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2 years ago

In the last 5 meters of braking, you lose ___ of your speed.

Engineering

1 answer:

expeople1 [14]2 years ago

5 0

Answer:

answer is

3/4

Explanation:

In the last 5 meters of braking, you lose 3/4 of your speed.

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Situation: Peter is designing a new hybrid car that functions on solar power. He is currently working on sketches of his design

givi [52]

Answer:

whats the question?

Explanation:

8 0

2 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

What is the power of a parallel circuit with a resistance of 1,000 omh and current of 0.03a

Sergeeva-Olga [200]

Answer: P = I2R = 0.032 x 1000 =0.9 W

Explanation: The power will be the product of the square of the current and

the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.

4 0

3 years ago

Do the following addition exercises by translating the numbers into 8-bit 2's complement binary numbers, performing the arithmet

Ganezh [65]

Answer:

I am attaching a file with the solution and explanation as the number character limit is exceeding.

Explanation:

Download docx

8 0

3 years ago

A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and

Igoryamba

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL / COP r where 75 Btu/min = 1.76856651 hp where W actual = 0.70 hp

a) W rev = 1.76856651 hp / 8.73 = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual = 0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

8 0

3 years ago