A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s
1 answer:
Answer:
His final velocity is 15.8 m/s.
Step-by-step explanation:
Given:
Initial velocity of the driver is, m/s
Acceleration of the driver is, m/s²
Time taken to reach final velocity is, s.
The final velocity is given using the Newton's equations of motion as:
, where,
is the final velocity.
Now, plug in the given values and solve for .
Therefore, his final velocity is 15.8 m/s.
You might be interested in
Answer:
Explanation:
According to Coulomb's law, the magnitude of the electric force between two equals charges (q) is given by:
Here k is the coulomb constant and d is the distance between the charges. For two electrons we have:
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
