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MrRissso [65]
3 years ago
13

There are 22 gloves in a drawer: 5 pairs of red gloves, 4 pairs of yellow, and 2 pairs of green. You select the gloves in the da

rk and can check them only after a selection has been made. What is the smallest number of gloves you need to select to have at least one matching pair in the best case?
Engineering
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

Best case = 2 gloves

Given Information:

Red gloves = 5 pairs

Yellow gloves = 4 pairs

Green gloves = 2 pairs

smallest number of gloves you need to select to have at least one matching pair in the best case = ?

Explanation:

How many gloves do you need to make one matching pair?

2

Yes, you are right. 2 gloves makes a matching pair and it is the smallest number of gloves you need to select to have at least one matching pair.

But what about worst case?

lets say

you tried all 5 red gloves either all of them were left or right

then you tried 4 yellow gloves either all of them were left or right

then you tried 2 green gloves either all of them were left or right

Now all the left or right gloves are tried (5 + 4 + 2 = 11) and the 12th one will definitely be either matching red, yellow or green.

Therefore, in the worst case scenario, the smallest number of gloves you need to select to have at least one matching pair is 12.

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A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

8 0
2 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0
3 years ago
What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances
gavmur [86]

Explanation:

Air fuel ratio:

 Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

Air\ fuel\ ratio=\dfrac{mass\ of\ air}{mass\ of\ fuel}

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.

When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.

When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.

5 0
3 years ago
Which best describes the body in terms of simple machines?
alex41 [277]

Answer:B

Explanation:

5 0
3 years ago
A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0
2 years ago
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