Hybrid
<u>Hybrid</u> modified the concept by adding an internal combustion engine and marketing hybrids that were part electric and part gas powered.
- The driving wheels of hybrid vehicles receive power from their drivetrains.
- A hybrid car has numerous sources of propulsion.
- There are numerous hybrid configurations.
- A hybrid vehicle might, for instance, get its energy from burning gasoline while alternating between an electric motor and a combustion engine.
- Although they have primarily been employed for rail locomotives, electrical vehicles have a long history of integrating internal combustion and electrical transmission, like in a diesel-electric power-train.
- Because the electric drive transmission directly substitutes the mechanical gearbox rather than serving as an additional source of motive power, a diesel-electric powertrain does not meet the definition of a hybrid.
- Only the electric/ICE hybrid car type was readily accessible on the market as of 2017.
- One type used parallel operation to power both motors at the same time.
- Another ran in series, using one source to supply power solely and the other to supply electricity.
- Either source may act as the main driving force, with the other source serving to strengthen the main.
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The diameter of the wire is 2.8 * 10^-3 m.
<h3>What is the length?</h3>
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
A = 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:

It will take 10 seconds to travel 200m at a speed of 10m/s
Explanation:
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Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
The correct one is that the force on B is half of the force on A
Explanation:
Because radius for the inside of the curve is half the radius for the outside and Car A travels on the inside while car B, travels at equal speed on the outside of the curve. Thus force on B will be half on A