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slavikrds [6]
3 years ago
7

Independent auto lots usually have ____ finance rates than dealerships

Engineering
1 answer:
wel3 years ago
8 0

Independent auto lots usually have <u>higher</u> finance rates than dealerships

<u>Explanation:</u>

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

But this negotiation and lowering of the finance rates is not possible with the independent auto lots and thus they charge higher rates compared to the dealerships.

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An empty metal can is heated to 908C and sealed. It is then placed in a room to cool to 208C. What is the pressure inside the ca
Natali5045456 [20]

The pressure inside the can upon cooling is 0.4 atm.

<u>Explanation:</u>

Given -

Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K

Final Temperature, T2 = 208°C = 208 + 273 K = 481 K

Pressure upon cooling, P2 = ?

Using Gay Lussac's law:

P1/T1 = P2/T2

P2 = P1 X T2 / T1

P2 = 1 atm X 481 / 1181

P2 = 0.4 atm

Therefore, the pressure inside the can upon cooling is 0.4 atm.

3 0
2 years ago
You could be sued if you injure someone while rescuing them if...
navik [9.2K]
I believe number 4 I could be wrong but I think it’s 4
3 0
2 years ago
Read 2 more answers
A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to
Arada [10]

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by  = 6.22

Therefore he heat removed is given by , $Q_H = 6.22 \times 0.434$

                                                                     $Q_H = 2.7 \ kW $

Now if the electricity is valued at  0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

                                                                                            = 6.48

Therefore the operating cost = $ 6.48 for 24 hours.

3 0
2 years ago
A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi
saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, \tau_{max}, is given by the following formula;

\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )

t_w = 1 cm = 0.01

h = 29 cm = 0.29 m

h_w = 25 cm = 0.25 m

b = 15 cm = 0.15 m

I_c = The centroidal moment of inertia

I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )

I_c = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right )  = 1.2257083\bar 3 \times 10^{-4}

I_c = 1.2257083\bar 3 × 10⁻⁴ m⁴

From which we have;

4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )

Which gives;

W = 11,416.6879 N

\sigma _{b.max} = \dfrac{M_c}{I_c}

\sigma _{b.max} = 1500 N/cm² = 15,000,000 N/m²

M_c = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

M_{max} = \dfrac{W \cdot L}{4}

L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417

L ≈ 0.64417 m ≈ 64.417 cm.

4 0
2 years ago
Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w
soldi70 [24.7K]

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

4 0
2 years ago
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