Answer:
14 secs and velocity will be 48
Explanation:
- Mass=m=44.9kg
- Distance=s=20.5m
- initial velocity=u=0m/s
- Final velocity=v=62m/s
- Force=F
- Acceleration=a
Using third equation of kinematics
Apply Newton's second law
Well, just for fun, let's see hat the form of the limit is
<span>v = dx/dt = b + 3 c t^2 </span>
<span>= 1.5 + 1.92 t^2 </span>
<span>========================= </span>
<span>now </span>
<span>from 1 to 3 </span>
<span>x(3)= 1.5*3 +.64*27=4.5+17.28 </span>
<span>= 21.78 </span>
<span>x(1) = 1.5+.64 = 2.14 </span>
<span>so </span>
<span>x(3)-x(1) = 19.64 </span>
<span>delta t = 3 - 1 = 2 seconds </span>
<span>so </span>
<span>v during interval = 19.64/2 </span>
<span>= 9.82 m/s </span>
<span>now check that with calculus v in the middle of that range </span>
<span>v(2) =1.5+ 1.92 (4) = 9.18 m/s </span>
<span>that makes sense, the parabola is headed up so the instant speed i the middle should be a bit less than average. </span>
Answer:(a)does not change
Explanation:
Given
Submarine is 50 m below the surface
Now submarine descends to 100 m and stops
As we know that pressure difference at a depth of h is given by
where =Density of fluid
g=acceleration due to gravity
h=height below free surface of liquid
buoyant force is equal to volume displaced by object
as the submarine is already in water therefore buoyant force will remain same.
the neutrons through the nucleus?
Explanation: