Mark each one as either Potential or Kinetic Energy?

If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?

sleet_krkn [62]

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

7 0

2 years ago

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

2 years ago

Which of the following is not a relationship set between elements in a sketch

Svet_ta [14]

Answer:

The entity relationship (ER) data model has existed for over 35 years. It is well suited to data modelling for use with databases because it is fairly abstract and is easy to discuss and explain. ER models are readily translated to relations. ER models, also called an ER schema, are represented by ER diagrams.

6 0

2 years ago

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

$E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\$

E=52000Hp.h

$E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\$

E=38724920Wh

$E=52000Hph(\frac{1977378.4 ft lb}{1Hph}$

E=1.028x10^11 ftlb

3 0

3 years ago

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni

arlik [135]

Answer:

The population size would be $p' = 5000$

The yield would be $MaxYield = 2082 \ fishes \ per \ year$

Explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

$\frac{dN}{dt} = \frac{2000}{1 \ year }$

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

$r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}$

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

$r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833$

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

$Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year$

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by $p'$ would be

$p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes$

7 0

3 years ago

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