000 favorable which means

$20000 / $1=0.2 = 100000 additional hours.

Total Standard Machine Allowance Allowed for output = 500,000 + 100,000 = 600,000 hours.

the use of one machine running for an hour as a basis for cost estimation and operating effectiveness evaluation. In order to determine the contribution margin per machine hour for a specific product: a. Total cost per unit divided by the quantity of machine hours required to produce each unit of the target product. The manufacturing overhead cost divided by the activity driver yields the predetermined overhead rate. For instance, if machine hours were the activity driver, you would divide overhead costs by the anticipated number of machine hours.

Learn more about machine hours here:

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7 0
1 year ago
A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 83,051 N , and experien
umka2103 [35]

Answer:

the answer is attached with required units.

Explanation:

5 0
3 years ago
A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles
tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

(c) μ = 39.07

Explanation:

We are given:

L  60  70  80  90  100  110  120  130  140  150  160  170  180  190  200  210

f    2     1    3     5     8     12     6     10     8     5     2      3      2      1       0      1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

         = 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

          = 8480/69

Mean = 122.9  

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

   =7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

     = 1104600/69 - (122.9)²

     = 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

              = P(z<-0.26)

Using the normal distribution probability table, we can compute the value of  P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0
3 years ago
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Mars2501 [29]
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3 years ago
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Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,
solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

8 0
3 years ago
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