Answer:
A) micro defects are left behind on the surface of metal components during the manufacturing process. These defects, in the form of micro-cracks or pits, becomes initiation sites for crack propagation or corrosion. Removing these imperfections on the surface of metal parts by electroplating greatly improves the life of metal components.
B) it will reduce fatigue crack growth.
Dispersion hardening involves the inclusion of small, hard particles in the metal, thus restricting the movement of dislocations, and thereby raising the strength properties. In dispersion hardening it is assumed that the precipitates do not deform with the matrix and that a moving dislocation bypasses the obstacles (precipitates) by moving in the clean pieces of crystal between the precipitated particles.
C) stress concentrations such as changes in section with sharp corners caused yielding, which will typically occur first at a stress concentration. For ductile materials localised plastic deformation can cause a redistribution of stress, enabling the component to continue to carry load. Brittle materials will typically fail at the stress concentration. Repeated loading may cause a fatigue crack to initiate and slowly grow at a stress concentration leading to the failure of even ductile materials. Fatigue cracks always start at stress raisers, so removing such defects increases the fatigue strength.
Answer:
Due to touch of teflon, its charge will reduce but will not go to zero. Some amount of its initial charge will be transferred to Aluminum rod. So, aluminum rod will have a non-zero negative charge.
Explanation:
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Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂