1) 2) In the United States, a doll house has the scale of 1:12 of a real house (
ID: 2138450 • Letter: 1
Question
1)
2)In the United States, a doll house has the scale of 1:12 of a real house (that is, each length of the doll house is 1/12 that of the real house) and a miniature house (a doll house to fit within a doll house) has the scale of 1:144 of a real house. Suppose a real house has a front length of L = 21 m, a depth of 12 m, a height of 6.0 m, and a standard sloped roof (vertical triangular faces on the ends) of height h = 3.4 m.
The Earth has a mass of 5.98 times 1024 kg. The average mass of the atoms that make up the Earth is about 38 u. Based on this average mass, how many atoms are in the Earth? In the United States, a doll house has the scale of 1:12 of a real house (that is, each length of the doll house is 1/12 that of the real house) and a miniature house (a doll house to fit within a doll house) has the scale of 1:144 of a real house. Suppose a real house has a front length of L = 21 m, a depth of 12 m, a height of 6.0 m, and a standard sloped roof (vertical triangular faces on the ends) of height h = 3.4 m. What is the volume of the corresponding miniature house in cubic meters? Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 3.00t + 18.00, with t 0, m in grams, and t in seconds. At what time is the water mass greatest? What is that greatest mass? In kilograms per minute, what is the rate of mass change at t = 2.00 s? In kilograms per minute, what is the rate of mass change at t = 5.00 s? The description of a certain brand of house paint claims a coverage of 415 ft2/gal. Express this quantity in square meters per liter. Express this quantity in SI base units (see Appendix A and Appendix D). What is the inverse of the original quantity? gal/ft2What is its physical significance?Explanation / Answer
a)
number of atoms = mass of earth/mass of each atom = 5.98*10^24/(38*1.67*10^-27) = 9.4*10^49
2)
volume = (12*6+0.5*12*3.4)*21 = 1940.4 m3
3)
a)differentiate (6.00t 0.8 - 3.00t + 18.00) with respect to t and equate to 0
So, d(6*t^0.8-3*t+18)/dt = 0
So, 6*0.8*t^(0.8-1)-3 = 0
So, t = (3/(6*0.8))^(-5) = 10.49 s <--------answer
b)
greatest mass = 6*10.49^0.8-3*10.49+18 = 25.86 s <-----answer
c)
rate of change = dm/dt = d(6*t^0.8-3*t+18)/dt
= 6*0.8*t^(0.8-1)-3
=4.8*t^-0.2 - 3
So, at t = 2, rate = 4.8*2^(-0.2) -3
= 1.18 kg/s
= 1.18*60 kg/min = 70.7 kg/min <----------answer
d)
at t = 5s
rate = 4.8*5^(-0.2)-3
=0.478 kg/s
=0.478*60 = 28.7 kg/min <----------answer
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