Which equation is most likely used to determine the acceleration from a velocity vs:time graph?
2 answers:
Answer: Hello there!
in this case, you have a graph of velocity vs time, then in this graph is represented the function v(t).
The acceleration is defined as the integral with respect to the time of the velocity v(t), then we could calculate the mean acceleration between two times in a next way:
this is the slope between the times t2 and t1, where t2>t1 and V(t2) is the velocity at the time t2 (the final velocity) and v(t1) is the velocity at the time t1 (the initial velocity). In this case, a is the mean acceleration between these times.
You might be interested in
a) Average power: 1425 W
b) Instantaneous power at 3.0 sec: 2850 W
Explanation:
a)
The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation
where
s = 18 m is the displacement along the ramp
u = 0 is the initial velocity
t = 3.0 s is the time taken
a is the acceleration of the object along the ramp
Solving for a,
Now we can apply Newton's second law to find the net force on the object:
This net force is the resultant of the applied force forward () and the component of the weight acting backward (
), so we can find what is the applied force:
where
m = 24 kg is the mass of the object
is the acceleration of gravity
Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:
where s = 18 m is the displacement.
Therefore the average power needed is:
b)
The instantaneous power at any point of the motion is given by
where
is the force applied
v is the velocity of the object
We already calculated the applied force:
While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:
And therefore, the instantaeous power at 3.0 sec is:
Learn more about power:
#LearnwithBrainly
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find: