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Tamiku [17]
2 years ago
10

Please helppppppppppppp

Physics
1 answer:
sergejj [24]2 years ago
8 0

Answer:

First 8 if wrong then correct me. Second it's direction correct me if wrong.

Explanation:

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To get the best possible answer. (sorry if im wrong)
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2 years ago
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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

6 0
3 years ago
PLEASE HELP ME!
PilotLPTM [1.2K]

Answer:

a) A = 3 cm,  b)  T = 0.4 s,   f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

          A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

          T = 0.4 s

the period is the inverse of the frequency

         f = 1 /T

         f = 1 /, 0.4

         f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

8 0
2 years ago
What is the formula that represents the simplest ratio of the atoms in the components
Dafna1 [17]

Answer:

Empirical formula

<em>Hope</em><em> </em><em>it'll</em><em> </em><em>help</em><em>!</em>

<em>stay</em><em> </em><em>safe</em><em>:</em><em>)</em>

5 0
3 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

\theta(t) = \theta_0 cos(5\pi t)

Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
3 years ago
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