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1 answer:
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1) 12 cm
2) 3 N
Explanation:
1)
The relationship between force and elongation in a spring is given by Hooke's law:
where
F is the force applied
k is the spring constant
x is the elongation
For the spring in this problem, at the beginning we have:
So the spring constant is
Later, the force is tripled, so the new force is
Therefore, the new elongation is
2)
In this second problem, we know that the elongation of the spring now is
From part a), we know that the spring constant is
Therefore, we can use the following equation to find the force:
And substituting k and x, we find:
So, the force to produce an elongation of 6 cm must be 3 N.
Answer:
electric field in the wide wire is
E₂ =
Explanation:
given
length of the copper wire = L
radius of the copper wire r₁ = b
length of the second copper wire = L
radius of the second copper wire r₂ = 2b
electric field in the narrow wire = E₁=E
recall
resistance R = ρL/A
where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.
Resistance of narrow wire, R₁
R₁ = ρL/A
where A = πb²
R₁ = ρL/πb²---------- eqn 1
Resistance of wide wire, R₂
R₂ = ρL/A
where A = π(2b)²
R₂ = ρL/π(2b)²
R₂ = ρL/4πb²-------------- eqn 2
R₂ = ¹/₄(ρL/πb²)
comparing eqn 1 and 2
R₁ = 4R₂
calculating the current in the wire,
I = E/(R₁ + R₂)
recall
R₁ = 4R₂
∴ I = E/(4R₂ + R₂)
I = E/5R₂
calculating the potential difference across R₁ & R₂
V₁ = IR₁
I = E/5R₂
∴ V₁ = ER₁/5R₂
R₁ = 4R₂
V₁ = 4ER₂/5R₂
∴V₁ = ⁴/₅E
potential difference for R₂
V₂= IR₂
I = E/5R₂
∴ V₂ = ER₂/5R₂
V₂ = ER₂/5R₂
∴V₂ = ¹/₅E
so, electric field E = V/L
for narrow wire E₁ = V₁/L ----------- eqn 3
for wide wire, E₂ = V₂/L------------ eqn 4
compare eqn 3 and 4
E₂/E₁ = V₂/V₁( L is constant)
E₂/E₁ = ¹/₅E/⁴/₅E
E₂ = E₁/4
note E₁ = E
∴E₂ =