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Svetllana [295]
3 years ago
6

A classroom that normally contains 40 people is to be air-conditioned with window air-conditioning units of 5 kW cooling capacit

y. A person at rest may be assumed to dissi- pate heat at a rate of about 80 Watts. There are 10 light bulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h.
a. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window air-conditioning units required.
b. What is the source of heat being generated for the light bulbs and for the students?
Engineering
1 answer:
kap26 [50]3 years ago
4 0

Answer:

Explanation:

a ) Heat energy given out by 40 people

= 40 x 80 = 3200 J in one second

Heat energy given out by 10 light bulbs

= 10 x 100 = 1000 J in one second

Heat transfer by wall = 15000 x 1000 J in one hour

= 15000 x 1000 / 60 x 60

= 4166.67 J in one second

Total energy

= 3200 + 1000 + 4166 .67 J in one second

= 8366.67 J

To drive this much of heat , we need 2 air -conditioner of 5000 W each.

b ) Source of heat for bulb is electrical energy of bulb

For the students , source of heat by student is the bio- chemical reactions taking place in the body of the student which is exothermic.

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Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I
Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0
3 years ago
A solid circular cylinder of mass m, radius r, and length l is pivoted about a transverse axis
SashulF [63]
To be honest i have no idea
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3 years ago
Question 6: Which of the following is NOT true if you are changing into a
baherus [9]

Answer:

You should activate your signals 100 feet before you turn, becuase of that you should not wait until you are turning to turn on your signals. If you do not have working lights use your hands. So the answer that is inccorect is B.

You should activate your signals when you get into the turn lane is <u>NOT true</u>

Because you are supposed to single beforhand not while you're in the lane.

Im taking my drivers ed too. :)

6 0
3 years ago
9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space &amp; n(s)
zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0
3 years ago
g Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of he
AVprozaik [17]

Answer:

38kJ

Explanation:

Final Energy= Total Energy at the beginning + Total energy added - energy lost

Final Energy = 12.5 + 500/1000 + 30- 5

                   = 38kJ

8 0
3 years ago
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