A car with front wheel drive, a car with rear wheel drive, and a car with 4 wheel drive all line up on a starting line backwards. They take off on a full tanks of gas (50 litres) in reverse (with drivers). A man on a donkey, a man on a camel and a midget on an ostrich then begin travelling in the same direction the cars went.
After 24 hours which one will be the farthest away from the starting line?
Originally posted by RamnedA swimmer has .820 liters of dry air in his lungs when he dives in a lake. Assuming the atmospheric pressure at the surface is 1.013e5 Pa and that the pressure of the dry air is 95% of the external pressure at all times, find the volume of the dry air at a depth 10 m.
I will post a problem tomorrow - one with no numbers 😉
Originally posted by uzlessnot sure if this is the maximal area, but he could connect the 1+2 yard sides to make a straight piece, and then form a 3yd-3yd-4yd isoceles triangle... this has height of sqrt(5) and thus an area of 2*sqrt(5) square yards. is there a better solution?
A farmer has four straight pieces of fencing: 1, 2, 3, and 4 yards in length. What is the maximum area he can enclose by connecting the pieces? Assume the land is flat.
Originally posted by AetheraelThe area of a quadrilateral is maximized when it is a cyclic quadrilateral. The formula for the area of a cyclic quadrilateral is given by Brahmagupta's formula:
not sure if this is the maximal area, but he could connect the 1+2 yard sides to make a straight piece, and then form a 3yd-3yd-4yd isoceles triangle... this has height of sqrt(5) and thus an area of 2*sqrt(5) square yards. is there a better solution?
A = SQRT((s-a)(s-b)(s-c)(s-d))
where a,b,c and d are the side lengths and s is the semi-perimeter (a+b+c+d)/2. Given the side lengths 1,2,3 and 4, the semi-perimeter is 5. Therefore, the maximum area is:
A = SQRT((5-1)(5-2)(5-3)(5-4)) = SQRT(4*3*2*1) = SQRT(24) = 2*SQRT(6)
EDIT: Oops! My first answer given above SQRT(10) is what you get when you add the numbers up...d'oh! 😕
Originally posted by RamnedSorry uzless - theres the problem for you
A swimmer has .820 liters of dry air in his lungs when he dives in a lake. Assuming the atmospheric pressure at the surface is 1.013e5 Pa and that the pressure of the dry air is 95% of the external pressure at all times, find the volume of the dry air at a depth 10 m.
Originally posted by uzlessRamned, i'll answer yours if you answer mine...
A car with front wheel drive, a car with rear wheel drive, and a car with 4 wheel drive all line up on a starting line backwards. They take off on a full tanks of gas (50 litres) in reverse (with drivers). A man on a donkey, a man on a camel and a midget on an ostrich then begin travelling in the same direction the cars went.
After 24 hours which one will be the farthest away from the starting line?
Originally posted by uzless
A farmer has four straight pieces of fencing: 1, 2, 3, and 4 yards in length. What is the maximum area he can enclose by connecting the pieces? Assume the land is flat.
rectangle: 2.5^2=6.25 = 6.25 ...
the max area circumfered by 10 units of fencing is a circle with radius 5/pi
Area = pi*r^2= 25/pi = 7.96 square yds
Edit: Assuming err... he has a hammer and some pieces of rope.... what?
FLUID PROBLEMS
1. A solid platic cube .5m on each side floats in water. If it has a mass of 100 kg, how much of the cube is below the surface? (B) An object of volume 2e-3 m^3 and weight of 6.0 N is placed into water and floats. What percentage of the objects volume is above the surface of the water? (C) An inflated baloon is submerged to 3 different depths. At which depth is the buoyant force on it the greatest, 5 meters, 10 meters, or 15 meters?
2. A Helium filled balloon is tied to a string (mass of string is 0.050 kg, length is 2.00 m). The ballloon is speherical and has a radius of .40 m. When released it lifts a length of string h abd tgeb renains in equilibrium and with a portion of the string still resting on the ground. When deflated the balloon has a mass of .25 kg. Determine ethe value of h.
3.Water and then oil are poured into a U-shaped tube that is open at both ends. They come to an equilibirum such that on the left side of the U tube the oil rises 27.2 cm (sitting on top of water). On the right side, the water's maximum height is 9.41 cm less than the oil's height on the left side, and this water is pushing down through the center to the left side (so there is no oil in the center of the tube, only on the left side) Find the density of the oil.
4. A tank is open to the atmosphere and is filled to a depth D with a liquid of density P. A scale rests on the bottom of the tank and upon that scale sits a block of denstiy Pblock. The block is at a depth h below the liquids surface (measured to the top of the block). The block has a length c, height z, and width y. In the simplest form using these variables, (A) Find an equation for the force due to pressure on the top and bottom surfaces of the block. (B) What are the average forces due to pressure on the other 4 surfaces of the block? (C) Find an expression for the buoyant force. (D) What is the reading on the scale?
Those are the tougher problems I've gotten on fluids.
CAN ANYONE HELP ME PLS...
1. A 0.40 kg iron horseshoe that is initially at 500 degrees Celsius is dropped into a bucket containing 20 kg of water at 22 degrees Celsius. What is the final equilibrium temperature? Neglect any energy transfer to or from the surrounding.
2. An aluminum cup contains 225 g of water and a 40 g coffee stirrer, all at 27 degrees Celsius. A 400 g sample of silver at an initial temperature of 87 degrees Celsius is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature at 32 degrees Celsius. Calculate the mass of the aluminum cup.
3. A 50 g ice cube at 0 degrees Celsius is heated until 45 g has become water at 100 degrees Celsius and 5 g has become steam at 100 degrees Celsius. How much energy was added to accomplish this?
4. What mass of steam that is initially at 120 degrees Celsius is needed to warm 350 g of water and its 300 g aluminum container from 20 degrees Celsius to 50 degrees Celsius?
5. Steam at 100 degrees Celsius is added to ice at 0 degree Celsius.
a. Find the amount of ice melted and the final temperature when the mass of the steam is 10 g and the mass of ice is 50 g.
b. Repeat with steam of mass 1.0 g and ice of mass 50 g.
Originally posted by LingXThe first two questions assume that the thermal energy in the system is initially partitioned amongst all the objects, but that this energy will flow from one partition into another until all the objects in the system are in thermal equilibrium. The trick is that the temperature rise in each object as a result of this energy transfer is proportional to its heat capacity. The last three questions involve the same principle, with the added wrinkle that "extra" energy is stored or released during a phase change.
CAN ANYONE HELP ME PLS...
1. A 0.40 kg iron horseshoe that is initially at 500 degrees Celsius is dropped into a bucket containing 20 kg of water at 22 degrees Celsius. What is the final equilibrium temperature? Neglect any energy transfer to or from the surrounding.
2. An aluminum cup contains 225 g of water and a 40 g coffee stirrer, all at 27 degrees ...[text shortened]... m is 10 g and the mass of ice is 50 g.
b. Repeat with steam of mass 1.0 g and ice of mass 50 g.
1. A 0.40 kg iron horseshoe that is initially at 500 degrees Celsius is dropped into a bucket containing 20 kg of water at 22 degrees Celsius. What is the final equilibrium temperature? Neglect any energy transfer to or from the surrounding.
Let:
T_iron = initial temperature of the iron horseshoe (C)
m_iron = mass of the horseshoe (kg)
c_iron = heat capacity of the horseshoe (J/kg*K)
T_water = initial temperature of the water (C)
m_water = mass of the water (kg)
c_water = heat capacity of the water (J/kg*K)
T_eq = the final equilibrium temperature
The final equilibrium temperature of the horseshoe/water system will be somewhere between the two initial temperatures, so by inspection we see that energy will flow out of the hot iron horseshoe and into the cold water. The energy flow out of the horseshoe is:
Q_iron = m_iron * c_iron * (T_iron - T_eq)
The energy flow into the water is:
Q_water = m_water * c_water * (T_eq - T_water)
By the assumptions made in the question, no energy flows happen between the horseshoe/water system and the surroundings, and there are no phase changes, so the two energy flows above equate:
m_iron * c_iron * (T_iron - T_eq) = m_water * c_water * (T_eq - T_water)
To make things simpler, let k = (m_iron * c_iron)/(m_water * c_water). Now we have:
k * (T_iron - T_eq) = (T_eq - T_water)
Solving for T_eq we get:
T_eq = (k*T_iron + T_water)/(1+k)
Plugging the given numbers back in, along with c_iron = 450 J/kg*K and c_water = 4181 J/kg*K, we get a final temperature of:
T_eq = 23 C
Now that you've seen one example, try workout out the rest. The procedure is basically the same for the last three questions, but with the added wrinkle of a phase change. Phase changes are interesting in that energy will flow in or out of the system during a change but the temperature will stay the same. You can account for this energy flow by including a term for "latent heat" where Q_latent heat = k_latent heat of object * m_object.
Good luck!