When a car is weighed, it is driven slowly on a horizontal floor over a scale that records a reading as the front wheels go over
1 answer:
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For an AC circuit:
I = V/Z
V = AC source voltage, I = total AC current, Z = total impedance
Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.
For a resistor R, inductor L, and capacitor C, their impedances are given by:
= R
R = resistance
= jωL
ω = voltage source angular frequency, L = inductance
= -j/(ωC)
ω = voltage source angular frequency, C = capacitance
Given values:
R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s
Plug in and calculate the impedances:
= 217Ω
= j(220)(0.875) = j192.5Ω
= -j/(220×6.75×10⁻⁶) = -j673.4Ω
Add up the impedances to get the total impedance Z, then convert Z to polar form:
Z = +
+
Z = 217 + j192.5 - j673.4
Z = (217-j480.9)Ω
Z = (527.6∠-1.147)Ω
Back to I = V/Z
Given values:
V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)
Z = (527.6∠-1.147)Ω (from previous computation)
Plug in and solve for I:
I = (30.0∠0+220t)/(527.6∠-1.147)
I = (0.0569∠1.147+220t)A
To get the voltages of each individual component, we'll just multiply I and each of their impedances:
= I×
= I×
= I×
Given values:
I = (0.0569∠1.147+220t)A
= 217Ω = (217∠0)Ω
= j192.5Ω = (192.5∠π/2)Ω
= -j673.4Ω = (673.4∠-π/2)Ω
Plug in and calculate each component's voltage:
= (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V
= (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V
= (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V
Now we have the total and individual voltages as functions of time:
V = (30.0∠0+220t)V
= (12.35∠1.147+220t)V
= (10.95∠2.718+220t)V
= (38.32∠-0.4238+220t)V
Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:
V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V
= 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
= 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
= 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V
Answer:
Giant impacts in its early history may have torn away much of its original crust and mantle.
So Mercury has a giant iron core, <em>but why?</em>
There are three general ideas.
1. Mercury was once the core of a gas giant that vaporized when the Sun became a fully-fledged star. This gas giant was probably more like Uranus or Neptune than Jupiter or Saturn.
There are significant issues here, especially that Mercury's existing surface has been exposed to space since the solar system's initial beginnings. On Mercury, remnants of the first massive bombardment have been found.
2. Mercury formerly had a deeper mantle and was a bigger terrestrial planet. During the early history of the solar system, a dwarf planet made a massive impact that robbed the planet of its mantle.
The existing surface of Mercury has been exposed to space since the very beginning of the solar system, which raises major issues once more.
Due to the factors in explanation 1, characteristics from the first significant bombardment have been found on Mercury. The quantity of low temperature volatiles that are trapped and sometimes sublimate out of the crust, causing hollows, is also a major issue. By such an impact, these low temperature volatiles—such as sodium, sulphur, magnesium, etc.—would have been completely pushed out.
3. A mostly iron-rich body developed, and "rock steam"—a gaseous mixture of atomic oxygen and silica—condensed and accreted onto it from the protoplanetary disk near the Sun. Small rock particles with lower temperature volatiles were subsequently formed into a crust. Additionally, it appears that iron was concentrated in the inner region of the protoplanetary disk by the Sun's early, extremely powerful magnetosphere.
Whilst there are a few minor issues with this, this appears by far the most likely scenario.
