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3 years ago

In the case of an automobile electric system, the chassis of the vehicle is the ___________ side of the circuit.

Engineering

2 answers:

raketka [301]3 years ago

6 0

Answer:

ground

Explanation:

ankoles [38]3 years ago

5 0

Answer:

ground

Explanation:

In modern cars, the chassis is connected to the negative terminal of the battery, and provides the "ground" or "return" for the circuit.

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In crash tests, a shock absorber is used to slow the test car. The shock absorber consists of a piston with small holes that mov

Lesechka [4]

Answer:

The heat transferred to water equals 1600 kJ

Explanation:

By the conservation of energy we have

All the kinetic energy of the moving vehicle is converted into thermal energy

We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Thus

$K.E_{car}=\frac{1}{2}\times 2000\times 40^{2}=1600\times 10^{3}Joules$

Thus the heat transferred to water equals $1600kJ$

3 0

3 years ago

0 - 1" -20 -15 -10 5 0 1 2 3 0

faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0

3 years ago

A heat engine uses fuel of energy content 43.1 MJ/kg and produces 17.4 kW of useful power. The heat rejection rate (through the

solmaris [256]

Answer:

a)27.9%

b) $\dot{m}=3.06 \frac{Kg}{h}$

Explanation:

Given that

Fuel energy content = 73.1 MJ/kg

Useful power = 17.4 KW

Heat rejection rate = 44.8 KW

From first law of thermodynamics

Heat addition rate =Heat rejection rate + Power out put

Now by putting the values in the above formula

Heat addition rate = 44.8 + 17.4

Heat addition rate =62.2 KW

We know that efficiency is given as follows

$\eta =\dfrac{power\ out\ put}{Heat\ addition\ rate}$

$\eta =\dfrac{17.4}{62.2}$

$\eta =0.279$

So the efficiency is 27.9%.

Now to find usage rate of fuel

Lets take usage rate is $\dot{m}$

Fuel energy content x usage rate of fuel = Heat addition rate

Now by putting the values

$73100\times \dfrac{\dot{m}}{3600}=62.2$

$\dot{m}=3.06 \frac{Kg}{h}$

7 0

3 years ago

A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic

DiKsa [7]

Answer:

The entropy change of nitrogen during this process. is - 0.32628 kJ/K.

Explanation:

Solution

Given that:

A piston cylinder device contains =0.78 kg of nitrogen gas

Temperature = 37°C

The nitrogen gas constant of R = 0.2968 kJ/kg.K

At room temperature cv = 0.743 kJ/kg.K

Now,

We assume that at specific condition the nitrogen can be treated as an ideal gas

Nitrogen has a constant volume specific heat at room temperature.

Thus,

From the polytropic relation, we have the following below:

T₂/T₁ =(V₁/V₂)^ n-1 which is,

T₂ = T₁ ((V₁/V₂)^ n-1

= (310 K) (2)^1.3-1 = 381.7 K

So,

The entropy change of nitrogen is computed as follows:

ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)

= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))

= 0.57954 * 0.2080 + (-0.2057)

= 0.12058 + (-0.2057) = -0.32628

Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.

6 0

3 years ago

It was found experimentally that a certain material does not change in volume when subjected to an elastic state of stress. Calc

lora16 [44]

How are you? because i’m great

5 0

3 years ago