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2 years ago

* Without heat you would be hot warm none of these cold

Physics

1 answer:

Arturiano [62]2 years ago

4 0

Answer:

It depends on the environment you're in but I'd go with warm

Explanation:

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What are the intercepts of the graphed function?

EastWind [94]

Answer:

the intercepts of a graph are point at which the graph crosses the axes

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3 years ago

The voltage across the diode indicates the energy given to charge carriers (electrons and holes, but more about that later in th

Elis [28]

Answer:

the charge carriers have an energy 2.8 10⁻¹⁹ J

Explanation:

The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.

The energy of a photon is given by the Planck expression

E = h f

the speed of light, wavelength and frequency are related

c = λ f

we substitute

E = $\frac{h \ c}{\lambda}$

a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m

we calculate the energy

E = 6.626 10⁻³⁴ 3 10⁸/700 10⁻⁹

E = 2.8397 10⁻¹⁹J

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2 years ago

A speaker vibrates at a frequency of 200 hz. What is it's period ?

Katen [24]

200 Hz = 200 cycles per sec

<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>

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3 years ago

A car traveling at 50 m/s comes to a stop in 5 seconds. What is the<br> acceleration of this car?

Alexandra [31]

Answer:10

Explanation:

You have to do speed divided by time so your answers 10

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2 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

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3 years ago