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Viktor [21]
4 years ago
6

One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight l

ast?
Engineering
1 answer:
Effectus [21]4 years ago
7 0

Answer:

time = 5.22 hr

Explanation:

Given data:

Energy of battery = 9400 J

Power consumed by three led bulb is 0.5 watt

we know Power is give as

Power = \frac{ energy}{time}

plugging all value and solve for time

time = \frac{energy}{power}

time = \frac{9400}{0.5}

time = 18,800 sec

in hour

1 hour = 3600 sec

therefore in 18,800 sec

time = \frac{18800}{3600} = 5.22 hr

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Which material would cause a more severe burn if equal masses of two distinct metals are heated to a temperature of 100 °C: the
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Answer:

The material with the higher specific heat capacity would cause a more severe burn.

Explanation:

Quantity of heat (Q) = mass of material (m) × specific heat capacity (C) × temperature difference (∆T)

From the formula above, the relationship between Q and C is direct in which increase in one quantity (C) leads to a corresponding increase in the other quantity (Q)

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3 years ago
The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
4 years ago
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