It's a Newton Meter Those two are multiplied to get a joule.
Answer:force equals to rate of change of momentum
Explanation:
F=force
t=time
m=mass
v=final velocity
u=initial velocity
(mv-mu)/t=rate of change of momentum
Force=rate of change of momentum
F=(mv-mu)/t
Answer:
The best estimate of the depth of the well is 2.3 sec.
Explanation:
Given that,
Record time,





We need to find the best estimate of the depth of the well
According to record time,
We can write of the record time





Here, all time is nearest 2.3 sec.
So, we can say that the best estimate of the depth of the well is 2.3 sec.
Hence, The best estimate of the depth of the well is 2.3 sec.
Answer:
t = 2.58*10^-6 s
Explanation:
For a nonconducting sphere you have that the value of the electric field, depends of the region:

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the sphere = 10.0/2 = 5.0cm=0.005m
In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

with this values of a you can use the following formula:

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>