Question

Problem2 (a) Suppose outside temperature is 55 °F while the temperature inside a building is 72 F. The building is heated but not air conditioned, so the partial pressure of water vapor is the same both inside and outside the building. Finally, suppose the dew point outside is 40°F What is the dew point inside? What are the relative humidities inside and outside the building? (We are most comfortable if relative humidity is between 30% and 50%-This question shows why air in heated buildings can be uncomfortably dry.) (b) Inside the building, what is the root-mean-square velocity of the water molecules in the air? (c) Suppose a room of size 20" × 20" × 9"is at 72 F. and initially contains completely dry air. Now you place into this room a bucket containing 1.0L of liquid water at 72 F. After a long time (assuming the room has been kept at the same temperature and has been otherwise isolated), what will be the final relative humidity in the room? Now assume you bring into this room an unopened refrigirated bottle. You will quickly notice that water is starting to condense on the surface of the bottle. However, once the bottle heats up sufficiently, the condensation will disappear. What minimal temperature must the bottle reach in order for the condensation to start disappearing? You will need the following information. i. You may assume that the density of liquid water is 1 g/em at all temperatures of interest. ii. The following is a table of vapor pressure of water at different temperatures. Use linear interpolation of the values in this table in order to find the values you need. t[C05 10 20 25 PH,o hPal 6.11 8.7 (Note that I hPa 1 x 10-2Pa.) Problem 3 A metal marble is launched with a speed of vo = 30.0m/s at 50° relative to the horizontal, from a height of 1.0m above ground, and towards a very tall vertical wall, which is 6.0 m away from the launch position. The projectile may first hit the wall, or the ground. Whichever it turns out to be, assume that this is an elastic specular collision, so that the projectile bounces off. After some time, the projectile will have another collision, which again could be cither with the ground or with the wall. Find: (a) the location of the second collision; (b) the highest point (both the x- and the y-coordinates) that the projectile reaches between its launch and the second collision (c) the velocity (the magnitude and the angle relative to the vertical-make a sketch so tha it is clear which angle you are reporting) 0.50s after launch.

Answer 1

3.

(a) Forget the wall for now. Find the horizontal range first. For that we need time of flight which is calculated as follows

-1 = 30(sin 5°)t-49t2

assuming g=9.8.

T ะ 0.791 sec

So that the range is

R - 30(cos 5°)T ~ 23.65 m

Since the wall is 6 m, the projectile will hit the wall first.

So the time when it hit the wall is

t6/[30 (cos 5°)1 0.2 sec

height of the projectile at this time

$h_1=1+v\sin\theta t_1-4.9t_1^2=1.3275\textup{ m}$

vertical component of velocity just before collision

0.6468 m/s 2, sín θ _ 9.84

Elastic collision follows laws of reflection; vertical component of velocity changes symmetrically about the normal and horizontal just reverses.

Thus, now we have a new projectile motion, launched from

(6,1.328)

with vertical and horizontal velocities being

ul_ 0.65 m/s, ul = 29.89 m/s

towards -ve x direction.

This time the projectile just hits the ground. Proceeding as before:-

time of flight

1.328 = 0.65t-4.9

12 0.59 sec

range

$R_2=29.89\times0.59=17.67\textup{ m}$

so that the location of collision is

$(6-17.67,0)=(-11.67,0)$

(b) Note that the first collision (with the wall) occurred before it reached its maxima (compare t₁ and T). So the highest point occurs after first collision. Which can be calculated as follows

time to reach 0 vertical velocity

$v^{\perp}_1-gt_3=0\Rightarrow t_3=0.07\textup{ sec}$

the height (y coordinate) is

Hmar 1.328t3 4.91.35 m

the corresponding x coordinate is

$6-v^{\parallel}_1t_3=4.03\textup{ m}$

Thus, the highest point is at

(c)

$v^{\perp}(0.5)=v^{\perp}_1-g\times0.3=-2.29\textup{ m/s}$

So the magnitude of velocity is

$|v|=\sqrt{2.29^2+29.89^2}\approx29.98\textup{ m/s}$

and the angle relative to a horizontal line is

$\alpha=\tan^{-1}\left ( \frac{2.29}{29.89} \right )\approx4.39\degree$

downwards.