A car is driven 200 km west and then 80 km southwest. What is the displacement of the car from the point of o
Q. I don't know how to do this . . . there are no given angles that I can plug into some equation, how do I solve this? a car is driven 200 km west and then 80 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
Asked by Michael - Sat Feb 9 03:13:53 2008 - - 5 Answers - 0 Comments
A. You must draw a sketch to be able to do this problem. You also need to be familiar with the angles on a compass. This will result in you having a triangle : I can try and describe what it will look like: The car travels due west for 200km. On your paper draw a horizontal line from point A to point B which is to the left of A. Mark off some distance which is to represent 200km. Now draw a second line to represent the 80km south west of line AB. This will be at an angle of 135 to the original line. Mark this point C. Now the displacement is represented by the line CA. Calculate this length by cos rule b = a + c -2ac Cos B b = 80 +200 -2*80*200*Cos 135 b = 6400 + 40 000 - 22627.4 b = 69 027.4 b = 262.73m What is the direction?Calculate [cont.]
Answered by Trevor H - Sat Feb 9 06:11:57 2008
Q. I don't know how to do this . . . there are no given angles that I can plug into some equation, how do I solve this? a car is driven 200 km west and then 80 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
Asked by Michael - Sat Feb 9 03:13:53 2008 - - 5 Answers - 0 Comments
A. You must draw a sketch to be able to do this problem. You also need to be familiar with the angles on a compass. This will result in you having a triangle : I can try and describe what it will look like: The car travels due west for 200km. On your paper draw a horizontal line from point A to point B which is to the left of A. Mark off some distance which is to represent 200km. Now draw a second line to represent the 80km south west of line AB. This will be at an angle of 135 to the original line. Mark this point C. Now the displacement is represented by the line CA. Calculate this length by cos rule b = a + c -2ac Cos B b = 80 +200 -2*80*200*Cos 135 b = 6400 + 40 000 - 22627.4 b = 69 027.4 b = 262.73m What is the direction?Calculate [cont.]
Answered by Trevor H - Sat Feb 9 06:11:57 2008
intial velocity: 2.o m/s constant acceleration: 3.0 m/s sq. whats the magnitude of displacement after 4 sec?
Q. intial velocity: 2.o m/s constant acceleration: 3.0 m/s sq. whats the magnitude of displacement after 4 sec?
Asked by unknown - Wed Jun 24 21:44:23 2009 - - 2 Answers - 0 Comments
A. D=1/2at^2+v0t D=1/2(3)(4)^2+2(4)= D=24+8=32meters.
Answered by nozar nazari - Wed Jun 24 22:17:57 2009
Q. intial velocity: 2.o m/s constant acceleration: 3.0 m/s sq. whats the magnitude of displacement after 4 sec?
Asked by unknown - Wed Jun 24 21:44:23 2009 - - 2 Answers - 0 Comments
A. D=1/2at^2+v0t D=1/2(3)(4)^2+2(4)= D=24+8=32meters.
Answered by nozar nazari - Wed Jun 24 22:17:57 2009
I just had a knee displacement..liek the new cap moved out of its spot and back...HELP O_O??
Q. well i am a guy..did a split all through out the day, then at the end of the day, the leg that goes back popped out of its place with the knee cap and moved back..i almost fainted trying to walk and needed assistance. So its been a few hours..but my mom who is a nurse said I shouldn't go to school..this is not a option, i have to go! ] she said there is fluid flowing around there and the knee cap doesnt look like its in place.. so my question is: what will happen if i have to attend school for the next 2 days which will force me to move around campus..
Asked by unknown - Tue May 27 20:29:48 2008 - - 3 Answers - 0 Comments
A. Listen to good advice & rest your injury so it can heal. U can wear compressive thermal supports to improve circulation for improved healing & relieve pain.
Answered by Jamie - Tue May 27 20:51:30 2008
Q. well i am a guy..did a split all through out the day, then at the end of the day, the leg that goes back popped out of its place with the knee cap and moved back..i almost fainted trying to walk and needed assistance. So its been a few hours..but my mom who is a nurse said I shouldn't go to school..this is not a option, i have to go! ] she said there is fluid flowing around there and the knee cap doesnt look like its in place.. so my question is: what will happen if i have to attend school for the next 2 days which will force me to move around campus..
Asked by unknown - Tue May 27 20:29:48 2008 - - 3 Answers - 0 Comments
A. Listen to good advice & rest your injury so it can heal. U can wear compressive thermal supports to improve circulation for improved healing & relieve pain.
Answered by Jamie - Tue May 27 20:51:30 2008
A particle is moving in a straight line so that ts after leaving fixed point o, its velocity is v=pt^2+qt+4?
Q. p and q are constants, when t=1 the acceleration of particle is 8m/s. When t=2 the displacement of particle from o is 22m. What is the value of p and q
Asked by unknown - Mon Apr 20 16:05:48 2009 - - 1 Answers - 0 Comments
Q. p and q are constants, when t=1 the acceleration of particle is 8m/s. When t=2 the displacement of particle from o is 22m. What is the value of p and q
Asked by unknown - Mon Apr 20 16:05:48 2009 - - 1 Answers - 0 Comments
Velocity and displacement?
Q. A particle P moves in a straight line so that its displacement, S metres, from a fixed point O is given by S = 4 + 15t - t^3 where t is the time in seconds after passing through a point X on the line. Calculate the distance moved by the particle during the third second of its motion.
Asked by Bookworm 101 - Thu Aug 14 07:58:33 2008 - - 4 Answers - 0 Comments
A. 4 meters The work: After 2 seconds S(2) = 4 + 15*2 - 2^3 = 26 After 3 seconds S(3) = 4 + 15*3 - 3^3 = 22 Since the motion is in a straight line distance = |S(3) - S(2)| = |22 -26| = 4
Answered by m_zyoot - Thu Aug 14 08:03:42 2008
Q. A particle P moves in a straight line so that its displacement, S metres, from a fixed point O is given by S = 4 + 15t - t^3 where t is the time in seconds after passing through a point X on the line. Calculate the distance moved by the particle during the third second of its motion.
Asked by Bookworm 101 - Thu Aug 14 07:58:33 2008 - - 4 Answers - 0 Comments
A. 4 meters The work: After 2 seconds S(2) = 4 + 15*2 - 2^3 = 26 After 3 seconds S(3) = 4 + 15*3 - 3^3 = 22 Since the motion is in a straight line distance = |S(3) - S(2)| = |22 -26| = 4
Answered by m_zyoot - Thu Aug 14 08:03:42 2008
Acceleration when only given displacement and time?
Q. a particle p moves in a straight line through a fixed point o with a constant accleration of A ms^2. 3 seconds after passing through O, P has travelled a further 33m in the same direction. Calculate the value of A
Asked by Jezza - Tue Feb 6 18:45:20 2007 - - 1 Answers - 0 Comments
A. You cannot unless you know the initial velocity V when at O. You would use the equation D = 1/2 A t^2 + Vt 33 = 9/2 A + 3V A = (33-3V)2/9 If V = 0, then A = 66/9 = 7.3 m/s^2
Answered by catarthur - Wed Feb 7 13:37:53 2007
Q. a particle p moves in a straight line through a fixed point o with a constant accleration of A ms^2. 3 seconds after passing through O, P has travelled a further 33m in the same direction. Calculate the value of A
Asked by Jezza - Tue Feb 6 18:45:20 2007 - - 1 Answers - 0 Comments
A. You cannot unless you know the initial velocity V when at O. You would use the equation D = 1/2 A t^2 + Vt 33 = 9/2 A + 3V A = (33-3V)2/9 If V = 0, then A = 66/9 = 7.3 m/s^2
Answered by catarthur - Wed Feb 7 13:37:53 2007
what did you think of the physics regents june 2008?
Q. i thought it was O.K. Some of the questions were a little tricky. What did you put for the displacement in the writing part? and what was the one where you had a 10N block on a 30 degree incline and you had to find the force?
Asked by Starfish - Wed Jun 25 11:13:10 2008 - - 23 Answers - 0 Comments
A. I thought the test was pretty good and straightforward, pretty much know the formula, plug it in. There were, of course, a few that you had to watch, like the ones you mentioned. Amanda's right; the displacement is 10m, and the force is 8N. To go into it a little further, you need to find the component of the weight that's going to pull on the block in the direction of the force of friction. Since mg is 10N, and the angle of incline is 30, you do 10sin30 (although you normally use cosine for the x value, this is a different case), and get 5N. Add that to 3N, and there's your answer of 8N. Another question that's sure to have given people trouble is the multiple choice question that asked which type of force is seen between a proton and… [cont.]
Answered by Jaclyn W - Wed Jun 25 12:48:36 2008
Q. i thought it was O.K. Some of the questions were a little tricky. What did you put for the displacement in the writing part? and what was the one where you had a 10N block on a 30 degree incline and you had to find the force?
Asked by Starfish - Wed Jun 25 11:13:10 2008 - - 23 Answers - 0 Comments
A. I thought the test was pretty good and straightforward, pretty much know the formula, plug it in. There were, of course, a few that you had to watch, like the ones you mentioned. Amanda's right; the displacement is 10m, and the force is 8N. To go into it a little further, you need to find the component of the weight that's going to pull on the block in the direction of the force of friction. Since mg is 10N, and the angle of incline is 30, you do 10sin30 (although you normally use cosine for the x value, this is a different case), and get 5N. Add that to 3N, and there's your answer of 8N. Another question that's sure to have given people trouble is the multiple choice question that asked which type of force is seen between a proton and… [cont.]
Answered by Jaclyn W - Wed Jun 25 12:48:36 2008
Physics help with vector addition and vector subtraction?
Q. Well were doing diagrams that we must use either vector addition or vector subtraction and we try t o solve the resultant displacement. Well heres the question : A planes velocity changes from 200 km/h[N] to 300 km/h[N30W]. Find the change in Velocity. delta V ? how would i do this can anyone help me.
Asked by Bob I - Wed Oct 1 17:19:27 2008 - - 1 Answers - 0 Comments
A. first of all you need to split your vectors according to their direction...lets assume east and north are positive and west and south are negative. Initial velocity in the east/west direction=0km/h For the final velocity we have use trigonometry since we don't have a straight forward direction. Final velocity in the E/W= -300*cos60 degrees(West is negative therefore -300 and we get 60 degrees by subtracting 30 from 90,u always have to do this step if ur direction starts with N or S) Velocity0km/h or 150 km/h{W} Therefore total initial velocity= {0+150}=150 km/h{w} Initial velocity in the north south direction=+200km/h{n}(bec ause we assumed north is positive) Final velocity in the N/S direction=300sin60 =260 km/h {N} Total final… [cont.]
Answered by nandy a - Wed Oct 1 17:37:01 2008
Q. Well were doing diagrams that we must use either vector addition or vector subtraction and we try t o solve the resultant displacement. Well heres the question : A planes velocity changes from 200 km/h[N] to 300 km/h[N30W]. Find the change in Velocity. delta V ? how would i do this can anyone help me.
Asked by Bob I - Wed Oct 1 17:19:27 2008 - - 1 Answers - 0 Comments
A. first of all you need to split your vectors according to their direction...lets assume east and north are positive and west and south are negative. Initial velocity in the east/west direction=0km/h For the final velocity we have use trigonometry since we don't have a straight forward direction. Final velocity in the E/W= -300*cos60 degrees(West is negative therefore -300 and we get 60 degrees by subtracting 30 from 90,u always have to do this step if ur direction starts with N or S) Velocity0km/h or 150 km/h{W} Therefore total initial velocity= {0+150}=150 km/h{w} Initial velocity in the north south direction=+200km/h{n}(bec ause we assumed north is positive) Final velocity in the N/S direction=300sin60 =260 km/h {N} Total final… [cont.]
Answered by nandy a - Wed Oct 1 17:37:01 2008
Could you give me an two examples of each type of chemical reaction by writing their balanced equations?
Q. Chemical reactions like Direct Combination, Decomposition, Displacement or Substitution, Double Decomposition. Please! I really need two examples of each so that I'll know how to identify each. Show em with the uh...equation thingy like.. like this... H+O ---> HO
Asked by Cerisse Sinclair - Sun Apr 20 06:14:28 2008 - - 1 Answers - 0 Comments
Q. Chemical reactions like Direct Combination, Decomposition, Displacement or Substitution, Double Decomposition. Please! I really need two examples of each so that I'll know how to identify each. Show em with the uh...equation thingy like.. like this... H+O ---> HO
Asked by Cerisse Sinclair - Sun Apr 20 06:14:28 2008 - - 1 Answers - 0 Comments
Help with a mechanics maths question please?
Q. A particle moves on the x-axis. Its displacement, x m, from the origin O is given by: x= 3t^2 - 3t + 2, where t is time in seconds How far is the particle from the origin when it is instantaneously at rest? Many thanks for a quick answer.
Asked by The Tourist - Mon Apr 20 14:56:18 2009 - - 2 Answers - 0 Comments
A. dx/dt=6t-3 when grad =0 partical is stationary, 0=6t-3 6t=3 t=0.5 when t=0.5 x=3(0.5)^2 -3(0.5) +2 x=3(0.25)-3(0.5) +2 x=0.75-1.5+2 x=1.25m I think that is right Hope this helps
Answered by Edwin w - Mon Apr 20 15:02:06 2009
Q. A particle moves on the x-axis. Its displacement, x m, from the origin O is given by: x= 3t^2 - 3t + 2, where t is time in seconds How far is the particle from the origin when it is instantaneously at rest? Many thanks for a quick answer.
Asked by The Tourist - Mon Apr 20 14:56:18 2009 - - 2 Answers - 0 Comments
A. dx/dt=6t-3 when grad =0 partical is stationary, 0=6t-3 6t=3 t=0.5 when t=0.5 x=3(0.5)^2 -3(0.5) +2 x=3(0.25)-3(0.5) +2 x=0.75-1.5+2 x=1.25m I think that is right Hope this helps
Answered by Edwin w - Mon Apr 20 15:02:06 2009
Stress, Displacement, and Stiffness of Rod?
Q. This is a question from my text book that is giving me a run for my money and I need some help solving it. Thank you. Question:
Asked by uoitcheater - Wed Nov 4 01:17:18 2009 - - 1 Answers - 0 Comments
A. I can give you a hint only. 1. Use Integration by getting 1m strip from the displacement diagram and use the Flexural Moment Equation 2. Since you have the limit of x to be 18 inches as displacement. You can reverse the equation you have established in item 1. This are easy for college students who do strength of materials and calculus but i can only recall the principle. I hope this helps
Answered by orAnGE jULiuS - Wed Nov 4 04:55:11 2009
Q. This is a question from my text book that is giving me a run for my money and I need some help solving it. Thank you. Question:
Asked by uoitcheater - Wed Nov 4 01:17:18 2009 - - 1 Answers - 0 Comments
A. I can give you a hint only. 1. Use Integration by getting 1m strip from the displacement diagram and use the Flexural Moment Equation 2. Since you have the limit of x to be 18 inches as displacement. You can reverse the equation you have established in item 1. This are easy for college students who do strength of materials and calculus but i can only recall the principle. I hope this helps
Answered by orAnGE jULiuS - Wed Nov 4 04:55:11 2009
Classify the following reactions as single or double displacement, decomposition, synthesis, or combustion?
Q. a. 4Na(cr) + O2(g) 2Na2(subscript)O(cr) b. Pb(NO3)2(aq) + Na2CrO4(aq) 2NaNO3(aq) + PbCrO4(cr) c.NbI3(cr) + I2(cr) NbI5(cr) d. 2Li(cr) + 2H2O(l) 2LiOH(aq) + H2(g) e. 2C7H14(l) +21O2(g) 14CO2(g) + 14H20(g)
Asked by San Fran Kid - Mon Jan 22 19:13:56 2007 - - 2 Answers - 0 Comments
A. a. (X + Y - XY) it combines (or synthesises) so it's ___ b. (XY + AB - XB + AY) their are two (DOUBLE) compounds displaced their elements so it's ___ c. (X + Y - XY) It combines (or synthesises) so it's ___ d. (X + AB - XB + Y) a SINGLE element displaced in compound released a single element so it's ___ e. (a hydrocarbon + oxygen - always yields carbon dioxide and water) KaBOOM!!!
Answered by cdixon - Mon Jan 22 19:30:02 2007
Q. a. 4Na(cr) + O2(g) 2Na2(subscript)O(cr) b. Pb(NO3)2(aq) + Na2CrO4(aq) 2NaNO3(aq) + PbCrO4(cr) c.NbI3(cr) + I2(cr) NbI5(cr) d. 2Li(cr) + 2H2O(l) 2LiOH(aq) + H2(g) e. 2C7H14(l) +21O2(g) 14CO2(g) + 14H20(g)
Asked by San Fran Kid - Mon Jan 22 19:13:56 2007 - - 2 Answers - 0 Comments
A. a. (X + Y - XY) it combines (or synthesises) so it's ___ b. (XY + AB - XB + AY) their are two (DOUBLE) compounds displaced their elements so it's ___ c. (X + Y - XY) It combines (or synthesises) so it's ___ d. (X + AB - XB + Y) a SINGLE element displaced in compound released a single element so it's ___ e. (a hydrocarbon + oxygen - always yields carbon dioxide and water) KaBOOM!!!
Answered by cdixon - Mon Jan 22 19:30:02 2007
Please help with this physic problem. Direction and displacement?
Q. A scout troop is practicing its orienting skills with map and compass. First they walk due east for 1.1 km... A scout troop is practicing its orienting skills with map and compass. First they walk due east for 1.1 km. Next, they walk 45 west of north for 2.1 km. In what direction must they walk to go directly back to their starting point? How far will they have to walk? Ok for what I did: a: x=1.1, y=0; b:-1.5, 1.5 (using sin(45)=O/2.1) a+b= squareroot of 0.4^2 + 1.5^2 = 1.6 That's how far they have to walk. Angle = sin^-1=1.1/2.1 = 32degree southeast What did I do wrong?
Asked by FalennixonX - Mon Feb 11 23:22:33 2008 - - 1 Answers - 0 Comments
A. use components its easier
Answered by d g - Mon Feb 11 23:32:12 2008
Q. A scout troop is practicing its orienting skills with map and compass. First they walk due east for 1.1 km... A scout troop is practicing its orienting skills with map and compass. First they walk due east for 1.1 km. Next, they walk 45 west of north for 2.1 km. In what direction must they walk to go directly back to their starting point? How far will they have to walk? Ok for what I did: a: x=1.1, y=0; b:-1.5, 1.5 (using sin(45)=O/2.1) a+b= squareroot of 0.4^2 + 1.5^2 = 1.6 That's how far they have to walk. Angle = sin^-1=1.1/2.1 = 32degree southeast What did I do wrong?
Asked by FalennixonX - Mon Feb 11 23:22:33 2008 - - 1 Answers - 0 Comments
A. use components its easier
Answered by d g - Mon Feb 11 23:32:12 2008
how to find the magnitude and direction of the displacement vector OB when X=7,Y=6?
Q. In a desert exercise a tank travels x km on a bearing of 050degree from a base O then y km on a bearing of 140degree to a bunker B. please show me how to draw the diagram as well.
Asked by Isuru s - Fri Apr 6 08:59:15 2007 - - 4 Answers - 0 Comments
A. Bearings start with 0 at north and go clockwise around the circle. First find the magnitude. When the tank changed bearing from 050 to 140 it made a turn of (140 - 50) = 90 . So we have a right triangle. OB is the hypotenuse and x and y are the legs. Let d = distance OB d = (x + y ) = (7 + 6 ) = (49 + 36) = 85 h = horizontal distance v = vertical distance h = 7sin50 + 6sin140 9.2190368 v = 7cos50 + 6cos140 -0.0967533 Bearing = 90 - arctan(v/h) 90 - arctan(-0.0104949) Bearing = 90 + 0.6012946 = 90.6012946
Answered by Northstar - Sat Apr 7 20:51:13 2007
Q. In a desert exercise a tank travels x km on a bearing of 050degree from a base O then y km on a bearing of 140degree to a bunker B. please show me how to draw the diagram as well.
Asked by Isuru s - Fri Apr 6 08:59:15 2007 - - 4 Answers - 0 Comments
A. Bearings start with 0 at north and go clockwise around the circle. First find the magnitude. When the tank changed bearing from 050 to 140 it made a turn of (140 - 50) = 90 . So we have a right triangle. OB is the hypotenuse and x and y are the legs. Let d = distance OB d = (x + y ) = (7 + 6 ) = (49 + 36) = 85 h = horizontal distance v = vertical distance h = 7sin50 + 6sin140 9.2190368 v = 7cos50 + 6cos140 -0.0967533 Bearing = 90 - arctan(v/h) 90 - arctan(-0.0104949) Bearing = 90 + 0.6012946 = 90.6012946
Answered by Northstar - Sat Apr 7 20:51:13 2007
Calcium phosphide reacts w/water to give 701mL of PH3 collected by water displacement at 20 C and 745.0 mm Hg.?
Q. The vapor pressure of water is 17.5 mm Hg. What mass of Ca3P2 reacted? (Atomic weights: Ca = 40.08, P = 30.97, H = 1.008, O = 16.00). Ca3P2(s) + 6 H2O(l) --> 3 Ca(OH)2(s) + 2 PH3(g)
Asked by Caitlin C - Wed Mar 25 21:47:41 2009 - - 1 Answers - 0 Comments
A. pressure PH3 = 745 - 17.5 = 727.5 mm Hg => 0.957 atm moles PH3 = 0.957 x 0.701 L/ 0.08206 x 293 =0.0279 moles Ca3P2 = 0.0279/2=0.0140 mass Ca2P2 = 0.0140 mol x 182.174 g/mol=2.55 g
Answered by Dr.A - Thu Mar 26 07:35:24 2009
Q. The vapor pressure of water is 17.5 mm Hg. What mass of Ca3P2 reacted? (Atomic weights: Ca = 40.08, P = 30.97, H = 1.008, O = 16.00). Ca3P2(s) + 6 H2O(l) --> 3 Ca(OH)2(s) + 2 PH3(g)
Asked by Caitlin C - Wed Mar 25 21:47:41 2009 - - 1 Answers - 0 Comments
A. pressure PH3 = 745 - 17.5 = 727.5 mm Hg => 0.957 atm moles PH3 = 0.957 x 0.701 L/ 0.08206 x 293 =0.0279 moles Ca3P2 = 0.0279/2=0.0140 mass Ca2P2 = 0.0140 mol x 182.174 g/mol=2.55 g
Answered by Dr.A - Thu Mar 26 07:35:24 2009
A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 z?
Q. a mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm. If a timer is started when its displacement is a maximum (hence x = 4 cm when t =o), what is the speed of the mass when t= 3s?
Asked by Kelly C - Mon Feb 12 12:08:23 2007 - - 2 Answers - 0 Comments
A. t=1/f=1/4 d=4cm v=d/t> v=4*1/4> 1m/s I think it is constant, 1 m/s, It comes and goes but even if the resultant displacement is 0 the speed is 1 m/s
Answered by ( ) The Cat - Mon Feb 12 14:55:14 2007
Q. a mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm. If a timer is started when its displacement is a maximum (hence x = 4 cm when t =o), what is the speed of the mass when t= 3s?
Asked by Kelly C - Mon Feb 12 12:08:23 2007 - - 2 Answers - 0 Comments
A. t=1/f=1/4 d=4cm v=d/t> v=4*1/4> 1m/s I think it is constant, 1 m/s, It comes and goes but even if the resultant displacement is 0 the speed is 1 m/s
Answered by ( ) The Cat - Mon Feb 12 14:55:14 2007
please help me with this chemistry please i really need help?
Q. 69.Which class of elements will contain both solids and gases at room temperature? metals nonmetals metalloids inner transition metals 9. Based on its type of chemical bond, which of the following has the highest boiling point? isopropyl alcohol chloromethane sodium chloride oxygen 10. How many electrons are depicted in the electron dot diagram of an electrically neutral oxygen atom? two six eight five 11. What feature of a chemical equation represents the law of conservation of matter? chemical formulas balanced equation subscripts coefficients 52.Which of the following is a mixture of a gas dissolved in a liquid? soda pop liquid oxygen liquid hydrogen … [cont.]
Asked by little dragon - Wed Jan 28 11:19:14 2009 - - 1 Answers - 0 Comments
A. 69) nonmetals 9) I think it is chloromethane 10) 6 for sure 11) I think it is coefficient 52) soda pop 18) Ar I am sure of it 19) synthesis 20) 20 protons, 20 electrons, 24 neutrons.
Answered by unknown - Wed Jan 28 11:28:19 2009
Q. 69.Which class of elements will contain both solids and gases at room temperature? metals nonmetals metalloids inner transition metals 9. Based on its type of chemical bond, which of the following has the highest boiling point? isopropyl alcohol chloromethane sodium chloride oxygen 10. How many electrons are depicted in the electron dot diagram of an electrically neutral oxygen atom? two six eight five 11. What feature of a chemical equation represents the law of conservation of matter? chemical formulas balanced equation subscripts coefficients 52.Which of the following is a mixture of a gas dissolved in a liquid? soda pop liquid oxygen liquid hydrogen … [cont.]
Asked by little dragon - Wed Jan 28 11:19:14 2009 - - 1 Answers - 0 Comments
A. 69) nonmetals 9) I think it is chloromethane 10) 6 for sure 11) I think it is coefficient 52) soda pop 18) Ar I am sure of it 19) synthesis 20) 20 protons, 20 electrons, 24 neutrons.
Answered by unknown - Wed Jan 28 11:28:19 2009
Types of Chemical Reactions?
Q. Science Homework... My homework asks for 3 types of reactions that are important to life. Then my teacher wants me to describe reaction and show how it's important to life. So the only types of chemical reactions I know are synthesis reaction, decomposition reaction, single-displacement reaction, and double-replacement reaction. I know what they do, but the only reactions I find useful for everyday life are synthesis reaction and decomposition reaction. How are single-replacement reaction and double-replacement reactions important to everyday life? O yea, what breaks down into glucose? Is it sucrose or just food? Thanks!
Asked by DL It's Me! - Sun Dec 2 13:43:25 2007 - - 2 Answers - 0 Comments
A. I would suggest: Photosynthesis reaction is a good start Chlorophyll +CO2 makes O2 Then the other one Hemoglobin exchanging CO2 against O2 and the other way around. in the lungs O2 binds is then transported to the cells where it is used and replaced by CO2. Both these are Lewis acid base reactions. The last one I would suggest is a basic oxidation CH4+ 2O2 -> 2H2O + CO2 + energy This is important to everyday life as this is the basis for heating our houses. Ion reactions are important for your body too as you have a finely tuned balance if ions inside and outside your cells, but I haven't got a good example handy. [Ah, lastly the starch in your food is made into glucose.]
Answered by klimbim - Sun Dec 2 13:58:22 2007
Q. Science Homework... My homework asks for 3 types of reactions that are important to life. Then my teacher wants me to describe reaction and show how it's important to life. So the only types of chemical reactions I know are synthesis reaction, decomposition reaction, single-displacement reaction, and double-replacement reaction. I know what they do, but the only reactions I find useful for everyday life are synthesis reaction and decomposition reaction. How are single-replacement reaction and double-replacement reactions important to everyday life? O yea, what breaks down into glucose? Is it sucrose or just food? Thanks!
Asked by DL It's Me! - Sun Dec 2 13:43:25 2007 - - 2 Answers - 0 Comments
A. I would suggest: Photosynthesis reaction is a good start Chlorophyll +CO2 makes O2 Then the other one Hemoglobin exchanging CO2 against O2 and the other way around. in the lungs O2 binds is then transported to the cells where it is used and replaced by CO2. Both these are Lewis acid base reactions. The last one I would suggest is a basic oxidation CH4+ 2O2 -> 2H2O + CO2 + energy This is important to everyday life as this is the basis for heating our houses. Ion reactions are important for your body too as you have a finely tuned balance if ions inside and outside your cells, but I haven't got a good example handy. [Ah, lastly the starch in your food is made into glucose.]
Answered by klimbim - Sun Dec 2 13:58:22 2007
you dont have to solve it just tell me how?
Q. a motorist drives north for 35.o minutes at 85.km/h and then stops for 15 min. he then travels North at 130 km for 2 hours. find the total displacement and the average velocity
Asked by Chem88 - Tue Sep 16 06:37:45 2008 - - 3 Answers - 0 Comments
A. 1st part of the journey 35/60hours X 85km/h = 49/7/12 km 2nd part of journey = 130km Total displacement = 49/7/12 km + 130 =179/7/12km Average velocity = 179/7/12km divided by (15/60 + 2 + 35/60) = 63.38235294 = 63.4km/h (3s.f.)
Answered by Brandon - Tue Sep 16 06:42:55 2008
Q. a motorist drives north for 35.o minutes at 85.km/h and then stops for 15 min. he then travels North at 130 km for 2 hours. find the total displacement and the average velocity
Asked by Chem88 - Tue Sep 16 06:37:45 2008 - - 3 Answers - 0 Comments
A. 1st part of the journey 35/60hours X 85km/h = 49/7/12 km 2nd part of journey = 130km Total displacement = 49/7/12 km + 130 =179/7/12km Average velocity = 179/7/12km divided by (15/60 + 2 + 35/60) = 63.38235294 = 63.4km/h (3s.f.)
Answered by Brandon - Tue Sep 16 06:42:55 2008
Acceleration as a function of time.?
Q. Okay i have two questions here actually. I tried figuring out the whole night but to no avail. So if you guys could help, it'll be a great thing. Q1: A particle moves along a straight line and t seconds after passing through an origin O the velocity of the body is given by v=kt^2 - ct, where k and c are constants. When t=2 the body is again at O and has an acceleration of 6 ms^-2. Find the values of k and c. Q2: A body moves along a straight line with acceleration given by a=7t/36, where t is the time in seconds. When t=0 the body is at rest at an origin O. The acceleration continues until t=6, whereupon it ceases and the body is retarded to rest. During this retardation the acceleration is given by a=-t/4. Find the value of t when the… [cont.]
Asked by maths_dumbass - Sun Aug 13 01:46:39 2006 - - 6 Answers - 0 Comments
A. Okay, first question. At t=0, the particle's at O; at time t>0, it has a velocity v = kt^2 - ct; and at t=2, the particle's back at O with an acceleration of 6 m/s^s. With that velocity, the acceleration is not constant. So we take the derivative: v(t) = kt^2 - ct a(t) = dv/dt = 2kt - c a(2) = 6 = 4k - c (Eq 1) That takes care of the acceleration. Now for the displacement, we integrate v(t): s(t) = (1/3)kt^3 - (1/2)ct^2 + so, where so, the initial displacement, is zero. Now find s(2): s(2) = 0 = (8/3)k - 2c (Eq 2) Solve equations (1) and (2) for k and c. Now for the second question. Looks like there's two parts to this one. Let's do the first part first. a(t) = 7t/36 v(t) = (7/72)t^2 (initial velocity vo is zero) s(t) = (7/ [cont.]
Answered by bpiguy - Sun Aug 13 02:45:26 2006
Q. Okay i have two questions here actually. I tried figuring out the whole night but to no avail. So if you guys could help, it'll be a great thing. Q1: A particle moves along a straight line and t seconds after passing through an origin O the velocity of the body is given by v=kt^2 - ct, where k and c are constants. When t=2 the body is again at O and has an acceleration of 6 ms^-2. Find the values of k and c. Q2: A body moves along a straight line with acceleration given by a=7t/36, where t is the time in seconds. When t=0 the body is at rest at an origin O. The acceleration continues until t=6, whereupon it ceases and the body is retarded to rest. During this retardation the acceleration is given by a=-t/4. Find the value of t when the… [cont.]
Asked by maths_dumbass - Sun Aug 13 01:46:39 2006 - - 6 Answers - 0 Comments
A. Okay, first question. At t=0, the particle's at O; at time t>0, it has a velocity v = kt^2 - ct; and at t=2, the particle's back at O with an acceleration of 6 m/s^s. With that velocity, the acceleration is not constant. So we take the derivative: v(t) = kt^2 - ct a(t) = dv/dt = 2kt - c a(2) = 6 = 4k - c (Eq 1) That takes care of the acceleration. Now for the displacement, we integrate v(t): s(t) = (1/3)kt^3 - (1/2)ct^2 + so, where so, the initial displacement, is zero. Now find s(2): s(2) = 0 = (8/3)k - 2c (Eq 2) Solve equations (1) and (2) for k and c. Now for the second question. Looks like there's two parts to this one. Let's do the first part first. a(t) = 7t/36 v(t) = (7/72)t^2 (initial velocity vo is zero) s(t) = (7/ [cont.]
Answered by bpiguy - Sun Aug 13 02:45:26 2006
From Yahoo Answer Search: 'o displacement'
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poetryfoundation.org
Joel Brouwer
Wed, 23 Sep 2009 20:18:33 GM
In terms . of. content, . of. particular note is the poet's (wholly understandable) obsession with deracinations both geographical and lingual. The poems . of displacement. from the homeland and from the mother tongue are . . . well, here, ...
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