Name three possible end stages of a star

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

$\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}$

$Ln(\frac{1}{2}) = -\frac{0.01}{RC}$

$- RC = \frac{0.01}{Ln(1/2)}$

$RC = 0.014s$

We know that RC is equal to the time constant, then

$T = RC = 0.014s = 14ms$

Therefore the time constant for the process is about 14ms

5 0

3 years ago

This instrument can used to visualize atoms

Lisa [10]

Answer:

can someone follow me

Explanation:

Plqae

3 0

2 years ago

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I am one of the most dynamic processes of Earth. I help recycle Earth's materials. I move Earth's crustal plates, creating many

Sergio039 [100]

I Can only think of tectonic plates or gravity

7 0

3 years ago

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A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass

Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

$\omega = \frac{v}{R}$

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

$2.2388= (\frac{7}{5}) (v)^2$

$v=1.26m/s$

7 0

3 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Nutka1998 [239]

Answer:

$6.99535\times 10^{-6}\ V/m$

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

Intensity of Electric radiation is given by

$I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2$

Intensity of Electric radiation is given by

$I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m$

The amplitude of the electric field vector is $6.99535\times 10^{-6}\ V/m$

6 0

3 years ago

Name three possible end stages of a star​

Name three possible end stages of a star