Sunday

April 20, 2014

April 20, 2014

Number of results: 272,511

**Physics**

OK, it just came to me. Time rising= altitude/veloictyrising=202/1.1= XXXseconds. during that time, it travels this horizontal distance: distance=xxxxseconds*2.7m/s Then add to that, the distance when at constant altitude distance= 10.3*2.7m/s
*Tuesday, November 9, 2010 at 6:53pm by bobpursley*

**Algebra1**

They could not have left at the same time. Jacob-you left part of the problem out. When you get it figured out, the distances are the same d = rate * time rate jet1 * time jet 1 = rate jet2 *(time jet 1 - delay)
*Tuesday, January 29, 2008 at 10:45pm by Damon*

**Uniform Motion**

Use the formula distance = rate times time (d = r*t).
*Thursday, December 6, 2007 at 4:50pm by Michael*

**algerbra 116**

distance=rate*time= 6m/min*7min
*Tuesday, April 6, 2010 at 9:13pm by bobpursley*

**physics**

The rate at which distance changes in a time period is called acceleration
*Wednesday, October 20, 2010 at 2:33am by Bhai*

**physics**

distance= rate*time = 2.2m/s*9.7min*60sec/min
*Tuesday, September 6, 2011 at 9:08pm by bobpursley*

**maths**

distance=rate*time=500km/.5yr * 3 years=3000km
*Tuesday, November 19, 2013 at 5:44pm by bobpursley*

**maths**

rate=amount/time rate combined=sum of rates=1tank/3.5hr+1tank/4.2hrs rate combined=(4.2+3.5)/(3.5*4.2) tanks/hr work this out. time= 1tank/(above rate)
*Friday, October 21, 2011 at 9:09am by bobpursley*

**math-word problem- PLEASE CHECK MY ANSWER**

distance apart at station =100mph*1/4hr relative velocity= distance/time time= 25mi/20mph=5/4 hr Add that to 5:35
*Sunday, March 28, 2010 at 8:06pm by bobpursley*

**Physics**

The time is the distance divided by the velocity. The total time is the time at 88km/h + the time at the unknown speed v. Half the distance is run at each speed: 0.8/88 + 0.8/v = 1.6/67 .009091 + 0.8/v = .02388 .8/v = 0.01365 v = 58.6 km/h
*Thursday, September 22, 2011 at 7:51pm by Steve*

**Math - Ratio or Proportion (I think??)**

figure food at rate per person day. amount of food=rate/person*persons*time rate= 6/(3*5) so amount of meat= (rate)people*time = 6/15 * 5*7 = you do it. 2. ratio problem; 1.5yards/2inches=1yard/Xinches xinches=1*2inches/1.5= 4/3 inch
*Monday, January 27, 2014 at 7:15pm by bobpursley*

**physics**

The time that the ball is in the air just depends on how far it has to fall down to the ground. Since the ball was thrown HORIZONTALLY the initial VERTICAL velocity is 0. So, the ball immediately starts to drop. This happens at the same rate as if you just dropped it from the ...
*Wednesday, April 2, 2008 at 5:46pm by Quidditch*

**algebra**

distance = rate * time One way is to make a table and fill it in. The same approach works for doing mixture problems. Your example does not work as you made it up because if they both start together car B will always be ahead. Let me change your example as follows: Car A and ...
*Tuesday, May 27, 2008 at 8:55pm by Damon*

**Physics**

A. stopping distance = (stopping time) x (average speed) = (V/a)*(V/2) = V^2/(2a) B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.
*Sunday, December 27, 2009 at 7:29pm by drwls*

**math**

Distance covered by me --- x distance covered by friend -- x- 1/2 time taken by me = x/3.4 time taken by friend = (x-1/2)/3 but they are the same x/3.4 = (x-1/2)/3 3.4x - 1.7 = 3x .4x = 1.7 x = 4.25 miles time = x/3.4 = 4.25/3.4 = 1.25 hrs or 75 minutes check: my distance ...
*Tuesday, October 22, 2013 at 7:12pm by Reiny*

**physics**

You need to memorize some formulas: distance= Vi*time+ 1/2 * acceleration*time^2 or distance=average velocity*time =(1/2)(Vf+Vi)*time
*Sunday, September 5, 2010 at 8:49pm by bobpursley*

**math**

A swimmer going downstream takes 1hour, 20 minutes to travel a certain distance. If takes the swimmer 4 hours to make to return trip against the current. If the river flows at the rate of 1.5mph. Find the rate of the swimmer in still water and the distance traveled one way.
*Saturday, July 31, 2010 at 11:30am by joy*

**physics**

distance fell= 1/2 g t^2 .809=4.9 t^2 solve for t, the time it was in the air. howfast=distance/time= 18.3m/time
*Thursday, October 20, 2011 at 4:17pm by bobpursley*

**Physics**

power = work/time = force*distance/time = mg * distance / time = 9 * 9.8 * 3.5 / 600 = 0.51 watt
*Thursday, December 1, 2011 at 11:17am by Steve*

**Physics**

distance = rate * time height = 5.2 * 340 = 1768m It should be as simple as that.
*Tuesday, September 4, 2007 at 8:33pm by Michael*

**Chemistry**

distance=rate(time) = 60mi/hr * 3.5 days ( 24hrs/day)
*Monday, August 17, 2009 at 7:11pm by bobpursley*

**physics**

distance=rate*time change 30 min to seconds first.
*Tuesday, August 9, 2011 at 11:57pm by bobpursley*

**math**

1 mile = 5280 feet 5 days * 24h * 60 min * 60 sec = 432,000 sec D = RT, Rate = Distance/Time Rate = 5280/432,000 = 0.0122 ft/sec
*Tuesday, February 8, 2011 at 4:05pm by helper*

**Not so fast, grasshopper**

First race: let Yolanda' time to finish the 100 yds be t sec then Yolanda's rate = 100/t yds/sec Yoko's rate = 90/t yds/sec Second Race Distance for Yolanda is 110 yrd Distance for Yoko is 100 yrd time to run for Yolanda to run 110 yrs = 110/(100/t) = 1.1 t time for Yoko to ...
*Thursday, January 26, 2012 at 7:53pm by Reiny*

**Algebra**

Same time? Distance=rate*time timeClark= distanceclark/rateclark timeRourke=distancerourke/raterourke times are the same, and distanceClark=distancerourke-20km You know rates. set the two equations equal... distanceclark/rateclark=distancrourke/raterouke (distancerourke-20)*...
*Wednesday, August 5, 2009 at 4:35am by bobpursley*

**math**

Let the rowing rate be x mph let the current's rate by y mph going with the current, will be a rate of x+y mph going against the current would be x-y mph so you get two equations, using Distance = rate x time 1.5(x+y) = 12 4(x-y) = 12 1st: divide by 1.5 ---> x+y = 8 2nd: ...
*Saturday, April 24, 2010 at 8:05pm by Reiny*

**probability**

3)If the random variable T is the time to failure of a commercial product and the values of its probability density and distribution function at time "t" are f(t) and F(t), then its failure rate at time t is given by f(t) / 1-F(t) Thus, the failure rate at time t is the ...
*Monday, November 12, 2007 at 11:22pm by chinnu*

**Algebra II**

time it took Joan = X time it took Buggs = X+2 hrs. distance each traveled is the same. distance Joan = r*time = 50t distance for Buggs = r*time = 40*(X+2) 40*(X+2) = 50X solve for X = time for Joan. X+2 = time for Buggs.
*Wednesday, January 2, 2008 at 5:02pm by DrBob222*

**math**

Let M be marias rate C be Carlos rate. M-8=C M*time-72=C*time but time=M/48 putting all that together.. M*M/48 -72=(M-8)*M/48 solve for M first.
*Wednesday, July 21, 2010 at 8:46pm by bobpursley*

**math**

can your help me with my homework question? Gildas family goes on vacation. They travel 125 miles in the first 2.5 hours.If Gildas family continues to travel at this rate.how many miles will they travel in 6 hours? distance= rate + time
*Wednesday, February 17, 2010 at 4:00pm by pavleen*

**Maths**

1) You start with the definition of average speed. avg speed= distance/time 60km/hr= 1km/min= distanceab/time distanceab= 1km/min*time but also distanceab= 100km/60min*(time-10) set these two equations equal, solve for time, then, find distance from the original equation. 2) ...
*Wednesday, April 21, 2010 at 8:57am by bobpursley*

**physics**

If you assume the body is stretched similarly, with one arm upward, before, at the peak of and at the end of the jump, then the vertical jumping distance can be related to hang time. This is not the way most people would jump. The foot-to-center-of mass-distance has to remain ...
*Thursday, September 11, 2008 at 11:03pm by drwls*

**math**

please help me to solve this problem A swimmer going downstream takes 1hour,20minutes to travel a certain distance. It takes the swimmer 4hours to make the return trip against the current. If the river flows at the rate of 1.5mph. Find the rate of the swimmer in still water ...
*Saturday, July 31, 2010 at 4:46am by joy*

**math**

Let time taken by smaller pipe be t hrs then time takes by larger pipe is t-6 hrs rate of smaller pipe = 1/t rate of larger pipe = 1/(t-6) combined rate = 1/t + 1/(t-6) = (2t-6)/(t(t-6)) time at combined rate = 1/ [ (2t-6)/(t(t-6)) ] = t(t-6)/(2t-6) then t(t-6)/(2t-6) = 4 t^2...
*Tuesday, June 21, 2011 at 3:28pm by Reiny*

**math**

John sets off 4 seconds after Tom does, and finishes the lap 2.5 seconds sooner than Tom does. So, John's total time will be 6.5 sec less than Tom's time. Do you understand? Also, If you solved correctly, R = 20 and T = 32.5. Remember, the problem wants each racers' speed and ...
*Monday, January 24, 2011 at 9:24pm by helper*

**Algebra**

I used to encourage my students to make a chart for these kind of problems the rows of the chart are the different situations and the columns would be titled D(distance) R(rate) and T(time) ............D ........R ......T 1st train: 75t ------ 75 ---- t 2nd train: 125(t-2)-- ...
*Monday, February 23, 2009 at 4:29pm by Reiny*

**distance problems**

two planes left Seattle at the same time, both traveling due east. One had the rate of 185 mi/h and the other had a rate of 225 mi/hr. After how many hours were they 220 miles apart?
*Wednesday, March 24, 2010 at 11:13pm by sky*

**Algebra**

distance=42*time distance=48(time-.5) set them equal, solve for time, then solve for distance.
*Wednesday, February 19, 2014 at 5:11pm by bobpursley*

**6th grade math**

I can much help making a graph on this. distance = speed*time, so on a distance time graph, you will have a sloping line upwards..
*Thursday, April 17, 2014 at 4:49pm by bobpursley*

**Math**

time taken to travel by bus= 30 min = 1/2 hour time taken to travel by train = 20min = 1/3 hour let the speed of bus be = x mph then speed of the train = (x+15) mph speed = distance/time let the distance to work be y metres then, distance= speed*time 1/2*x=1/3*(x+15) x=30mph ...
*Thursday, May 9, 2013 at 12:39am by saujanya*

**Math**

time taken to travel by bus= 30 min = 1/2 hour time taken to travel by train = 20min = 1/3 hour let the speed of bus be = x mph then speed of the train = (x+15) mph speed = distance/time let the distance to work be y metres then, distance= speed*time 1/2*x=1/3*(x+15) x=30mph ...
*Thursday, May 9, 2013 at 12:39am by saujanya*

**math**

First, how far did Marla walk? The library is 4 miles. So, the halfway point woud be? Let D be the halfway distance. Time=distance/speed Time = D/(3 mi/h)
*Tuesday, October 23, 2007 at 8:42pm by Quidditch*

**physics**

It goes half a wavelength(.65meter) in 1/115 seconds distance = rate * time
*Monday, February 22, 2010 at 2:53pm by Damon*

**Science**

Speed is the rate of change in _____. A.velocity B.time C.direction D.distance I think it is D...?
*Tuesday, March 12, 2013 at 1:14pm by Cassie*

**algebra1**

distance = rate times time. Use the formula and plug in the constants and variables.
*Thursday, April 11, 2013 at 9:52pm by Knights*

**math**

the el paso middle school girls baxketball team is going from el paso to san antonio for the texas state championship game.the trip will be 560 miles.their bus travels at an average speed of 60 miles per hour. estimate the distance the bus travels in 2 hours 2and 3 4...2 and a...
*Monday, October 2, 2006 at 8:54pm by chemiii*

**Chemistry**

Hydrogen iodine is a strong acid that is used in chemical synthesis to produce some common drugs as well as a source for iodine, I2. Data for the formation of hydrogen iodine was obtained as shown. The reaction was found to be first order with respect to hydrogen gas and ...
*Saturday, May 19, 2007 at 12:10pm by Christian*

**math**

Let the time at 40 mph be t distance covered at 40 mph = 40t distance covered at 60 mph = 60t total distance = 40t+60t total time = 2t avg speed = total distance/total time = 100t/(2t) = 50 No trick here, had it said, Chris travels the same distance at 40, and then at 60, you ...
*Wednesday, September 28, 2011 at 12:09pm by Reiny*

**algebra**

Jose traveled to San Diego at an average rate of 50 mph. Carlos made the same trip in one hour less time at an average rate of 60 mph. Find the distance of the trip.
*Wednesday, September 7, 2011 at 8:06pm by Brianne*

**physics**

Horizontal distance=horzvelocity*time time= distance/velocity= 25/(75cos64.7) figure time, then put it here. Vertical: h=75*sin64.7t-4.9t^2 how much higher is h than 11 m.
*Sunday, October 24, 2010 at 2:10pm by bobpursley*

**Math**

laura's rate --- 1/4 lucky's rate ---- 1/2 combined rate = 1/4 + 1/2 = 3/4 time at combined rate = 1/(3/4) = 4/3 hours (painting a house in 2 hours ??? , WOW)
*Monday, August 5, 2013 at 10:15pm by Reiny*

**MATH**

Jack's rate alone = 1/3 Marilyn's rate = 1/5 combined rate = 1/3 + 1/5 = 8/15 time at combined rate = 1/(8/15) = 15/8 = 1.875 hrs or 1 hour and appr 53 minutes
*Tuesday, April 2, 2013 at 9:51pm by Reiny*

**Algebra**

An easier way than what? In these kind of problems you have to find which of the three variables, either time, distance or rate, are equal. In this case, at the moment the 2nd train catches up to the first, they will have gone the same distance. let the time taken for the 2nd ...
*Monday, January 3, 2011 at 9:14pm by Reiny*

**Physics**

One step at a time. First compute the net force acting: F = 1153 - 939 = 214 N Then compute the acceleration rate a using F = m a; a = F/m Then compute the time t required to reach velocity V: t = V/a The distance travelled is X = (1/2) a t^2
*Saturday, January 22, 2011 at 8:27pm by drwls*

**math**

Since Time = Distance/rate time at 65 mph = 220/65 = 3.385 hours = 3 hours and .385(60) minutes = 3 hours 23 minutes , to the nearest minute
*Thursday, April 28, 2011 at 6:50am by Reiny*

**chemistry**

I would convert 5.9 miles/hour to feet/hour (there are 5,280 feet in a mile) , then use distance = rate x time solve for time in hours and add to 11:15 A. M.
*Friday, September 3, 2010 at 7:43pm by DrBob222*

**8th grade**

constant rate of change. constant speed is constant change of distance/time so if you are going 25mph, your distance is changing at a constant 25miles each hour.
*Wednesday, December 1, 2010 at 7:50pm by bobpursley*

**physics**

Two cars are initially moving at constant speeds of 50 m/s (first car) and 30 m/s (second car) along a straight track with the first car being behind. At a certain point (e.g. point A) on the track the first car overtakes the other one. At this very moment of time the first ...
*Thursday, September 29, 2011 at 6:33pm by Chris*

**Math**

distance = rate x time d for 40 mph vehicle = 50 mi = 40 mi/hr x time. solve for time for 40 mph car = 50/40 = 5/4 hour. d for 60 mph car = 50 mi = 60 mi/hr x t solve for t for 60 mph car = 50/60 = 5/6 hr. Add the times together. 5/4 + 5/6 = 1.250 + 0.833 = 2.083 hours total ...
*Tuesday, October 2, 2007 at 6:15pm by DrBob222*

**Math**

Not quite, Fiona. When you finally got an answer, you ought to have done just a sanity check. The average speed was somewhere near 200 mph, and the flew for 7 hours, a distance of 57 mile is ludicrous. since time = distance/speed, and the total time was 7 hours, then we want ...
*Monday, March 17, 2014 at 12:18am by Steve*

**Math**

To help you understand the solution, let's say the track is a typical 400 m track, Eillen can run 4 m/sec and Malia can run 8 m/sec let t be the time when they first meet Eileen has run 40t m, and Malia has run 80t m (distance = rate x time) but when they meet: 4t + 8t = 400 ...
*Tuesday, October 23, 2012 at 7:54pm by Reiny*

**FREE BODY DIAGRAM calculus**

A car is traveling north towards an intersection at 60mph at the same time a truck is headed east toward the same intersection at 45mph. Find the rate of change of the distance between the car and truck when the car is 3 miles south of the intersection and the truck is 4 miles...
*Sunday, October 7, 2012 at 11:57pm by erica*

**Math**

Joey's rate = 1/4 Luigi's rate = 1/5 combined rate = 1/4 + 1/5 = 9/20 time at combined rate = 1/(9/20) = 20/9 = 2.2222.. hrs = 2 hours and appr. 13 minutes
*Friday, February 17, 2012 at 9:46am by Reiny*

**correction - math**

Brooke's answer makes no sense Frank's rate = 1/4 James rate = 1/6 combined rate = 1/4 + 1/6 = 5/12 time at combined rate = 1/(5/12) = 12/5 = 2.4 hours
*Monday, July 23, 2012 at 12:11am by Reiny*

**Math (Earning Money)**

I know that calculating overtime rate for time and a half is 1.5 and double time is 2 and triple is 3 and so on... however my HW math question said "he is paid DOUBLE TIME AND A HALF on Saturdays..." So what number will I multiply with the number of hours and normal pay rate ...
*Friday, February 3, 2012 at 6:37am by CrissyXD*

**algebra**

The height of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?
*Saturday, March 29, 2014 at 10:24am by Anonymous*

**pysics**

Simple. How long does it fly in the air? time= distance/combinedspeed= distance/20 You didn't say the distance, but put that in, and figure time. Now how far does the bug fly? distance=speedbug*timeflying you didn't put the speed of the bug in the problem either.
*Sunday, October 31, 2010 at 4:27pm by bobpursley*

**Maths**

Use the equation distance = rate * time. Don't forget to convert everything to the same units.
*Monday, November 26, 2007 at 3:03pm by Michael*

**Science**

x = time (independent) y = distance (dependent) horizontal: y isn't changing diagonal: y is increasing at a constant rate
*Saturday, November 15, 2008 at 6:37pm by Albert*

**math**

Interest = Principal*rate*time I=PRT , so R = I/(PT) = 60/(3000*1) = .02 so the annual rate is 2% and the annual rate would be 24% (WOW, get rid of that card!)
*Wednesday, May 7, 2008 at 11:36pm by Reiny*

**Science**

Thanks for your reponse, it helped greatly. Can you please also tell me if the braking distance is 24metres and it decelerates at rate of 6m/s, does that mean that the time taken to stop after braking would be 4 seconds or do I need to consider anything else to calculate time.
*Friday, May 16, 2008 at 4:58am by Melody*

**math**

rate of pump A = 1/6 rate of pump B = 1/x combined rate = 1/6 + 1/x = (x+6)/(6x) time at combined rate = 1/( (x+6)/6x ) = 6x/(x+6) but 6x/(x+6) = 4 6x = 4x+ 24 2x = 24 x = 12 Pump B alone would take 12 hours check: combinedrate =1/12 + 1/6 = 3/13 = 1/4 time at combined rate = ...
*Thursday, January 31, 2013 at 11:51pm by Reiny*

**Physics**

(a) distance = speed x time or time = distance/speed The two times can be found by dividing the distance by each speed. Subtract the two times. (b) Let t = time it takes the slower student. 5.50 min = 330 s The distance is 0.90t = 1.90(t-330) Solve for t, then substitute into ...
*Sunday, August 31, 2008 at 10:12pm by GK*

**10th grade algebra**

The cars are 500 miles apart. They are driving toward each other with the combined speed of 55+60 = 115 mph. distance = rate*time time = d/r = 500/115 = ??
*Thursday, November 12, 2009 at 1:23pm by DrBob222*

**physics**

A car accelerates from rest at a steady rate of 1 ms-2. Calculate: (a) the time taken to reach 15 ms-1 (b) the distance travelled during this time (c) the velocity of the car when it was 100 m from the start point
*Wednesday, February 8, 2012 at 4:44pm by LK*

**physics**

A car accelerates from rest at a steady rate of 1 ms-2. Calculate: (a) the time taken to reach 15 ms-1 (b) the distance travelled during this time (c) the velocity of the car when it was 100 m from the start point
*Wednesday, February 8, 2012 at 4:44pm by LK*

**maths --plse help me..**

Distance is equal to rate times time Therefore time is equal to distance divided by rate. 1586/39= 40 2/3 hours That means it takes 40 2/3 hours to go 1586 km when you are going 39 km per hour So if you start at 10:30 and go for 40 and 2/3 hours, which is 40 hours and 40 mins ...
*Friday, December 14, 2012 at 4:20am by Mac*

**physics**

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 14.0s. It runs at a constant speed for 70.0 s and then slows down at a rate of -3.50 m/s2 until it stops at the next station. a) Sketch a position vs. time graph for this problem. b) Sketch...
*Monday, June 4, 2012 at 5:57pm by cody*

**physics**

an athlete runs at a distance of 1500m in the following manner:1.starting from the rest he acceralate himself uniformly 2m/s*s till he covers a distance of 900m 2.he then runs the remaining distance of 600m at the uniform speed developed Calculate the time taken by the athlet ...
*Sunday, July 31, 2011 at 9:20pm by abid*

**Math**

Is elevation constant? distance=velocty*time you do not have time. Figure the distance for each. You have on the triangle SAS you can use the law of consines to figure the last side. distance^2=d2^2+d1^2 -2*d2*d1*cos48
*Saturday, February 23, 2013 at 8:05pm by bobpursley*

**Math**

Jill distance = x Matt distance = (400-x) Jill speed = j Matt speed = (j-2) distance = rate * time x = j (40) (400-x) = (j-2)(40) so (400 -40 j) = 40 j - 80 I guess you can do it from there.
*Wednesday, October 22, 2008 at 6:34pm by Damon*

**Physics**

(a) acceleration=(final velocity-initial velocity)/time distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time) (b) 100m - _____ (<-answer from (a)) find time using this formula: distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time) finally, add ...
*Thursday, September 16, 2010 at 1:25am by TutorCat*

**Physics**

Wave equation: frequency*wavelength=speed of light Distance equation: Distance=velocity*time where distance it distance there+return distance.
*Wednesday, May 12, 2010 at 12:36pm by bobpursley*

**physics**

the horizontal speed never changes, so you can figure out how long it is in the air by distance = rate times time then figure out how far it fell in that amount of time starting with zero vertical velocity.
*Tuesday, January 1, 2008 at 8:16pm by Damon*

**RATE, DISTANCE, TIME**

TIM PADDLED HIS KAYAK 12 KM UPSTREAM AGAINST A 3 KM/H CURRENT AND BACK AGAIN IN 5 HR 20 MINUTES. iN THAT TIME HOW FAR COULD HE HAVE PADDLED IN STILL WATER 12/(v+3) + 12/(v-3) = 5.33 hours Solve for this speed in still water, v. 12(v-3) + 12(v+3) = 5.33 (v^2 -9) 24 v = 5.33 v^2...
*Thursday, December 21, 2006 at 2:16am by CINDY*

**Science**

The reaction distance is how far the car travels before brakes are applied. That equals (reaction time)*(intial velocity) The initial velocity (V) satisfies the equation V = sqrt(2 a X), where a is the deceleration rate and X is the distance travelled while decelerating. ...
*Friday, May 16, 2008 at 4:58am by drwls*

**chemistry**

I suppose you could work the 2-4 combinations out and memorize them; however, it's relatively easy to do one. For example, for the rate = k(x)^1(y)^2 for a third order reaction, you have k = rate/(x)^1(y)^2 = rate is M/t where M is molar and t is time. Tben k = (M/t)/(M)(M)2 k...
*Sunday, December 12, 2010 at 5:41pm by DrBob222*

**physics**

consider the mug: F=ma you know F, you know the time of force (time=distance/velocitycloth), so Force*time=massmug*velocitymug from that,you caculate the averagevelocity of the mug (1/2 velocityfinalmug), then distance=avgvelocity*time
*Thursday, February 24, 2011 at 10:12am by bobpursley*

**Science**

The distance will rise as time increases, at an increasng rate (slope). The curve is called an upward parabola.
*Wednesday, October 24, 2007 at 5:05pm by drwls*

**Chemistry**

The speed of light is, for all practical purposes, 3 x 10^8 meters/second. Distance = rate x time.
*Thursday, November 1, 2007 at 9:32pm by DrBob222*

**physics**

do distance=rate*time for each leg, then add the distances. Change min to hour by dividing by 60.
*Sunday, October 4, 2009 at 9:32pm by bobpursley*

**calc**

A____________B 9 miles is between A and B? at 3pm Andy began travelling South at a rate of 4mi/hr from point A at the same time Bill began travelling West at a rate of 5mi/hr from point B to A both stop wlking when Bill reaches point A What time will the distance between Bill ...
*Thursday, April 5, 2012 at 7:49am by angel*

**math**

You cant trust the average always. I assume you are using a rate equation, amount=rate*time Consider in 28 hours, shelby can paint 7, and Zoe can paint 4. So the combined rate = (7+4 rooms)/28 hrs= 11/28 rooms/hr Now the time to paint one room amount= rate*time time= amount/...
*Saturday, September 12, 2009 at 12:31pm by bobpursley*

**Physics**

force= mass*acceleration average velocity= distance/time final velocity= 2*distance/time acceleration= (finalvelocity-0)/time = 2*distance/time^2 Or you can do it this way: Vf^2=Vi^2 + 2 a d where a is Force/mass
*Tuesday, November 2, 2010 at 7:17pm by bobpursley*

**physics**

They both travel the same distance, in the same time. truck: distance=10t car: distance=1/2 2 t^2 10t=t^2 10=t so time is ten seconds. distance=10*t velocitycar=2*t=20m/s
*Thursday, February 7, 2013 at 9:21am by bobpursley*

**Math**

rate of 1st tap = 1/4 rate of 2nd tap = 1/6 rate of both taps = 1/4 + 1/6 = 5/12 time at combined rate = 1/(5/12) = 12/5 or 2.4 minutes
*Monday, February 4, 2013 at 5:27pm by Reiny*

**math**

Well The person typing is going at a rate of 90 mintutes per rate and is getting payed 2$ per minute. Is 10 dollars for 1800 words a better pay than the rate of the time taken???
*Monday, September 22, 2008 at 4:15pm by Carly*

**Physical Science**

What is the hiker's total displacement? part a of the hike: distance:4km east start time: 9:00 am end time: 9:45 am part b: distance:6 km south start time: 9:45 am end time: 10:45 am part c: distance:4 km west start time:10:45 am end time: 11:15 part d: distance 6 km north ...
*Thursday, February 20, 2014 at 1:23pm by dark1*

**Math**

Distance = Rate * Time or D = RT for Pete : D = 9x for Jan : D = 18(1.5) so 9x + 18(1.5) = 42 solve for x (I got 1 hour, 40 minutes)
*Monday, May 26, 2008 at 7:41am by Reiny*

**physics**

Question: 15 of 30: The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible for this situation? Time (s) A. Distance (m) B. Distance (m) C. Distance (m) D. Distance (m) E. Distance (m) 0 0 2.00 9.00...
*Thursday, May 23, 2013 at 10:48pm by ....*

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