Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
2 answers:
Answer:
Explanation:
diameter of 24 gauge wire = 0.51 mm
diameter of 12 gauge wire = 2.05 mm
A.
length of wire, l = 1 m
diameter of wire, d = 0.51 mm
radius of wire, r = 0.51 /2 = 0.255 mm
resistivity of copper, ρ = 1.7 x 10^-8 ohm metre
The formula for the resistance of wire is given by
where, l is the length of wire and A be the area of crossection of wire
A = πr² = 3.14 x 0.255 x 10^-3 x 0.255 x 10^-3 = 2.04 x 10^-7 m²
The resistance of wire is
R = 0.083 ohm
B.
R = 0.083 ohm
V = 12 V
Let the current is I.
Use ohm's law
V = I R
12 = I x 0.083
I = 144.6 A
C.
I = 2 A
V = 12 V
Area of wire, A = 2.04 x 10^-7 m²
Resistivity of wire, ρ = 1.7 x 10^-8 ohm metre
Let the length of the wire is l.
V = I R
12 = 2 x R
R = 6 ohm
l = 72 m
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19.8 N force is tending to lift Rover vertically off the ground.
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According to the question,
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Answer:
θ = 13.7º
Explanation:
- According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
- The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
- As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:
- ΔK, is the change in kinetic energy, as follows:
- ΔU, is the change in the gravitational potential energy.
- If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:
- Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:
- Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:
- Replacing by the givens and the knowns, we can solve for sin θ, as follows:
⇒ θ = sin⁻¹ (0.236) = 13.7º