Learn what power means and how we use it to describe the rate of energy transfer.
Log in adityasarkar2706 8 years agoPosted 8 years ago. Direct link to adityasarkar2706's post “how is P=m.a.v instantane...” how is P=m.a.v instantaneous velocity? • (6 votes) Alexis 8 years agoPosted 8 years ago. Direct link to Alexis's post “V= instantaneous velocity...” V= instantaneous velocity; Since acceleration and velocity are in the same equation, then the velocity is changing (accelerating) over time, so it would only make sense to pluck in the velocity of a specific time. However, Pavg= m . a . 1/2 (Vf+Vi) because you are using the AVERAGE velocity. So when using P = m . a . v , this is using the instantaneous velocity at a specific time and actually results in Pi (instantaneous power) not just P; you will need to use the average velocity in order to get the average power. In other words, to find Pavg you must specify that you are also using Vavg, and when using V (velocity at a specific time) you get Pi. (12 votes) kpobievans12 8 years agoPosted 8 years ago. Direct link to kpobievans12's post “How can we measure power” How can we measure power • (5 votes) Mark Zwald 8 years agoPosted 8 years ago. Direct link to Mark Zwald's post “If you know how much ener...” If you know how much energy is being converted per unit time, that is all you need to calculate power. Ex. measure the current from the battery in a circuit and multiply by voltage to get power. (11 votes) Yashika 8 years agoPosted 8 years ago. Direct link to Yashika's post “In "Can the concept of po...” In "Can the concept of power help us describe how objects move?", in the last few steps, why is Pavg = m a (1/2 vfinal). Isn't it 1/2 ( vfinal + vinitial) ? Thanks. • (4 votes) Cameron Garbutt 8 years agoPosted 8 years ago. Direct link to Cameron Garbutt's post “Vinitial is 0, therefore ...” Vinitial is 0, therefore it is 1/2Vfinal (9 votes) EvilBlood 7 years agoPosted 7 years ago. Direct link to EvilBlood's post “i cant understand the ex...” i cant understand the exercise 2(average power output) please someone explain me • (8 votes) deka 2 years agoPosted 2 years ago. Direct link to deka's post “1. power is all about con...” 1. power is all about converting whatever your work into the work with 1 second of window 2. in most cases, you do work for more than 1 sec. thus you have to do divide them by the time it take to do the work 3. but in exercise 2 (getting average power output of the laser device), the direction is opposite (converting less than 1 sec into 1 sec) hope this to help clear some of your confusion (1 vote) Marvin Pangilinan 5 years agoPosted 5 years ago. Direct link to Marvin Pangilinan's post “how does finding the area...” how does finding the area under, of a power vs time graph gives me an average power? shouldn't that be finding the slope? • (3 votes) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “Area under a power vs. ti...” Area under a power vs. time graph is the equivalent of integrating power with respect to time, which gives work or change in energy. To find average power, you could divide the integral by the time interval. (5 votes) Alex Sánchez 8 years agoPosted 8 years ago. Direct link to Alex Sánchez's post “Power can be expressed as...” Power can be expressed as P = Work/Change in Time, instead of P = Change in Energy/Change in Time, because the change in energy is basically the work done. Is this correct? • (4 votes) APDahlen 8 years agoPosted 8 years ago. Direct link to APDahlen's post “Hello Alex,In general I...” Hello Alex, In general I would agree. In both cases you end up with joules / time. Watch out for context. It seems each field of study has it's own vocabulary. Regards, APD (3 votes) Rize7 7 years agoPosted 7 years ago. Direct link to Rize7's post “About the last exercise, ...” About the last exercise, the first thing I thought about was 4.17/8 = 0.52, therefore 48% of the time is being spent on something else (fighting air drag) (if I sound somewhat weird, well,English is not my first language,sorry) • (3 votes) Quinn Ciccoretti 7 years agoPosted 7 years ago. Direct link to Quinn Ciccoretti's post “This is actually a very g...” This is actually a very good way to think about it. Its good to know equations sometimes, but you just solved a rather tricky problem in a very elegant way. Also, there were no English errors in your question in case you were wondering :) (4 votes) Hari Krishnan 7 years agoPosted 7 years ago. Direct link to Hari Krishnan's post “an engine of 4.9 kW power...” an engine of 4.9 kW power is used to pump water from a well 50 m deep.calculate the quantity of water in kilolitres which it can pump out in one hour. • (2 votes) redmermaid2006 4 years agoPosted 4 years ago. Direct link to redmermaid2006's post “In exercise 1 where did t...” In exercise 1 where did they get the number 116.5 squares? • (2 votes) deka 2 years agoPosted 2 years ago. Direct link to deka's post “sorry to say thisbut you...” sorry to say this another way to do it roughly is cancelling out areas of uphill and downhill if you hold the uphill area above ~0.45kW from hour 14 to 22.5, flip and split it into two, and then fit them into the downhill areas below ~0.45kW from hour 0 to 6 and 22.5 to 24, you may see a flat line along the 0.45W point from 0 to 24, and 0.45kW is not far from 0.485kW, the real solution it's not mathematical (1 vote) Lightning Driver, Esq. a year agoPosted a year ago. Direct link to Lightning Driver, Esq.'s post “If horsepower determines ...” If horsepower determines acceleration and not velocity, why do cars have a maximum speed? Why can't they accelerate to infinity? I know it's impossible, but why? • (2 votes) Gavankar Yashvi 4 months agoPosted 4 months ago. Direct link to Gavankar Yashvi's post “1. Exponential Power Requ...” 1. Exponential Power Requirements: As a car accelerates, the power needed to go faster increases exponentially. To double a car's top speed, its engine must be eight times as powerful. Aerodynamic lift over the car's body also escalates exponentially. Simultaneously, the traction of the tyres must improve to deliver the higher power required to the road. Consequently, cars capable of extreme speeds are significantly more expensive, and only a select few can afford them. 2. Tyre Limitations: Designing tyres capable of speeds over 300 kilometers per hour (186 miles per hour) is both difficult and expensive. Even tyres designed for such speeds need to be in pristine condition to operate safely. After a few weeks of normal use, they might not be safe at high speeds. 3. Human Reaction Time: At high speeds, human reaction times become critical. When travelling at 300 kilometres per hour, you cover more than 80 meters per second. Reacting to anything 100 meters ahead, such as debris on the road, becomes nearly impossible due to varying human reaction times. 4. Aerodynamic Limits: A powerful car, unrestricted by an onboard computer, eventually reaches a speed where it can't compress any more wind in front of itself. This compression increases with the square of speed, requiring exponentially more energy to overcome. Additionally, drag at the rear of the car from the induced vacuum becomes significant. Spoilers and air dams are used to increase downward force and prevent the car from trying to fly as wind speed under the vehicle increases. 5. Safety and Practicality: Speed limits exist primarily for safety. Unrestricted high speeds would make roadways significantly more dangerous. Cars become aerodynamically unstable due to wind resistance at extreme speeds, making them harder to handle. Running an engine at maximum speed constantly reduces its lifespan and puts additional strain on the car. Beyond physics, practical considerations like tyre safety, driver reaction times, and road conditions play a crucial role. In summary, while there's no fundamental reason why a road car couldn't go faster, practical limitations and safety concerns keep them well below infinity. 🚗 (1 vote)Want to join the conversation?
e.g. work_of_pushing_a_box_right = 30J, time = 3s
power = work/time = 30J/3s = 10J/1s = 10W
meaning you do 10J amount of work per 1 sec on the box
1) let's simplify the example
work_of_making_laser( pulse)s = 30J, time = 0.3s
power = work/time = 30/0.3 = 30 / 3/10 = 30 * 10/3 = 300/3 = 100W
or (more scientifically)
30/0.3 = 3*10^1 / 3*10^-1 = 3/3 * 10^1/10^-1 = 1 * 10^1*10^1 = 1 * 10^(1+1) = 1 * 10^(2) = 100W
2) the numbers in exercise 2 are simply having more 0's in them than this simple case
3) all you need to do is to plug different numbers (in scientific notation, preferably) into the equation of work/time
Then i applied the same ratios to the engine power
Is this a wrong way to think about it?
but you can simply count them with your hand (or eyes)
but faster than counting
and i do love estimating with eyes and mind-calculation, diving into actual calculation, and then comparing the two answers. and this case is one of the examples
