a=change in velocity/time, is this the formula i use for this question:
a car takes 5 seconds to speed up from 10 m/s to 60 m/s. what is the acceleration?
yes!
so what do i do from there?
If it goes from 10 to 60, what is it's CHANGE in velocity (speed)?
is it 50
Yes...so now you know the change in velocity, and the time, so you can plug them into your formula.
i have 50/5 i divide right or wrong
Right! you divide.
so a=10
You got it!
ok thx now can u help me through the steps to this question?
what is the momentum of a 500kg object moving at 10 m/s?
is it p=mv
That is the correct formula, and you have all the information. 500 kg is the mass, m, and 10 m/s is the velocity, v.
so its p=500x10
correct.
so p=5000
correct
this is were i get stuck:
how much force is neede to stop in 20 seconds?
Please post this as a new question, and complete the question...how much force is needed to stop what in 20 seconds?
To calculate the force needed to stop an object in a given time, you need to know the change in momentum and the time it takes to stop. The formula to calculate force is:
Force = (Change in Momentum) / (Time)
If you have the mass of the object and the initial velocity, you can calculate the initial momentum. Then, if you know the final velocity (which would be zero in this case since the object is stopping), you can calculate the final momentum. The change in momentum is the difference between the initial and final momentum.
So, the steps to calculate the force needed to stop an object in 20 seconds would be:
1. Determine the mass of the object.
2. Calculate the initial momentum using the formula p = mv, where p is momentum, m is mass, and v is initial velocity.
3. Calculate the final momentum (which would be zero in this case).
4. Find the change in momentum by subtracting the final momentum from the initial momentum.
5. Divide the change in momentum by the time of 20 seconds to get the force needed.
If you provide the mass and initial velocity, and confirm that the object is stopping, I can assist you with the calculations.