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.
X =(3.00x4.00 x3-1.00t x 2.00) x m
x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00 times m
Answer:
The brightness of bulb 1 dies because it is switched off.
A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:
v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour
Therefore, the average velocity is 1.257 cm per hour.
For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.
Answer:
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