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irga5000 [103]
2 years ago
12

If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in

calories and joules)
Physics
1 answer:
Anit [1.1K]2 years ago
6 0
For the answer to the question above, 
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>
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The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and invers
soldier1979 [14.2K]

Answer:

L' = 555.95 lb

Explanation:

Analyzing the given conditions in the question, we get

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 also,

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combining the above conditions, we get an equation as:

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 now, for the first case we have been given

w = 3 in

d = 6 in

l = 11 ft

L = 1213 lbs

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1213 lb = k ((3 × 6²)/11)

or

k = 123.54 lbs/(ft.in³)  

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Using the calculated value of k to calculate the value of L in the second case  

in the second case, we have

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6 0
3 years ago
A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr
dmitriy555 [2]

Answer:

1.69515 seconds

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t = Time taken

u = Initial velocity

v = Final velocity

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a = Acceleration

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m

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\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds

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4 0
2 years ago
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Explanation:

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oksian1 [2.3K]

Answer:

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