Acceleration = (change in speed) / (time for the change)
= (49 m/s) / (5 seconds)
= (49 / 5) m/s / s
= 9.8 m/s²
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
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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?
Answer:
+2m/s
Explanation:
average velocity = displacement traveled / total time taken
= +12m/ 6s
= +2 m/s
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **
Ans: The vertical distance = y = M/(ρA)
Explanation:Support the vertical distance = y
Object's density = M/(A*h) (since A*h = volume)
By applying the condition,
(M/(Ah))/ρ = y/h
M/(ρAh) = y/h
y = M/(ρA)
