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Okay, that's wrong, I only have one more chance to put in the correct answer, so how would I do this?

So would it be: \epsilon+(1.69x3.54)+(9.34x3.54)+(5.66x3.54)+(1.74x3.54)+11.88v=0 \epsilon=-77.44

Homework Statement What is the emf E of the battery at the lower left in the figure? Answer in units of V. http://img508.imageshack.us/img508/6/82852928.png [Broken] Can someone guide me? I'm confused as to where to start. I think I am supposed go the route of the loop, but I am...

The answer I had gotten (2.03e6) is wrong. Why?

Homework Statement This is a two part problem: 3- A conducting spherical shell of radius 19 cm carries a net charge of −8.15 μC uniformly distributed on its surface. Find the electric field at points just outside the shell. (Take the radially outward direction to be positive.) Answer in...

I have a similar problem here's my work but its wrong cause its not one of the answer choices: Problem: Consider the setup of the two-lens system shown in the figure, where the separation of the two lenses is denoted by d = 1.5 a. Their focal lengths are respectively f1 = a and f2 = 2 a. An...

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Thank you! My work: 3\frac{1}{f}-\frac{1}{-3f} = \frac{3-1}{3f} = \frac{4}{3}f (inverse) = \frac{3}{4}f

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Homework Statement This is a 2 part problem but I figure out the first part. Heres the 1st problem and the solution: 7- Given a convergent lens which has a focal point f. An object is placed at distance p = \frac{4}{3}f to the left of the lens. See the sketch. Solution: q1 = 4f, and is a...

Homework Statement This is a 2 part problem: 4- A person is to be fitted with bifocals. She can see clearly when the object is between 49 cm and 1.5 m from the eye. The upper portions of the bifocals should be designed to enable her to see distant objects clearly. What power should they...

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I figured it out ^thanks

So it would be: \theta3=180-(26+26+90)=38\circ n3sin\theta3=n4sin\theta4 1.69sin38=1.33sin\theta4 \frac{1.0405}{1.33}=sin\theta4 sin-10.7823=\theta4 51=\theta4 And #3 is right? TIA

*BUMP* Anybody?!

Anybody?!

Anybody?!

Homework Statement This is a two part problem: 11) The index of refraction for violet light in silica flint glass is 1.66, and that for red light is 1.62. What is the angle of deviation for the red ray passing through a prism of apex angle 59.6◦ if the angle of incidence is 54.8◦...

Homework Statement This is a two part problem: 2) As shown in the figure, a light ray is incident normally on one face of a 26◦–64◦–90◦ block of dense flint glass (a prism) that immersed in water. Find the exit angle θ4 of the light ray (Assume the index of grass is 1.69, and that of water...

mZnCZn(Tf-TZn i) + mH2OCH2O(Tf-TH2O i)=0 When I plug in my numbers: (10 kg)(388 J/kg C)(Tf-68 C) + (32 kg)(4180 J/kg C)(Tf- 14 C) = 0 3880 J/C (Tf-68 C) + 133760 J/C (Tf- 14 C) = 0 Than?!

Copper: Q=mc\DeltaT Q=(.186 kg)(387 J/kg C)(24 C -75 C) Q=-3671.1 J Aluminum: But I am not given Q, so I have to solve for 2 things: Q and m Q=mc\DeltaT Q=m(900 J/kg C)(24 C -4.6 C) Qm=(900 J/kg C)(19.4 C) Qm=17,460 J/kg Which doesn't make sense. So what do I do?

So is it 41 C?

Thank you!

Oh $h*t! 3.48 cm crap I didnt even realize it had asked in cm!

Homework Statement A 10 kg piece of zinc at 68◦C is placed in a container of water. The water had a mass of 32 kg and a temperature of 14◦C before the zinc was added. The specific heat of zinc and water are 388 J/kg ·◦C and 4180 J/kg ·◦C. What is the final temperature of water and zinc? Answer...

So: Q=mc\DeltaT Q=(7kg)(1000 cal/kg C)(100 C) Q= 700,000 cal heat of fusion: Q=mL Q=(7kg)(79500 cal/kg) Q= 556,500 cal heat of vaporization: Q=mL Q=(7kg)(5.4x105 cal/kg) Q=3,780,000 cal \sumQ= 700,000 + 556,500 + 3,780,000 \sumQ=5,036,500 cal = 5036.5kcal Is that how?!

^It came out to be wrong, so what did I do wrong?

Homework Statement I think this is more of a gen chem question, but this is for my physics lecture. A student drops two metallic objects into a 240 g steel container holding 62 g of water at 24◦C. One object is a 186 g cube of copper that is initially at 75◦C, and the other is a chunk of...

Thank you! So it would be: \DeltaL=Lo\alpha\DeltaT \DeltaL= (76 m)(1.2 x 10-5)(◦C)-1(20.5◦C+17.7◦C) \DeltaL=0.0348 m

Thank you gneill!

Homework Statement 7 kg of ice at 0◦C is converted to steam at 100◦C. How much heat is needed? The heat of fusion for water is 79500 cal/kg, its heat of vaporization is 5.4 × 105 cal/kg, and its specific heat is 1000 cal/kg ·◦C. Answer in units of kcal. Homework Equations What is the...

Homework Statement A bridge is made with segments of concrete 76 m long (at the original temperature). If the linear expansion coefficient is 1.2 × 10−5(◦C)−1, how much spacing is needed to allow for expansion for an increase in temperature of 69◦F? Answer in units of cm. Homework Equations...

Its correct! So what would be the usual units for specific gravity? TIA

Homework Statement A collapsible plastic bag contains a glucose solution. If the average gauge pressure in the vein is 15000 Pa, what must be the minimum height of the bag in order to infuse glucose into the vein? Assume that the specific gravity of the solution is 1.04. The acceleration of...

Thank you tiny-tim! :smile: It seems the second part was wrong, because I figured it had wanted me to 56-19=37, so it would've been from the bottoms up, but it was supposed to be 19 so: b) P2=P1+\rhogh P2-P1=\rhog P2-P1=(1000 kg/m3)(9.80 kg m /s2)(19 m) P2-P1= 186,200 Pa which...

So would it be: a) P2=P1+\rhogh P2-P1=\rhogh P2-P1=(1000 kg/m3)(9.80 kg m /s2)(56 m) P2-P1=548,800 Pa which would equal 548.8 kPa b) P2=P1+\rhogh P2-P1=\rhog P2-P1=(1000 kg/m3)(9.80 kg m /s2)(37 m) P2-P1= 362,600 Pa which would equal 362.6 kPa So is the work and...

^Are you assuming the reservoir is filled with water, so \rho would be 1000 kg/m3?

Homework Statement This is a two part problem: a) A reservoir behind a dam is 56 m deep. What is the gauge pressure at the base of the dam? The acceleration of gravity is 9.8 m/s 2 and atmospheric pressure is 10100 Pa . Answer in units of kPa. b) What is the gauge pressure 19 m from...

^It came out right with 929 Pa. The way the prof had shown it in class is P1-P0: P2=Po+\rhooilgh1 P1=P2+\rhoH2Ogh2 P1=Po+\rhogh1+\rhogh2 P1-Po=\rhogh1+\rhogh2

Thank you! :smile:

Homework Statement A test tube standing vertically in a test tube rack contains 3.8 cm of oil, whose density is 0.81 g/cm3 and 6.4 cm of water. What is the gauge pressure on the bottom of the tube? The acceleration of gravity is 9.8 m/s2. Answer in units of Pa Homework Equations...

Thank you! I put in 16,794 and it was right.

Thank you, oh I know the digits are long, I plan to answer it as: .0533 atm

Homework Statement This is for my physics homework and I haven't done gen chem in a while so I was wondering if this was right: You need to raise the temperature of 62 g of water from 8.2◦C to 73◦C. How much heat is needed to accomplish this? The specific heat of water is 4180 J/kg ·◦C. Answer...

Homework Statement This is for my physics homework and I haven't done gen chem in a while so I was wondering if this was right: One mole of an ideal gas at 0◦C is held at a constant volume of 94 liter. Find the change in pressure if the temperature increases by 61◦C. Answer in units of atm...

So it would be: \omega = \omega o + \alpha t \omega = \omega o + \alpha t than: \alpha = \Delta \omega / \Delta t

Homework Statement This is a two part problem, but I don't know what equation to use for part II. a) A record has an angular speed of 15.4 rev/min. What is its angular speed? Answer in units of rad/s. 1.61 rad/s b) Through what angle does it rotate in 1.1 s? Answer in units of rad...

I figured it out! Sorry for my messy work! :tongue:

Forget it, I realized my work is right :smile: