bob24 said:
3. The attempt at a solution
First I converted the 90km/h to m/s and got 25m/s/
OK
Then I did F=mg with m=1500 and g=9.81m/s^2
You can't use g hare. Find the acceleration. And then the displacement.
rl.bhat said:bob24 said:
3. The attempt at a solution
First I converted the 90km/h to m/s and got 25m/s/
OK
Then I did F=mg with m=1500 and g=9.81m/s^2
You can't use g hare. Find the acceleration. And then the displacement.
----
Ok I got acceleration=5m/s^2, and then I did Vf^2-Vi^2=2aS and got 62.5 meters for S.
So do I do
F=ma
F=1500 x 5
F=7500 N?
rl.bhat said:Now average power =ForceX average velocity
An average power problem is a type of physics problem that involves calculating the average power of an object or system over a given period of time. This is typically done by finding the rate at which energy is transferred or transformed within the system.
Average power is the average rate at which energy is being transferred or transformed over a period of time, while instantaneous power is the rate at which energy is being transferred or transformed at a specific moment in time. Average power takes into account the entire time period, while instantaneous power only considers a single moment.
Some common examples of average power problems include calculating the average power output of an engine, finding the average power consumption of an electrical appliance, and determining the average power generated by a wind turbine over a certain period of time.
The units for average power are watts (W) or joules per second (J/s). These units represent the rate at which energy is transferred or transformed over a given period of time.
To solve an average power problem, you will need to know the amount of energy transferred or transformed over a given period of time, as well as the time it took for the energy transfer or transformation to occur. Once you have this information, you can use the formula P = E/t, where P is the average power, E is the energy, and t is the time period.