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<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F - 
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F -
= m x a
F -
= 50 x 0
F -
= 0
F =
-------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force
is opposing the movement of the box.
∑F = 1.5F - 
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F -
= m x a
1.5F -
= 50 x 0.1
1.5F -
= 5 ---------------------(iii)
<em>Substitute </em>
<em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
Answer: A) Force = 3.841*10^-18 N.
B) force (f) is 4.30* 10^12 times greater than the weight (Fg).
Explanation: mass of electronic charge = 9.11*10^-31kg
v = final velocity = 6.80*10^5 m/s
u = initial velocity = 2.40 * 10^5 m/s
S= distance covered = 4.8cm = 0.048m
a = acceleration
Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.
Thus, recall that
v² = u² + 2aS
(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)
46.24 * 10^10 = 5.76 * 10^10 + 0.096a
46.24 *10^10 - 5.76* 10^10 = 0.096a
40.48* 10^10 = 0.096a
a = 40.48 * 10^10/0.096
a = 4.2167*10^12m/s².
Force = mass * acceleration
Force = 9.11*10^-31 * 4.2167*10^12
Force = 3.841*10^-18 N.
Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²
Fg= 9.11*10^-31 * 9.8
Fg = 8.9278* 10^-30 N
By comparing the force and the weight, we have that
F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.
This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).
To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.
This definition is described in the following equation as,

Where,
permeability of free space
Number of turns in solenoid 1
Number of turns in solenoid 2
Cross sectional area of solenoid
l = Length of the solenoid
Part A )
Our values are given as,





Substituting,



PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.