Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
Flux=q/e0
e0=8.85*10^-12
!!! Cube has 6 sides, so flux through one side is equal to total flux/6
Ans=9.11*10^(-6)/((8.85*10(-12))*6)=(approximately)1.72*10^5Nm^2/C
2.71 m/s fast Hans is moving after the collision.
<u>Explanation</u>:
Given that,
Mass of Jeremy is 120 kg (
)
Speed of Jeremy is 3 m/s (
)
Speed of Jeremy after collision is (
) -2.5 m/s
Mass of Hans is 140 kg (
)
Speed of Hans is -2 m/s (
)
Speed of Hans after collision is (
)
Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is
= 
Substitute the given values,
= 120 × 3 + 140 × (-2)
= 360 + (-280)
= 80 kg m/s
Linear momentum after the collision of Jeremy and Hans is
= 
= 120 × (-2.5) + 140 ×
= -300 + 140 ×
We know that conservation of liner momentum,
Linear momentum before the collision = Linear momentum after the collision
80 = -300 + 140 ×
80 + 300 = 140 ×
380 = 140 ×
380/140=
= 2.71 m/s
2.71 m/s fast Hans is moving after the collision.
Depends on how well built he is probably for the average American 8 MPH
