Answer:
Explanation:
As per Ampere's law the magnetic field at the surface of the wire is given as
here we have
so we will have
now again we use same value of current but wire with double the diameter
so the magnetic field at the surface is given as
so we have
Answer:
final kinetic energy of the hammer is 10 kJ
Explanation:
As we know that there is no non conservative force on the system
So here we can use the theory of mechanical energy conservation
So we will have
here we know that
from above expression now
so final kinetic energy of the hammer is 10 kJ
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
