Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The  angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;

(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
 
        
             
        
        
        
Answer:
a)  x(t) = 10t + (2/3)*t^3
b) x*(0.1875) = 10.18 m
Explanation:
Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.
Given:
- v(t) = 10 + 2*t^2 (radar gun)
- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)
Find:
-The position x of horse as a function of time t in radar system.
-The position of the horse at x = 2m in our coordinate system
Solution:
- The position of horse according to radar gun:
                               v(t) = dx / dt = 10 + 2*t^2
- Separate variables:
                               dx = (10 + 2*t^2).dt
- Integrate over interval x = 0 @ t= 0
                               x(t) = 10t + (2/3)*t^3
- time @ x = 2 :
                               2 = 10t + (2/3)*t^3
                               0 = 10t + (2/3)*t^3 + 2
- solve for t:
                               t = 0.1875 s
- Evaluate x* at t = 0.1875 s
                               x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3
                               x*(0.1875) = 10.18 m
 
        
             
        
        
        
C. A step-by-step process that takes time, and is essential for learning physics concepts.
        
                    
             
        
        
        
The resistance of the cylindrical wire is  .
.
Here  is the resistance,
 is the resistance,  is the length of the wire and
 is the length of the wire and  is the area of
cross section. Since the wire is cylindrical
 is the area of
cross section. Since the wire is cylindrical  .
 .
Comparing two wires,

Dividing the above 2 equations,

Since 
The above ratio is

We also have,

The current through the Silver wire will be 4.23 times the current through the original wire.