Yahoo Answers: Answers and Comments for Can you help me in physics? [Physics]
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From Samantha Messer
enUS
Thu, 24 Mar 2011 10:47:27 +0000
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Yahoo Answers: Answers and Comments for Can you help me in physics? [Physics]
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From Argent: 1.
According to the manual of a certain car, a...
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https://answers.yahoo.com/question/index?qid=20110324104726AAFGbQg
Thu, 24 Mar 2011 21:53:51 +0000
1.
According to the manual of a certain car, a maximum torque of magnitude 56.0 N·m should be applied when tightening the lug nuts on the vehicle. If you use a wrench of length 0.362 m and you apply the force at the end of the wrench at an angle of 75.00° with respect to a line going from the lug nut through the end of the handle, what is the magnitude of the maximum force you can exert on the handle without exceeding the recommendation?
Torque (L) = F·r, where F is the force employed and r is the distance to the point of application of the force.
Here, r = 0.362 m and the torque is 56.0 N·m (maximum),
so F = L/r = (56.0 N·m)/(0.362 m) = 154.7 N.
No more than this amount of force could be used if it were applied at an angle of 90°. Because the angle is less, however, the effective force (Fe) is less than the maximum force. To get the same effect at a smaller angle, the applied force must then be greater.
The effective force (Fe) is the product of the applied force (F) and the sine of the given angle:
Fe = F·sin(θ), so F = Fe/sin(θ) = (154.7 N) / sin(75.00°) = 160.2 N.
2.
Lake Erie contains roughly 4.00×10¹¹ m³ of water.
(a) How much energy is required to raise the temperature of that volume of water from 9.0°C to 11.0°C?
Water requires 1 cal = 4.186 J per mL to change temperature by 1°C.
In 4.00×10¹¹ m³ there are 4.00×10^17 mL, so to change this by 2°C requires 8.00×10^17 cal = 3.35×10^18 J.
(b) How many years would it take to supply this amount of energy by using the 1200MW exhaust energy of an electric power plant?
1200 MW = 1.2×10^9 J/s.
(3.35×10^18 J) / (1.2×10^9 J/s) = 2.79 Gs. Divide this by 31.56 Ms/y to get 88.4 y.
3.
An 85g sample of silicon is at 20°C. If 650 cal of energy is transferred to the silicon, what is its final temperature?
The specific heat capacity (C) of silicon (the amount of heat required to raise the temperature of 1 cm³ by 1°C) is 19.789 J·molˉ¹·Kˉ¹. Taking 1 cal = 4.186 J, this is 4.727 cal·molˉ¹·Kˉ¹.
The molar mass of silicon is 28.086 g·molˉ¹, so 85 g = 3.026 mol.
20°C = 293 K.
The amount of temperature change is calculated as the input energy (E) divided by the mass (M) and by the specific heat capacity (C):
ΔT = E/(MC)
= (650 cal) / ((3.026 mol)(4.727 cal·molˉ¹·Kˉ¹))
= 45.4 K.
The final temperature is then 293 K + 45.4 K = 338.4 K, or 338 K to 3 significant figures. (In Celsius, it's 65°C.)

From Anonymous: 56 = 0.362 . F cos 15˚
F = 160 N
Ene...
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Thu, 24 Mar 2011 21:29:37 +0000
56 = 0.362 . F cos 15˚
F = 160 N
Energy needed = Mass . specific heat . Temperature rise
E = 4 . 10^14 . 4.2 , 10^3 . 2 = 3.36 . 10^18 J
1.2 . 10^9 Joules per second = 3.78 . 10^16 J / year
It will take 88.89 = 88.9 years
What is the specific heat of Silicon? I have no idea.