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A train travel from station, a, to the the nest station, b, at a constant speed of 100 km/h and return at a constant speed of 150 km/h. Compare the average speed and the velocity from this journey.

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You can find the solution here;
Velocity and Speed (pdf file)

Good luck.

Thanks for using FixYa.

Posted on Jul 09, 2010

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How fast do the Japan bullet trains travel?


The maximum operating speed is 320 km/h (200 mph) (on a 387.5 km section of the T?hoku Shinkansen). Test runs have reached 443 km/h (275 mph) for conventional rail in 1996, and up to a world record 581 km/h (361 mph) for maglev trains in 2003.

Mar 02, 2015 | Crafts & Hobbies

1 Answer

A train leaves Middelburg at 09:00 and arrives in Johannesburg at 11:00 having traveled a distance of 180 km. Find the average speed of the train in m/s.


Journey time 2hrs = 2 x 60 mins = 2 x 60 x 60 seconds
Distance 180 km = 180,000 m
Average speed = 3 x 6 x 10000 / 2 x 6 x 6 x 100 m/s
= 6 x 6 x 100 x 100 / [2 x (2 x 6 x6 x100)] m/s
= 100 / 4 m/s
= 25 m/s

5.May.2014

May 05, 2014 | Office Equipment & Supplies

1 Answer

One train leaves a station heading west, while another leaves two hours later travelling 15km/h faster heading east. After 6 hours the trains are 580 km apart. How fast is each train travelling?


To start out, you need to figure out what the equation for the problem is. You know a few things that should help you along the way:
  • The trains are 580 km apart
  • one train traveled for 6 hours
  • the other train traveled for 4 hours
  • one train is going 15km/h faster than the other.
Knowing these facts you can write the equation 6x+4x+60=580.

The 6 at the front is how many hours the first train traveled. the 4 is the hours the second train traveled. the 60 is (4*15) which is how much faster the second train was going. the 580 is the total distance between them.

Solve for x

10x=520
x=52

So the westbound train was going 52 km/h and the east bound train was going 67km/h.

May 03, 2011 | Texas Instruments TI-30 XIIS Calculator

1 Answer

. A person meets a


If the train's still going the right way, up about ((5 *60)/25) km/h. Classical relativity doesn't change it much at that person's speed, since c >> 25kmph, but that's not to say quantum relativity will continue to favor the Standard Model with the information given for meeting in your OS.

Jan 03, 2011 | Computers & Internet

1 Answer

Train A and B are traveling in the same direction on parallel tracks. Train A is traveling at 100 miles per hour and Train B is traveling at 110 miles per hour. Train A passes a station at 9:25am If train...


The first thing to figure out is the amount of time between the trains is 12 minutes. That would mean for train B to pass the station at 9:37 it would have to travel 22 miles in 12 minutes. Because, traveling at 110 miles/hour in 12 minutes the train would have travel 22 miles. Shown below shows the interpolation to come up with 22 miles.

oneplusgh_0.jpg
Train A had to traveled 12 minutes so, it would have been 20 miles ahead of the station. Screened below shows the interpolation to equal 20 miles.

oneplusgh_1.jpg
Though this calculation of 20 miles isn't necessary for the parametric solution that is being shown. It was very helpful in my solution in successive approximation. This parametric solution's starting point is at 9:37, not 20 miles ahead.

Using successive approximation, when train A traveled 200 miles in 2 hours, and train B was only 2 miles behind train A. At that point the time varibles needed to be broke down to minutes and seconds which gave me a real accurrate answer. Both trains were side by side in 2 hours and 12 minutes, and had traveled 220 miles from the station. You wouldn't believe how far the significant bits needed to be carried out to get such an accurate number. That's what gave me the idea to substitute, and use a parametric solution. In this solution the t (aka. time) is going to be substituted for the y axis. Next, I broke miles/hour down to mile/minute because, I knew the answer was in minutes: 2 hours and 12 minutes.

110/60=1.833333 so, -20+1.833333 t and 100/60=1.6666667 so, 1.6666667 t

Now load the equation:

oneplusgh_2.jpg
Now let's look at the solution from the graph. Each line represents a train.

oneplusgh_3.jpg
Notice how the two lines are converging. Where the two lines meet is the exact point when the two trains are side by side.

oneplusgh_4.jpg
Substituting t (aka. time) for the y axis show that at the point when the trains were side by side equals 132 minutes. Or 2 hours and 12 minutes. The distance traveled x axis=220 mile from the station.

To a point successive approximation would have been messy, and hard to follow. Somethings can only be solved by using graphs, or better solved by using graphs.

Dec 20, 2010 | Office Equipment & Supplies

1 Answer

Obd2 system rediness status won't clear so I can run the smog test due to replacing the battery.


takes a long time to run the rediness test since there are 6 different tests to complete...

PROCEDURE 1
EVAPORATIVE EMISSION SYSTEM LEAK MONITOR
Inspection Conditions
Engine coolant temperature: 30°C (75°F) or less (The engine is stopped before the test drive is started).
Atmospheric temperature: 5 - 30°C (41 - 113°F).
Condition of A/T: Selector lever D range.
Fuel remained in fuel tank: 30 - 50% is recommended.
Time required: 16 minutes or less after started the engine.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position. Well be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start.
Accelerate until the vehicle speed is 89 km/h (55 mph) or more.
While keeping the accelerator pedal opening degree constant, keep the vehicle speed at 89 km/h (55 mph) or more and travel for 16 minutes.
Return the vehicle to the shop, then turn the ignition switch to "LOCK" (OFF) position.

PROCEDURE 2
FUEL TRIM MONITOR
Drive Cycle Pattern
Inspection Conditions
Engine coolant temperature: 100°C (212°F) or less.
Atmospheric temperature: -10 - 60°C (14 - 140°F).
Condition of A/T: Selector lever D range.
Time required: 30 minutes or more.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position). Will be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start.
Accelerate until the vehicle speed is 89 km/h (55 mph) or more.
While keeping the accelerator pedal opening degree constant, keep the vehicle speed at 89 km/h (55 mph) or more and travel for 30 minutes.
Return the vehicle to the shop, then turn the ignition switch to "LOCK" (OFF) position

PROCEDURE 3
CATALYTIC CONVERTER MONITOR
Drive Cycle Pattern
Inspection Conditions
Atmospheric temperature: -10°C (14°F) or more.
Condition of A/T: Selector lever D range.
Engine coolant temperature: Not set.
Time required: 16 minutes or more.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position). Will be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start.
Accelerate until the vehicle speed is 89 km/h (55 mph) or more.
Travel for 300 seconds or more while keeping the vehicle speed at 89 km/h (55 mph) or more.
Decelerate until the vehicle speed is within 80 km/h (50 mph) or less.
While traveling at 56 - 80 km/h (35 - 50 mph) for 10 minutes or more, fully close the throttle at least once in 2 minutes and decelerate for 10 seconds or more.
Do not repeat deceleration too often.
Vehicle speed may go below 56 km/h (35 mph) after the deceleration.
Stopping and braking are permitted. (If stopped or drive at 56 km/h (35 mph) or less for more than 5 minutes the monitoring may be stopped. In this case please restart monitoring from the beginning.)
After completing the above deceleration, bring the vehicle speed back to 56 - 80 km/h (35 - 50 mph) and keep it in the range until starting the deceleration again.
Repeat the above deceleration at least 5 times.
Return the vehicle to the shop, then turn the ignition switch to "LOCK" (OFF) position

PROCEDURE 4
HEATED OXYGEN SENSOR MONITOR
Drive Cycle Pattern
Inspection Conditions
Engine coolant temperature: 100°C (212°F) or more.
Atmospheric temperature: -10°C - 60°C (14 - 140°F) or more.
Condition of A/T: Selector lever D range.
Time required: 16 minutes or more.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position). Will be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start.
Accelerate until the vehicle speed is 56 - 80 km/h (35 - 50 mph).
While keeping the accelerator pedal opening degree constant, keep the vehicle speed at 56 - 80 km/h (35 - 50 mph) and travel for 16 minutes or more.
Stopping and braking during this operation are permitted. Keep the accelerator opening degree constant for 1 minute or more after each acceleration.
Return the vehicle to the shop, then turn the ignition switch to "LOCK" (OFF) position

PROCEDURE 5
EXHAUST GAS RECIRCULATION (EGR) SYSTEM MONITOR
Drive Cycle Pattern
Inspection Conditions
Engine coolant temperature: 100°C (212°F) or less.
Atmospheric temperature: 5 - 60°C (41 - 140°F).
A/C switch: OFF.
Condition of A/T: Selector lever D range.
Time required: 16 minutes or more.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position). Will be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start.
Accelerate until the vehicle speed is 56 - 80 km/h (35 - 50 mph).
While traveling 16 minutes or more at 56 - 80 km/h (35 - 50 mph) with engine speed 2,000 r/min or above, fully close the throttle at decelerate for 5 seconds or more until the engine speed reaches 1,000 r/min or under.
Do not repeat deceleration too often.
Stopping and braking are permitted (Rapid deceleration and sharp steering are not permitted).
After completing the deceleration, bring the vehicle speed back to 56 - 80 km/h (35 - 50 mph) and keep it in the range until starting the deceleration again.
Repeat the above deceleration at least 8 times by fully closing the throttle valve.
Return the vehicle to the shop, then turn the ignition switch to "LOCK" (OFF) position

PROCEDURE 6
OTHER MONITOR (Main Components, Sensors & Switches, Wire Breakage & Short Circuit)
Drive Cycle Pattern
Inspection Conditions
Engine coolant temperature:
100°C (212°F) or less (Except engine coolant temperature sensor monitoring).
30°C (86°F) or less (Engine coolant temperature sensor monitoring).
Atmospheric temperature:
5°C (41°F) or more (Except engine coolant temperature sensor monitoring).
-10°C (14°F) or less (Engine coolant temperature sensor monitoring).
Condition of A/T: Selector lever D range.
Time required: 21 minutes or more.
Drive cycle pattern: One trip monitor (from start to ignition switch to "LOCK" (OFF) position). Will be completed while traveling with the following drive cycle pattern.
Test Procedure
Engine: start
Accelerate until the vehicle speed is 56 km/h (35 mph) or more.
While keeping the accelerator pedal opening degree constant, keep the vehicle speed at 56 km/h (35 mph) or more and travel for 16 minutes or more.
Return the vehicle to the shop.
After stopping the vehicle, continue idling for 5 minutes, and then turn the ignition switch to the "LOCK" (OFF) position. Moreover, the vehicle should be set to the following conditions for idling.
A/C switch: OFF
Lights and all accessories: OFF
Transmission: Neutral

Sep 15, 2010 | 1998 Mitsubishi Eclipse

2 Answers

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car traveling 40 mph car stop in 90ft, how many feet will it take the same car to stop when...


The stopping distance is proportional to square of speed.

that is 90ft proportional to 40^2 mph.

We can write as 90 = X * 40^2. ( X is a constant of proportionality) .

Which implies X = 90 / 1600 = .05625.

Assume S be the distance at which car stops when the speed is 90 mph.

S = X * 90^2. ( since X does not vary as same car is used)

=> S =.05625 * 8100 = 455.625 ft.

Thank you.

Jun 12, 2010 | Bagatrix College Algebra Solved Full...

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