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jenyasd209 [6]
3 years ago
8

FAST PLLZZ!! Ideally, the backrest is tilted back slightly, so when you turn the wheel your shoulders are _______ the seat.

Engineering
1 answer:
exis [7]3 years ago
8 0

Answer:

touching

Explanation:

The backrest of the seat should be tilted back ever so slightly, and when turning the steering wheel your shoulders should remain in contact with the seat – rather than hunched forward.

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BlackZzzverrR [31]

Answer:

what are is ethiopia cultural ?

7 0
2 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

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

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

z1= \frac{\left(150-137\right)}{27.7}=0.4693

z2=\frac{\left(201-137\right)}{27.7}=2.3105

P(0.4693

B.)

z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289

\\P(2.9309

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0
3 years ago
Read 2 more answers
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
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
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