An object on a straight line travels from point A to point B at a constant speed of 60 kmh-1 and returns to point A at a constan
1 answer:
You might be interested in
Answer:
Explanation:
When activated, the receptor most likely prompting the production of saliva is the taste receptor. When food enters the mouth, the salivary glands produce the saliva upon the sensation of taste.
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
