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Sonbull [250]
2 years ago
13

How does beam spreading affect the temperature on earth and the seasons

Physics
1 answer:
navik [9.2K]2 years ago
8 0
Get to school and learn boi
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A has less energy and lower frequency, while B has greater energy and higher frequency.
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niobium metal becomes a superconductor when cooled below 9 k. its superconductivity is destroyed when the surface magnetic field
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Niobium wire with a 2.60 mm diameter has a maximum current capacity of 500 A while still remaining superconducting.

<h3>Describe the present.</h3>

Current is the rate at which charge passes from one point on a circuit to another. In a circuit, a significant current flows when several coulombs or charge pass over the cross section of a wire. When the charge carriers are firmly packed inside the wire, high currents can be generated at low speeds.

<h3>What do current and electron actually mean?</h3>

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For high and low tides differences would they be caused by the moon and how?
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2 years ago
How many Oxygen (O) atoms are in the following? H2O + CO2<br><br> 4<br><br> 3<br><br> 2
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3 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
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