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Hil
Questions (15)
Which equation can be used to find A, the area of the parallelogram shown?
A = 5 × 9 A = 4 × 9 A = 12 × 5 × 9 A = 12 × 4 ×
1 answer
614 views
How is a ratio table used to graph equivalent ratios?
1 answer
1,026 views
Helium balloons are used regularly in scientific research. A typical balloon would reach an altitude of 40.0 km with an air
1 answer
471 views
A closed container with a volume of 8000cm^3 is filled with Xenon gas. the temperature is 273K and the pressure is 2 atm. How
1 answer
641 views
A beam of electrons with momentum p=10^-23 kg m/s passed through a narrow slit of width d=10^-5 m. What is the uncertainty
1 answer
1,293 views
Rank the following three elements in order of increasing Young'sModulus: Ge, Sn, Si. My answer is as follows:
Sn<Ge<Si. I am not
1 answer
541 views
Is 7.27332*10^-11 meters/second the same as 0.07 nm per second. I believe it is after rounding off. Thanks for your help.
2 answers
417 views
a) Two pendulums consist of a very thin string at the end of which hangs a mass which is much larger than the mass of the rod.
1 answer
604 views
A stick of length l =70.0cm rests against the wall. The coefficient of static friction between stick and wall and between the
2 answers
612 views
Thanks to everyone who helped me out. I got through.
0 answers
317 views
Consider a rocket in space that ejects burned fuel at a speed of Vex = 1.5 km/s with respect to the rocket. The rocket burns 5%
13 answers
924 views
Hi, Greco. I am using the following formula to solve Ballistic Missile: sqrt(5*G*m_1/4*r_1). Given the following: m_1=5*10^24kg;
13 answers
640 views
MIT
A uniform disc of mass m and radius r is acted upon by three forces. find the magnitude of angular acceleration and the sense
3 answers
753 views
Greco I am having a problem with the turntable (gyroscope) problem. The angular speed is 300 radians per second. I tried a
2 answers
427 views
For this problem, assume the balls in the box
are numbered 1 through 8, and that an experiment consists of randomly selecting 3
0 answers
1,265 views
Answers (27)
Please Help Me guys!
No. 25 This is my answer: lowest density: cristobalite; next lowest: tridymite; next lowest: beta quartz; next lowest: alpha quartz; highest density: coesite. Please verify.
No.29 is correct. No. 18 can you check my math. ln(100)=H_v/8.314*(1/772.15-1/1109.15) ln(100)= H_v/8.314*0.000393 H_v= ln(100)*8.314/0.000393 H_v=ln(100)*21155.2163) H_v=97423.37 joules H_v=97.42337 kilo joules Thanks.
Thanks Tania. I was making calculation errors and unsure of what they were looking for on the gold problem. I am having he same problem with no 18. I know the formula but getting the wrong answer. Temp = 836 C and 499 C
On no. 25. I believe the highest density is Coesite. The others are tricky. Does anyone know. On 26, phases for Cu at 95wt.% I think is 2. At 15wt.% for Cu, I think is 2. For no. 29, I think linear structure has a larger molar volume. Branched structure a
Hi, KS and Kim. Can you help me with No. 18 electrical measurements. Metal is two orders of magnitude higher at 836 degrees than at 499 degrees C. What is the energy formation for this metal. Thanks. In(100)= H_v/8.314(1/772.15-1/1109.15)
Hi, KS and Kim. Can you help me with No. 18 electrical measurements. Metal is two orders of magnitude higher at 836 degrees than at 499 degrees C. What is the energy formation for this metal. Thanks.
Hi. KS, I have a few more to go. Only one chance left. The graphs, the one on gold, x-ray tube. If you can fill in what you have. 9b is correct.21. is b and both glasses are weaker. Thanks
9. b) I think the answer is "Both molecules are ionic but the higher melting point has a larger cohesive energy". I ruled out some of the choices, LiF has a strong lattice energy. There are a lot of choices to choose from. This is what I think.
28 is A and D
This is what I did with No. 12 Gallium Nitride. A) I said No. B) Wavelength = h*c/E_g = 6.626*10^-34 x 3*10^8/ 3.2 x 1.602 * 10^-19 = 0.388 *10^-6 This is a long shot. Maybe someone can pick up on it.
Hi KS, Thanks. 6b) 3.22*10^5
Hi, Menes. I an getting a very strange answer, which is incorrect. I must be using the wrong formula. Can you assist. Thanks.
a. for copper: 2.65; 1.6%; 0.014%; 0.0014% for invar: 0.00012%; 0.00072%; 0.38%; 0.58% This is what I have done so far: Change in L for copper= 17*10^-6*10^3*16= 0.272m=27.2cm Change in L for invar= 0.9*10^3*16= 0.0144m= 1.44cm b.For copper: 12 sec; 18sec;
Steve, thanks for your help.
Hi, koala. I am having trouble solving for u_s. So far, tan(theta) = 2*(u_s)/(1-u_s)*(u_s+1) u_s=?
Hi, Unknown. Please show formula for your answer on Venturi flow meter. I would like to compare with mine. Thanks.
Hi, Greco. To be honest Questions 1 and the ruler are by far the most difficult for me. when I started the test I knew I was in trouble. I saw this approach: Apply the conservation of energy U= m*g*h cm Ek = m*g* (L/2)+0 l=1/3*m*L Einitial= m*g*(L/2)+0
Hi Anonymous, I tried that problem and was unsuccessful. My answer was 8 radians. I knew it was wrong. I put that off to the side. I am just trying to get a passing grade. Thanks for your help. If I learn any thing I will let you know. Also Thanks Greco
Hi Anonymus, So my question is V= 76.939416 m/s and a=0.2747 m/s^2 correct. Is the other information just trying to through us off? Thanks.
I have tried to solve this problem the following way: 5% mass is burned in 280s. 1.5 km/s = 1500 m/s u=fuel speed m/s Solution: V= u*ln(1/(1-p)) 1500*ln(1/0.950) = 76.9399416 m/s a= V/t = 76.9399416/280 = 0.274785506 m/s^2 What is confusing to me what is
Hi Anon or Joe. Rocket Acceleration The following are the variables: Vex=1500m;rocket burns 5% of its mass in 280 seconds. speed v after 140 seconds? V= Vex*ln(1/(1-0.025)) V= 1500*ln(1/(1-0.025)) V= 1500*0.0253178 V=37.97612 m/s
Can some explain Part d of doppler shift. I used the following: f'= f*(1+speed of observer/speed of sound) f'=450*(1+20/340)=450*1.0588=476.46. This is my last chance. The wind blows toward the observer. Any ideas. Thanks.
Hi, Greco. Thanks I got the same answer on the second try.
Hi, Roma. I am having the same problem as you. If I find out I will let you know. I am still trying to solve the problem on the missile.
I am not sure about (a) I think it is us*d(theta). (b)T/R 1/e^(us*pi)-1 (c)T/R 1/1-e^(-us*pi) I am not 100% positive. Take a look maybe you feel the same way. The tension is opposite.
Thanks, Greco. I figured it out.
