ed_saifa's ambiguous physics questions =(

[ed_saifa]

Please help with these awful questions! Thanks

Space invaders land on earth. On the invaders' temperature scale, the ice point is at 24.0 °I (I = invader), and the steam point is at 156 °I. The invaders' thermometer shows the temperature on earth to be 66.6 °I. What would this temperature be on the Celsius scale?

Ans:32.3   


When you take a bath, how many kilograms of hot water (40.8 °C) must you mix with cold water (11.0 °C) so that the temperature of the bath is 31.5 °C? The total mass of water (hot plus cold) is 192 kg. Ignore any heat flow between the water and its external surroundings.

Ans:132kg

A woman finds the front windshield of her car covered with ice at -11.5 °C. The ice has a thickness of 2.18 x 10-4 m, and the windshield has an area of 1.32 m2. The density of ice is 917 kg/m3. How much heat in joules is required to melt the ice?

Ans:94500

A 13.3-kg block of ice has a temperature of -20.5 °C. The pressure is one atmosphere. The block absorbs 6.76 x 106 J of heat. What is the final temperature in degrees Celsius of the liquid water?

Ans: 31.6

[enwiabe]

1. Assuming a linear temperature scale, let C = aI + b

When C = 0, I = 24, and when C = 100, I = 156.

therefore, 24a + b = 0 (1)
and        156a + b = 100 (2)

Taking eqn 1,
24a = -b

it turns out that 156/24 = 6.5

so 6.5 * 24a = 6.5 * - b

156a = -6.5b

Going back to equation 2:
156a + b = 100
-6.5b + b = 100
-5.5b = 100
therefore b = -100/5.5 = -200/11 ~ -18.18

Subbing back into 24a = -b
a = -b/24 = -(-200/11)/24 = 25/33 ~ 0.76

So now we have our relationship!

C = 25*I/33 - 200/11

so when earth's temperature = 66.6 degrees = 333/5 degrees

333/5 = 25I/33 - 200/11

25I/33 = 333/5 + 200/11
          = (3663 + 1000)/55
          = 4663/55
I = 33*4663/(25*55)
  = 3*4663/(25*11)
  = 13989/275
  ~ 50.87 degrees Invader

[ed_saifa]

Thanks marsh. The ans is different though =(

[ed_saifa]

Please help with this question. thanks

Light in a vacuum is incident on a transparent glass slab. The angle of
incidence is 35.0°. The slab is then immersed in a pool of liquid. When the
angle of incidence for the light striking the slab is 20.3°, the angle of
refraction for the light entering the slab is the same as when the slab was in a
vacuum. What is the index of refraction of the liquid?

Ans: 1.65

A point source of light is submerged 2.2 m below the surface of a lake and
emits rays in all directions. On the surface of the lake, directly above the
source, the area illuminated is a circle. What is the maximum radius that this
circle could have?

Ans: 2.5m

[Mao]

1.

for slab in vacuum:



for slab in liquid:



, since the refractive index of vacuum is 1, the rest is trivial.


2.

finding the critical angle:

hence, the radius is (draw a picture of this)

[ed_saifa]

A row of seats is parallel to a stage at a distance of 11.1 m from it. At the center and front of the stage is a diffraction horn loudspeaker. This speaker sends out its sound through an opening that is like a small doorway with a width D of 6.89 cm. The speaker is playing a tone that has a frequency of 1.39 x 104 Hz. The speed of sound is 343 m/s. What is the separation (in terms of m) between two seats, located on opposite sides of the center of the row, at which the tone cannot be heard?

Ans: 8.515726717169m   

Cheers