The speed of the ball is 101miles/hr.
A mile is a unit of length that is exactly 1,609.344 metres long. Similarly, 5,280 feet or 1,760 yards make up one mile. The mile is an imperial and common US measurement of distance.
We just have to deal with unit conversions.
One mile is 5280 feet, or 1 ft = 0.000189
The speed of the ball in miles per hour is

So, the speed of the ball in miles per hour is 101miles/hr.
Learn more about miles here;
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Answer:
a)
b) 
Explanation:
a) The displacement of the first object is 22.5 m, so we can use the next equation:



positive acceleration.
b) Using the same equation we can find the second value of the acceleration:


positive acceleration.
I hope it helps you!
Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;

where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m