Answer:
200 N
Explanation:
Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.
So, E = σ/ε = FL/eA
Now, since at break extension = e.
So making e subject of the formula, we have
e = FL/EA = FL/Eπr² where r = radius of metal wire
Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r
So, e' = F'(2L)/Eπ(2r)²
e' = 2F'L/4Eπr²
e' = F'L/2Eπr²
Since at breakage, both extensions are the same, e = e'
So, FL/Eπr² = F'L/2Eπr²
F = F'/2
F' = 2F
Since F = 100 N,
F' = 2 × 100 N = 200 N
So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.
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Answer:
dt/dx = -0.373702
dt/dy = -1.121107
Explanation:
Given data
T(x, y) = 54/(7 + x² + y²)
to find out
rate of change of temperature with respect to distance
solution
we know function
T(x, y) = 54 /( 7 + x² + y²)
so derivative it x and y direction i.e
dt/dx = -54× 2x / (7 +x² + y²)² .........................1
dt/dy = -54× 2y / (7 + x² + y²)² .........................2
now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2
dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²
dt/dx = -0.373702
and
dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²
dt/dy = -1.121107
Answer:
look it up im not a sheaperd sorry
Explanation: