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IceJOKER [234]
4 years ago
13

QUESTION 29

Physics
1 answer:
siniylev [52]4 years ago
6 0

Answer:

a. Many strong friends

Explanation:

The Characteristics of Self-Actualization. They are comfortable with themselves and they appreciate other people for who they are. Self actualized people are also more likely to have what are known as peak experiences. These are profound moments of fulfillment to leave a person with a sense of awe.

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Solar collectors are parts of ______. a. passive solar heating systems b. active solar heating systems c. internal combustion en
Bad White [126]

Answer:

b. active solar heating systems

Explanation:

A solar collector is a device that collects and/or concentrates solar radiation from the Sun. These devices are primarily used for active solar heating and allow for the heating of water for personal use.

5 0
3 years ago
2. Is this chemical equation balanced?<br> 2C4H10 (g) + 13O2 (g) = 8CO2 (g) → 10H2O (l)
ehidna [41]
Yes that is a balaned equation
3 0
3 years ago
Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 K. Determine (a)
sergeinik [125]

Answer:

a) W=12166.20876 J

b) U= -12166.20876 J

Explanation:

No. of moles, n = 8.41

Change of temperature, ΔT = T1 - T2

                                         = 395 - 279

                                         = 116 K

For monatomic gas, γ = 5/3

γ -1 = 2 /3

Solution:

(a)

Work done,W= \frac{nR}{\gamma-1}(T_1-T_2)

plugging values we get

W= \frac{8.314\times8.41}{2/3}(116)

Ans:   12166.20876 J

Work done, W = + 12166.20876 J

(b)

From first law of thermodynamics, dQ = U + W

but, dQ = 0 ( adiabatic process)

Hence, U = - W

                 = - 12166.20876 J

Ans:

Change in internal energy, U = - 12166.20876 J

8 0
3 years ago
A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de
oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

\eta = \frac{W}{Q_{in}}

where

W is the useful work done by the engine

Q_{in} is the heat in input to the machine

In this problem, we have:

Q_{in}=600 kJ is the heat absorbed

W=600 kJ-175 kJ=425 kJ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)

So, the efficiency is

\eta = \frac{425 kJ}{600 kJ}=0.71 = 71\%

(a) 737^{\circ}C

The efficiency of an engine can also be rewritten as

\eta = 1-\frac{T_C}{T_H}

where

T_C is the absolute temperature of the cold sink

T_H is the temperature of the source

In this problem, the temperature of the sink is

T_C = 20^{\circ}C + 273=293 K

So we can re-arrange the equation to find the temperature of the source:

T_H = \frac{T_C}{1-\eta}=\frac{293 K}{1-0.71}=1010 K\\T_H = 1010 K - 273=737^{\circ}C

7 0
3 years ago
Cómo es útil el ejercicio Squat Side Kick para practicar deportes?? Dame dos ejemplos
svetlana [45]

Answer:

do you speak english

Explanation:

4 0
3 years ago
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