Answer:
The graph appears to be in error.
The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5
The work done (F * S) is the area of the rhombus
1/2 * (5 +15) * 5 = 50 J
Answer:
2.667m/s to the north and 3.333 m/s to the west
Explanation:
According to law of momentum conservation, the total momentum should be conserved before and after the explosion.
Before the explosion, the momentum was
0.5*2 = 1 kg m/s to the west
Therefore the total momentum after the explosion should be the same horizontally and vertically.
Vertically speaking, it was 0 before the explosion. After the explosion:
0.2*4 + 0.3v = 0
0.3v = -0.8
v = -0.8/0.3 = -2.667 m/s
So the vertical component of the 0.3kg piece is 2.667m/s to the north
Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:
0.2*0 + 0.3V = 1
0.3V = 1
V = 1/0.3 = 3.333 m/s
So the horizontal component of the 0.3kg piece is 3.333 m/s to the west
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?
