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amm1812
4 years ago
10

You can't mix conventional and syntheic oil

Engineering
1 answer:
elena-14-01-66 [18.8K]4 years ago
4 0

Answer:

There is no danger mixing synthetic and conventional motor oil. However, conventional oil will detract from the superior performance of synthetic oil and reduce its benefits.

Explanation:

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A tailgate may have a latch on both sides.<br> O True<br> O False
stepladder [879]

Answer:

true

Explanation:

3 0
3 years ago
Read 2 more answers
A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min
hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min (1)

The Laplace transform of this function is:

\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)    (3)

\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}    (4)

4 0
4 years ago
An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat
IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature T_L = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature T_H = 99° F = ( 99 + 460 )R = 559 R

Cooling Load Q_L = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

COP_{air-condition = Cooling Load Q_L  / power P

we substitute

COP_{air-condition = 30000 Btu/h / 8396.619 Btu/h

COP_{air-condition = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

Q_L / P_{required = T_L / ( T_H - T_L )

we substitute

30000 Btu/h / P_{required = 530 R / ( 559 R - 530 R )

30000 Btu/h / P_{required = 530 R / 29 R

we solve for P_{required

P_{required  = ( 30000 Btu/h × 29 R ) / 530 R

P_{required  = ( 870000 Btu/h / 530 )

P_{required  = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

P_{required  = ( 1641.5094 / 2544.43 ) hp

P_{required  = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

COP_{rev-air-condition = T_L / ( T_H - T_L )

we substitute

COP_{rev-air-condition = 530 R / ( 559 R - 530 R )

COP_{rev-air-condition = 530 R / 29 R

COP_{rev-air-condition = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0
3 years ago
Are connected to a larger parent organization that can help with marketing and provide will have the same
Setler [38]
Answer: D. Franchises

7 0
3 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
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