Explanation:
F = 20N m= m1 a=10m/s²
m=m2 a=5m/s²
F = ma
<u>for the first one</u><u>:</u><u> </u>
f=m1 × a
20 = m1 ×10
20=10m1
m1=20/10
m1=2
<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>
f=m2×a
20=m2×5
m2= 20/5
m2= 4
since F=ma
F=(m1+m2) ×a
F =(4+2)×a
F =6×a
F=20(from the question above )
20=6×a
a=20/6
a=3.33
By calculation, the diameter of the wire is 2.8 * 10^-3 m.
<h3>How do we obtain the length?</h3>
The following data are given in the question;
Mass of the wire = 1.0 g or 1 * 10^-3 Kg
Resistance = 0.5 ohm
Resistivity of copper = 1.7 * 10^-8 ohm meter
Density of copper = 8.92 * 10^3 Kg/m^3
V = m/d
But v = Al
Al = m/d
A = m/ld
Resistance = ρl/A
= ρl/m/ld =
l^2 = Rm/ρd
l = √ Rm/ρd
l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3
l = 1.82 m
A = πr^2
Also;
A = m/ld
A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3
Area of the wire = 6.2 * 10^-5 m^2
r^2 = A/ π
r = √A/ π
r = √6.2 * 10^-5 m^2/3.142
r = 1.4 * 10^-3 m
Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m
Learn more about resistivity:brainly.com/question/14547003
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Missing parts;
Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?
Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2
s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft
96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.
v(2) = 16
v(3) = -16
