What are some things that AREN’T manufactured

A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i

dmitriy555 [2]

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

${v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2}$

Now, we substitute v, u, and a in the formula

$v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds$

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6 0

3 years ago

what is the expected life 1 inch diameter bar machined from AISI 1020 CD Steel is subjected to alternating bending stress betwee

Alexeev081 [22]

Answer:

1.287 *10⁷ cycles.

Explanation:

See attached pictures.

3 0

2 years ago

A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a

Xelga [282]

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) $in^u$

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum

Ts is torque transmitted by steel

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅ $_{alu$ = TaLa/GaJa

for steel rod ∅ $_{steel$ = TsLs/GsJs

but we know that, ∅a = ∅s = ∅ $_B$

so

[TaLa/GaJa] = [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa] = [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

so

∅ $_{steel$ = TsLs / GsJs

we substitute

∅ $_{steel$ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅ $_{steel$ = 83.26 / 276460.1535

∅ $_{steel$ = 0.000301

∅ $_{steel$ = 3.01 × 10⁻⁴ rad

so

∅ $_{steel$ = ∅ $_B$ = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad

5 0

2 years ago

A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat

Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

$Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ$