Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
= 1.3l, and 
= 
So that;
= (ρ) ÷ = (ρ × 1.3l) ÷ ()
= (1.3lρ) ÷ ()
= 
× [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
= 1.69R
Where is the new resistance, is the new length, and is the new area after stretching the wire.
The change in resistance of the wire = - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = 
× 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
Charge dQ on a shell thickness dr is given by
dQ = (charge density) × (surface area) × dr
dQ = ρ(r)4πr²dr
∫ dQ = ∫ (a/r)4πr²dr
∫ dQ = 4πa ∫ rdr
Q(r) = 2πar² - 2πa0²
Q = 2πar² (= total charge bound by a spherical surface of radius r)
Gauss's Law states:
(Flux out of surface) = (charge bound by surface)/ε۪
(Surface area of sphere) × E = Q/ε۪
4πr²E = 2πar²/ε۪
<span>E = a/2ε۪
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Answer:
Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.
Explanation:
The given data is as follows.
Angular velocity (
) = 2.23 rps
Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
= 2.23 revolutions per second
= 
= 14.01 rad/s
Now, tangential speed will be calculated as follows.
Tangential speed, v = 
= 0.379 x 14.01
= 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.
