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4 years ago

Jump starting a car requires _____.

Engineering

2 answers:

pantera1 [17]4 years ago

6 0

Answer:

Explanation:

maria [59]4 years ago

5 0

Answer:

Explanation: you need a jump cable to hook on a power source because the jump cable has two color cables to hook up to the power source. the two jump cables have two colors to tell you which one is positive and negative. The red one is the positive and the black on is the negative one.

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Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1

steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

$z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}$

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

$50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$ 50 m + 30.704358 m + 20.4081633 m = 10.234786 m + $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

90.8777353 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0

3 years ago

Suppose the following two events occur at the same time: the Chicago Cubs win the World Series, and the workers who make Cubs me

Annette [7]

Answer:

the answer would be decrease decrease

3 0

3 years ago

Tion A. Classwork

pantera1 [17]

Answer:

ambot...

Explanation:

hahahahahhahahahahaha

5 0

3 years ago

A pump operating at steady state receives liquid water at 20°C, 100 kPa with a mass flow rate of 53 kg/min. The pressure of the

VARVARA [1.3K]

Answer:

Input Power = 6.341 KW

Explanation:

First, we need to calculate enthalpy of the water at inlet and exit state.

At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:

Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:

h₁ = hf at 20° C = 83.915 KJ/kg

s₁ = sf at 20° C = 0.2965 KJ/kg.k

At the exit state,

P₂ = 5 M Pa

s₂ = s₁ = 0.2965 K J / kg.k (Isentropic Process)

Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:

h₂ = 88.94 KJ/kg

Now, the total work done by the pump can be calculated as:

Pump Work = W = (Mass Flow Rate)(h₂ - h₁)

W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)

W = 4.438 KW

The efficiency of pump is given as:

efficiency = η = Pump Work/Input Power

Input Power = W/η

Input Power = 4.438 KW/0.7

<u>Input Power = 6.341 KW</u>

5 0

3 years ago

Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 20 C.?

SIZIF [17.4K]

Answer:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

Explanation:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

The saturation temperature of steam is 45.81⁰ C at the pressure of 10 KPa which is higher than 20⁰C of river water. So river water at 20⁰C can be used to maintain the condenser pressure to 10 KPa.

7 0

3 years ago