# PH220 Week 1 Homework

Chap. 1, #1. The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten in (a) years, (b) seconds.

$14 billion years = 14*10^9 years = 14*365*24*3600*10^9 seconds =$

$= 4.41504*10^8*10^9 seconds= 4.41504*10^{17} seconds = 44*10^{16} seconds$

Chap. 1, #9. Multiply $2.079*10^2$ , $0.082*10^{-1}$ taking into account significant figures.

$207.9*0.0082 = 1.70478 =1.705$

Chap. 1, #10. What is the area of a circle of radius $3.8*10^4$ cm?

$area = \pi*r^2 =\pi*(3.8*10^4) ^2 =4.5365*10^9 cm^2 =4.5*10^5 m^2$

Chap. 1, #38. One hectare is defined as  $10^4 m^2$. One acre is $4*10^4 ft^2$.  How many acres are in one hectare?

$1 meter = 3.28 ft$

$1 sq m =10.7584 sq ft$

$one hectare = 10.7584*10^4 sq ft= (10.7584*10^4)/(4*10^4) acres =2.6896 acres$

Chap. 2, #10.   A horse canters away from its trainer in a straight line, moving 116 m away in 14.0 s. It then turns abruptly and gallops halfway back in 4.8 s. Calculate (a) its average speed and (b) its average velocity for the entire trip, using “away from the trainer” as the positive direction.

Total distance traveled = 116+(116/2) = 174 m

total displacement = 116-(116/2) =58 m

total time = 14+4.8 =18.8 sec

average speed =total distance/time = 174/18.8 =9.255 m/s

average velocity = total displacement/time =58/18.8 = 3.085 m/s

Chap. 2, #19. A sports car moving at constant speed travels 110 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration in  ?

speed = distance/time =110/5 =  22 m/s

acceleration = (0 – speed)/time = -22/4 =-5.5 m/s/s

Chap. 2, #58. The acceleration due to gravity on the Moon is about one-sixth what it is on Earth. If an object is thrown vertically upward on the Moon, how many times higher

will it go than it would on Earth, assuming the same initial velocity?

at maximum height $v = 0 m/s$

$0 = v0^2 -2*g*h_{max}$

$h_{max} = v0^2/(2g)$

Therefore:

$h(moon)/h(earth) = g(earth) /g(moon) = 6$

On the moon for the same initial velocity the maximum height is 6 times bigger than on the Earth.

Chap. 2, #69. A stone is dropped from the roof of a high building. A second stone is dropped 1.50 s later. How far apart are the stones when the second one has reached a speed of 12.0m/s?

time t for the second stone to reach $v = 12 m/s$

$v = g*t$

$t = v/g = 12/9.81 =1.2232 seconds$

total time for the first stone

$t1 = t +1.5 =1.2232+1.5 =1.7232 seconds$

distance traveled by first stone

$D1 = (g*t1^2)/2 =9.81*1.7232^2/2 =8.094 m$

distance traveled by second stone

$V^2 =2*g*D2$

$D2 = V^2/(2g) =12*12/2/9.81 =6 m$

Distance between stones is

$Distance = 8.094-6 =2.094 m$