no vacancy Posted November 4, 2010 Share Posted November 4, 2010 Well if it's just one substance, it should just have 1 freezing point? Can you elaborate? Link to comment Share on other sites More sharing options...
Skks_King Posted November 4, 2010 Share Posted November 4, 2010 yeah, to me it sounded like it was for a lab so there could be various things they're asking for, either way i don't think there's enough info to answer 300 would technically be said to have one significant figure. zeros behind the last nonzero digit are not significant when there is no decimal point, unless otherwise stated. if there are zeros written after a decimal point those are significant (along with everything in front of the decimal point), but trailing zeros when there is not a decimal point aren't significant, they are just placeholders to indicate the order of magnitude of the figures that are significant. really 300 shouldn't ever be written, it should be written as 3 x 102 because then there is no ambiguity. EDIT: bah, i went to make coffee and now you already have it... Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted November 4, 2010 Share Posted November 4, 2010 lol screw the freezing point question, i have it second block ill just copy someone else. But for physics, i've got myself in some trublee. A tennis ball is dropped from 1.20m above the ground, it rebounds to a height of 1.00m. a)with what velocity does it hit the ground? b)with that velocity does it leave the ground? I have the answers, I just need to know what formulas to use, and how to get em. Answers: -4.85 m/s for a 4.43 m/s for b Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted November 4, 2010 Share Posted November 4, 2010 Anyone Link to comment Share on other sites More sharing options...
hr_cephei Posted November 4, 2010 Share Posted November 4, 2010 A tennis ball is dropped from 1.20m above the ground, it rebounds to a height of 1.00m. a)with what velocity does it hit the ground? b)with that velocity does it leave the ground? Link to comment Share on other sites More sharing options...
Minion Posted November 6, 2010 Share Posted November 6, 2010 I have two physics 11 questions . 1) When you drop a 400-g apple, Earth exerts a force on it that accelerates it at 9.8m/s2 towards Earth's surface. According to Newton's third law, the apple must exert an equal and opposite force on Earth. If the mass of Earth is 5.98 x 1024 kg, what's the magnitude of Earth's acceleration? 2) A 60-kg boy and a 40-kg girl use an elastic rope while engaged in a tug-of-war on an icy frictionless surface. If the acceleration of the girl toward the boy is 3.0 m/s2, determine the magnitude of the acceleration of the boy toward the girl. Link to comment Share on other sites More sharing options...
no vacancy Posted November 6, 2010 Share Posted November 6, 2010 I have two physics 11 questions . 1) When you drop a 400-g apple, Earth exerts a force on it that accelerates it at 9.8m/s2 towards Earth's surface. According to Newton's third law, the apple must exert an equal and opposite force on Earth. If the mass of Earth is 5.98 x 1024 kg, what's the magnitude of Earth's acceleration? 2) A 60-kg boy and a 40-kg girl use an elastic rope while engaged in a tug-of-war on an icy frictionless surface. If the acceleration of the girl toward the boy is 3.0 m/s2, determine the magnitude of the acceleration of the boy toward the girl. Link to comment Share on other sites More sharing options...
Minion Posted November 6, 2010 Share Posted November 6, 2010 OH! Thank you no vacancy Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted November 19, 2010 Share Posted November 19, 2010 Physics 11 question. Need help please: An archer stands 40.0m from the target. If the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bulls-eye must he aim to compensate for gravity pulling his arrow downward? je ne comprends pas Link to comment Share on other sites More sharing options...
The Situation Posted November 19, 2010 Share Posted November 19, 2010 Physics 11 question. Need help please: An archer stands 40.0m from the target. If the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bulls-eye must he aim to compensate for gravity pulling his arrow downward? je ne comprends pas Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted November 19, 2010 Share Posted November 19, 2010 Well the time it takes for the arrow to get there is determined from the horizontal velocity and distance. 40.0m/90.0m/s=.444s There was some formula of d=vit + 1/2at^2 vit=0 so d=1/2(9.8m/s^2)(.444s)^2=something He has to be that distance above because gravity will pull the arrow down. He fires horizontally so don't worry about initial velocity. Link to comment Share on other sites More sharing options...
GoaltenderInterference Posted November 21, 2010 Share Posted November 21, 2010 Could someone please explain how to do this question? 3. The demand function for a company's product is p=60 000(x + 3000)-1, with demand 3000 units weekly. a) Compute the elasticity of demand at the current demand. Should the company raise or lower the price if it wants to increase revenue? c) What price will maximize revenues? What will be the demand at this price? Link to comment Share on other sites More sharing options...
no vacancy Posted November 21, 2010 Share Posted November 21, 2010 Could someone please explain how to do this question? 3. The demand function for a company's product is p=60 000(x + 3000)-1, with demand 3000 units weekly. a) Compute the elasticity of demand at the current demand. Should the company raise or lower the price if it wants to increase revenue? c) What price will maximize revenues? What will be the demand at this price? Link to comment Share on other sites More sharing options...
GoaltenderInterference Posted November 22, 2010 Share Posted November 22, 2010 i don't know a lot about economics but i think i can take a shot at this. elasticity is defined as Ed = (p/x)*(dx/dp), where p is price and x is demand. so solve the equation you have above for x in terms of p and take the derivative: Ed = (p/x)*(dx/dp) = (p/x) * d/dp[ 60000/p -3000 ] = (p/x) * (-60000/p2) = -60000/(px) = -(x + 3000)/x now, evaluate for x = 3000 and you have Ed = -2. as expected it's negative since demand goes down as price goes up, and it's magnitude is larger than 1 which means that fractional change in demand is larger than the fractional change in price when a change in price is made. hence an increase in price leads to a decrease in revenue, so the company should decrease the price to increase revenue. now, revenue r = x*p. so that gives r = 60 000x*(x + 3000)-1 now, this doesn't have a maximum for any value of x. it starts approaches 60000 asymptotically ( x/(x+3000) is always less than 1). so i suppose for part © the answer would be that no such price exists? or is there something i'm missing? Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted November 29, 2010 Share Posted November 29, 2010 I have a math unit test tomrw, and need some help guys! Thanks Solve for f(f(x)): a ) f(x)= (1)/(1-x) where x cannot equal 1 b ) f(x)= (x-1)/(x+1) where x cannot equal -1 The answers are: a ) (x-1)/x b )-(1/x) Link to comment Share on other sites More sharing options...
no vacancy Posted November 29, 2010 Share Posted November 29, 2010 I have a math unit test tomrw, and need some help guys! Thanks Solve for f(f(x)): a ) f(x)= (1)/(1-x) where x cannot equal 1 b ) f(x)= (x-1)/(x+1) where x cannot equal -1 The answers are: a ) (x-1)/x b )-(1/x) Link to comment Share on other sites More sharing options...
Conservative Posted December 17, 2010 Share Posted December 17, 2010 What is (OH-) in pure water if the ph is 7.4 Link to comment Share on other sites More sharing options...
-Goose- Posted December 17, 2010 Share Posted December 17, 2010 What is (OH-) in pure water if the ph is 7.4 Link to comment Share on other sites More sharing options...
Minion Posted January 12, 2011 Share Posted January 12, 2011 A motor having an efficiency of 88% operates a crane having an efficiency of 42%. With what constant speed does the crane lift a 410kg crate of machine parts if the power supplied to the motor is 5.5kW? So I figured out that the output energy of the motor is the input of the crane. Also that I have to somehow use P=W/t except I don't have any time or displacement? Help? Link to comment Share on other sites More sharing options...
Denguin Posted January 30, 2011 Share Posted January 30, 2011 Chemistry help much needed! A sample of sodium hydroxide, NaOH, is known to have become contaminated with sodium carbonate, Na2CO3, by reaction with CO2 in the air. A 1.00 g sample is titrated with 0.500M HCl and it is found that 48.6mL of HCl are required for neutralization. Calculate the percentage by mass of Na2CO3 in the sample. ( Hint: Let mass of Na2CO3 = x g. ) So far I have: HCl + NaOH --> NaCl + H2O 2 NaOH + CO2 --> Na2CO3 + H2O 0.500M HCl x 0.0486 L HCl = 0.0234 mol HCl. Since mol HCl = mol NaOH, mol NaOH = 0.0234 mol. 0.0234 mol NaOH x (40.0g/1 mol) = 0.972g From here, is it simply just 1.00g - 0.972g? Or do I need to do a little more work with the stoichiometric ratios to achieve an answer? Also, with regards to the "hint" the question gave, where does that fit in to the question? Thanks in advance! Link to comment Share on other sites More sharing options...
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