Answer:
B) 10 minutes
Explanation:
When the size of the problem is 100 then it takes 1 minute to solve
So, when the size of the problem is 1 then it takes the following minutes
For when the size of the problem is 1000 we have
So, it will take 10 minutes to solve the algorithm of size 1000
Answer:
(a)Look at the attached graphic
(b)
(b)-1 Equation 1 : m1= 5kg
50-F1= 5 *a
(b)-2 Equation 2 : m2= 3kg
F1-F2= 3 *a
(b)-3 Equation 3 : m3= 2kg
F2 = 2*a
(c) F1 =25 N
(d) F2 =10 N
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) Draw the free-body diagrams for each of the boxes
Look at the attached graphic
(b) Write Newton’s equation for each mass along the horizontal direction.
Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg
<em>Look</em> <em>m1 free-body diagram:</em>
∑Fx = m1*a
50-F1= 5 *a Equation 1
<em>Look</em> <em>m2 free-body diagram:</em>
∑Fx = m2*a
F1-F2= 3 *a Equation 2
<em>Look</em> <em>m3 free-body diagram:</em>
∑Fx = m3*a
F2 = 2*a Equation 3
(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?
<em>Look</em> <em>Free body diagram of the mass set</em>
∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg
50 = 10*a
a= 50/10 = 5 m/s²
We replace a = 5 m/s² in the equation 1:
50-F1= 5 *5
50-25= F1
F1 = 25 N
<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>
We replace a= 5 m/s² in the equation 3
F2 = 2*5 = 10 N
