W = F•dx
F = 65 N,
dx = xf - xi = 0.03 m - (-0.03 m) = 0.06 m
W = 65 N × 0.06 m = 3.9 J
Here are a list of items I found.
some brushes
a armature
a permanent magnet
some slip rings
Answer: 
Explanation:
Given
Cross-sectional area of wire 
Extension of wire 
Extension in a wire is given by
where, 
for same force, length and material
Divide (i) and (ii)
Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
Answer:
54.3N
Explanation:
The normal force is perpendicular to the slope, so:
Normal Force = cos(37.2)(9.8*65).......507.39N
F(friction)=mu*F(normal)
F(friction)=(0.107)(507.39)
F(friction)=54.3N
