Newton’s 2nd law states that Force is equal to
the product of mass (m) and acceleration (a):
F = m a --->
1
While in magnetic forces, force can also be expressed as:
F = q v B --->
2
where,
q = total charge
v = velocity = 45 cm / s = 0.45 m / s
B = the magnetic field = 85 T
First we solve for the total charge, q:
q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 ×
10^-19 C / electron)
q = 1.594 × 10^-19 C
We equate equations 1 and 2 then solve for acceleration a:
m a = q v B
a = q v B / m
a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg
a = 160,437,862.2 m/s^2
Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.
<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
<span>Your answer isStratified drift</span>
Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector 
is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, 
must be in the north direction.
Let 
is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector 
and head of the vector must lie on the north-south line.
Now, for this condition, from the triangle OAB
Hence, the kayaker must paddle in the direction of in the north of east direction.
