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3 years ago

Facts about Alessandro Volta?

Physics

2 answers:

Mila [183]3 years ago

8 0

Invented the battery
Italian
Has 3 kids

Firdavs [7]3 years ago

4 0

<u>Alessandro Volta :</u>

- He was born on February 18, 1745 in Côme (Italia).
- He died on March 5, 1827 in Côme (Italia).
- Inventor of the electric battery. (voltaic pile)
- He was an Italian scientist.

<em>Do not forget to vote for a "best answer" to recover a quarter of your points! =)</em>

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I really need help on this question!

steposvetlana [31]

Answer:

option "c"

Explanation:

because in gases molecules are further apart and move very quickly

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3 years ago

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

Alexxandr [17]

Answer:

magnitude of A − B = 15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = $\sqrt{(-15)^{2}+(-5)^{2}}$

|A - B| = $\sqrt{250}$

|A - B| = 15.81 km

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3 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

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3 years ago

Suppose that a star has a spectrum that includes red, blue, and violet lines spaced in the pattern of the lines from hydrogen bu

ladessa [460]

Answer:

It can be concluded that the star is moving away from the observer.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest for this case is 434 nm and 410 nm ( $\lambda_{0} = 434nm$ , $\lambda_{0} = 410nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Since, $\lambda_{measured}$ (444nm) is greater than $\lambda_{0}$ (434 nm) and $\lambda_{measured}$ (420nm) is greater than $\lambda_{0}$ (410 nm), it can be concluded that the star is moving away from the observer

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3 years ago