- Ennio Morricone's Da Uomo A Uomo 22-Jan-2021 - Beat Records will present on CD Da Uomo A Uomo (aka Death Rides A Horse), a Western film directed in 1967 by Giulio Petroni, with music by Ennio Morricone, who perfectly managed to describe all the emotions of Bill's revenge against those who massacred his family by creating a recurring main theme in an aggressive ride for guitars and rhythm.
- The UV Collector. 1,876 likes 2 talking about this. Your Place to Buy and Sell Vudu, Movies Anywhere, and iTunes. Fantastic and cheap prices.
50:750:203 General Physics
I'm a private guitar collector interested in vintage instruments by Gibson, Fender, Martin, Gretsch, Epiphone, National, Dobro, Rickenbacker made from 1920 to 1969. I collect information about these guitars, and of course the guitars themselves. (This includes both electric and acoustic vintage guitars.).
Test 3. July 1999Attempt all questions. You must show some working. Credit will not be given for an answer without working.
1. A 65-kg mountain climber hangs freely on a nylon rope (radius = 6.0 mm). If the rope stretches by 6.0 × 10-2 m, what is the unstretched length of the rope? (Young's modulus for nylon is 3.7×109 N/m2)
Unite 1 0 6 download free. F = Y(DL/L0)A
or
L0 = Y(DL)A/F
The force in this case is the weight of the climber
L0 = (3.7×109)(6.0×10-2) p(6.0×10-3)2/(65×9.8) = 39 m
2. A spring stretches by 0.016 m when a 3.0-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 2.0 Hz?
First find the spring constant, k
F = kx
k = F/x = (3.0 kg × 9.8 m/s2)/(0.016 m) = 1837.5 N/m
The vibration frequency is
f = (1/2p)Ö(k/m)
Solve this for m:
m = k/(4p2f2) = (1837.5)/(4p2(2.0)2) = 11.6 kg
3. (Conceptual question) On a distant planet the acceleration due to gravity is less than it is on earth. Would you float more easily in water on this planet than on earth? Account for your answer.
The answer is that you would float just as well on the other planet as you would on earth, but no better. Your weight is mg, and, since g is less on the other planet, you would weigh less there. However, Archimedes principle is that you float when your body displaces a weight of water equal to your own weight. The weight of a given volume of water also contains the same factor of g, so its weight is also lower on the distant planet. You will displace exactly the same volume of water as on earth.
4. A pirate in a movie is carrying a chest (0.40m × 0.20m × 0.25m) that is supposedly filled with gold. To see how ridiculous this is, determine the weight (in newtons) of the gold. To judge how large this weight is, remember that 1 N = 0.225 lb. Give the weight is pounds also.
The volume of the chest is 0.40×0.20×0.25 = 0.020 m3. The mass of this volume of gold is
m = rV = (19300 kg/m3)(0.020 m3) = 386 kg
The weight of the gold is (386 kg)(9.8 m/s2) = 3780 N = 850 lb.
5. Find the approximate length of the Golden Gate bridge if it is known that the steel in the roadbed expands by 0.56 m when the temperature changes from +2 to +33o. (Linear thermal expansion coefficient of steel is 12 × 10-6 (Co)-1.)
DL = aL0DT
so
- Ennio Morricone's Da Uomo A Uomo 22-Jan-2021 - Beat Records will present on CD Da Uomo A Uomo (aka Death Rides A Horse), a Western film directed in 1967 by Giulio Petroni, with music by Ennio Morricone, who perfectly managed to describe all the emotions of Bill's revenge against those who massacred his family by creating a recurring main theme in an aggressive ride for guitars and rhythm.
- The UV Collector. 1,876 likes 2 talking about this. Your Place to Buy and Sell Vudu, Movies Anywhere, and iTunes. Fantastic and cheap prices.
50:750:203 General Physics
I'm a private guitar collector interested in vintage instruments by Gibson, Fender, Martin, Gretsch, Epiphone, National, Dobro, Rickenbacker made from 1920 to 1969. I collect information about these guitars, and of course the guitars themselves. (This includes both electric and acoustic vintage guitars.).
Test 3. July 1999Attempt all questions. You must show some working. Credit will not be given for an answer without working.
1. A 65-kg mountain climber hangs freely on a nylon rope (radius = 6.0 mm). If the rope stretches by 6.0 × 10-2 m, what is the unstretched length of the rope? (Young's modulus for nylon is 3.7×109 N/m2)
Unite 1 0 6 download free. F = Y(DL/L0)A
or
L0 = Y(DL)A/F
The force in this case is the weight of the climber
L0 = (3.7×109)(6.0×10-2) p(6.0×10-3)2/(65×9.8) = 39 m
2. A spring stretches by 0.016 m when a 3.0-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 2.0 Hz?
First find the spring constant, k
F = kx
k = F/x = (3.0 kg × 9.8 m/s2)/(0.016 m) = 1837.5 N/m
The vibration frequency is
f = (1/2p)Ö(k/m)
Solve this for m:
m = k/(4p2f2) = (1837.5)/(4p2(2.0)2) = 11.6 kg
3. (Conceptual question) On a distant planet the acceleration due to gravity is less than it is on earth. Would you float more easily in water on this planet than on earth? Account for your answer.
The answer is that you would float just as well on the other planet as you would on earth, but no better. Your weight is mg, and, since g is less on the other planet, you would weigh less there. However, Archimedes principle is that you float when your body displaces a weight of water equal to your own weight. The weight of a given volume of water also contains the same factor of g, so its weight is also lower on the distant planet. You will displace exactly the same volume of water as on earth.
4. A pirate in a movie is carrying a chest (0.40m × 0.20m × 0.25m) that is supposedly filled with gold. To see how ridiculous this is, determine the weight (in newtons) of the gold. To judge how large this weight is, remember that 1 N = 0.225 lb. Give the weight is pounds also.
The volume of the chest is 0.40×0.20×0.25 = 0.020 m3. The mass of this volume of gold is
m = rV = (19300 kg/m3)(0.020 m3) = 386 kg
The weight of the gold is (386 kg)(9.8 m/s2) = 3780 N = 850 lb.
5. Find the approximate length of the Golden Gate bridge if it is known that the steel in the roadbed expands by 0.56 m when the temperature changes from +2 to +33o. (Linear thermal expansion coefficient of steel is 12 × 10-6 (Co)-1.)
DL = aL0DT
so
L0 = (DL)/( aDT) = (0.56 m)/(12×10-6 (Co)-1 × 31oC) = 1500 m
(Is this a reasonable answer?)
6. How much heat must be added to 0.40 kg of aluminum to change it from a solid at 120oC to a liquid at 660oC (its melting point)? The specific heat capacity of aluminum is 9.00 × 102 J/(kg-Co) and its latent heat of fusion is 4.0 × 105 J/kg.
There are two contributions to the total. Let Q1 be the amount of heat needed to increase the temperature of the solid aluminum from 120oC to 660oC, and Q2 be the heat needed to convert the solid aluminum into liquid. Since 660oC is the melting temperature, there is no heat needed to change the temperature of the liquid.
Q1 = mcDT = (0.40)(9.00×102)(660 - 120) = 1.94 × 105 J
Q2 = mL = (0.40)(4.0×105) = 1.6 × 105 J
Then Q1 + Q2 = 3.54 × 105 J
7. One end of a brass bar is maintained at 300oC, while the other end is kept at a constant but lower temperature. The cross-sectional area of the bar is 4.0 × 10-4 m2. Because of insulation, there is negligible heat loss through the sides of the bar. Heat flows through the bar, however, at the rate of 2.8 J/s. What is the temperature of the bar at a point 0.20 m from the hot end?
Apply the heat conduction equation just to the 0.20 m of bar.
Q = kA(DT)t/L
Solve for DT:
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DT = QL/(kAt)
= (2.8 J)(0.20 m)/(110 J/(s-m-Co)×(4.0×10-4 m2)×1 s)
= 12.7oC
The temperature of the point 0.20 m from the hot end is therefore
300 - 12.7 = 287oC
Movie Collector 20 0 40 Oz
8. A solar collector is placed in direct sunlight where it absorbs energy at the rate of 840 J/s for each square meter of its surface. The emissivity of the solar collector is e = 0.70. What equilibrium temperature does the collector reach? Assume that the only loss is due to the emission of radiation.
Vitamin r 2 52 – personal productivity tool free. The rate of loss of heat by emission, for each square meter of surface is
esAT4 = (0.70)(5.67×10-8)(1)(T4)
In equilibrium, this is equal to the rate at which the collector absorbs heat, so
(0.70)(5.67×10-8)(1)(T4) = 840
T4 = 2.116 × 1010
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T = 381 K = 108oC
Movie Collector 20 0 40 Mm
Movie Collector 20 0 404
