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Llana [10]
3 years ago
11

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 13 A, while that

of the vacuum cleaner is 4.8 A. Determine the power consumed by (a) the blow-dryer and (b) the vacuum cleaner. (c) Determine the ratio of the energy used by the blow-dryer in 15 minutes to the energy used by the vacuum cleaner in 40 minutes.
Physics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

(a) 1560 W

(b) 576 W

(c) 1.01

Explanation:

Voltage, V = 120 V

Current in dryer, I = 13 A

current in vacuum cleaner, i' = 4.8 A

(a) Power consumed by dryer,

P = V I = 120 x 13 = 1560 W

(b) Power consumed by vacuum cleaner

P' = V I' = 120 x 4.8 = 576 W

(c) Energy consumed by dryer

E = P x t = 1560 x 15 x 60 = 1404000 J

Energy consumed by the vacuum cleaner

E' = P' x t' = 576 x 40 x 60 = 1382400 J

the ratio of energies is

E : E' = 1404000 : 1382400 = 1.01

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Discuss the correlation or connection between stars with a higher mass and the amount of fuel they have to work with
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3 years ago
A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many
jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

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5 0
3 years ago
A signal source that is most conveniently represented by its Th´evenin equivalent has vs = 10 mV and Rs = 1 k. If the source fee
gogolik [260]

Answer:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

Explanation:

Here we have a power source in serie with a resistor of 1K and  RL, in order to obtain the Vo voltage we have to apply the voltage divider rule, that states:

Vo=Vin*\frac{RL}{RL+R1} \\R1=1kOhm

Substituing the resistor values of RL we obtained the following results:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to find the lowest value that gives us 80% of the source voltage we have to use the voltage divider rule again and make the Vo equal to 0.8 Vin:

0.8*Vin=Vin*\frac{RL}{RL+R1}

The result of the last equation is 4000, so in order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

8 0
3 years ago
5. What's the temperature -10°C in kelvins? ​
konstantin123 [22]

Answer:

263.15

. . . . . . . . . ..  ... . . . . . .. . . . . . . . ... .

7 0
2 years ago
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