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coldgirl [10]
3 years ago
14

What is Kirchoff law​

Physics
1 answer:
Juli2301 [7.4K]3 years ago
6 0

Answer:

.

Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff.[1] This generalized the work of Georg Ohm and preceded the work of James Clerk Maxwell. Widely used in electrical engineering, they are also called Kirchhoff's rules or simply Kirchhoff's laws. These laws can be applied in time and frequency domains and form the basis for network analysis.

Both of Kirchhoff's laws can be understood as corollaries of Maxwell's equations in the low-frequency limit. They are accurate for DC circuits, and for AC circuits at frequencies where the wavelengths of electromagnetic radiation are very large compared to the circuits.

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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.20 nm?
Nat2105 [25]
The momentum of a photon is given by the following relationship:
p= \frac{h}{\lambda}
where
h is the Planck constant
\lambda is the photon wavelength

For the photon in our problem, \lambda=0.20 nm = 0.20 \cdot 10^{-9}m, so its momentum is
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{0.20 \cdot 10^{-9} m}=3.3 \cdot 10^{-24} kg m/s

The electron must have the same momentum of this photon, and its momentum is given by (in the non-relativistic approximation)
p=mv
where
m is the electron mass
v is its speed

Re-arranging this formula, we can calculate the electron speed:
v= \frac{p}{m}= \frac{3.3 \cdot 10^{-24} kg m/s}{9.1 \cdot 10^{-31} kg} =3.6 \cdot 10^6 m/s

And this velocity is quite small compared to the speed of light (c=3 \cdot 10^8 m/s), so the non-relativistic approximation that we used for the electron's momentum is valid.
8 0
3 years ago
Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are
Doss [256]

Answer:

r=0.127

Explanation:

When  connected in series

Current = I

When connected in parallel

Current = 10 I

We know that equivalent resistance

In series  R = R₁+R₂

in parallel  R= R₁R₂/(R₂+ R₁)

Given that voltage is constant (Vo)

V = I R

Vo = I (R₁+R₂)  ------------1

Vo = 10 I (R₁R₂/(R₂+ R₁)) -------2

From above equations

10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)  

10  R₁R₂ =  (R₁+R₂) (R₂+ R₁)

10  R₁R₂  = 2  R₁R₂  + R₁² + R₂²

8 R₁R₂  =     R₁² + R₂²

Given that

r =  R₁/R₂

Divides by R₂²

8R₁/R₂  = ( R₁/R₂)²+ 1

8 r = r ² + 1

r ² - 8 r+ 1 =0  

r= 0.127 and r= 7.87

But given that R₂>R₁  It means that r<1 only.

So the answer is r=0.127

3 0
3 years ago
Surviving a ship wreck, what is the minimum mass of wood (density 60% that of sea water) necessary to support a 70kg woman stand
Shkiper50 [21]

To solve this problem we will apply the concepts related to the principle of archimedes. for which we will summarize that the bearing force must be equivalent to the mass of the individual and the mass of the way. Said mass of wood will be expressed in terms of density and volume. Finally with the values found we will proceed to find the Volume of the wood and thus find the mass.

(M_{woman}+M_{wood})g=F_B

(M_{woman}+M_{wood})g=V\rho_{water} g

(M_{woman}+V \rho_{wood})g=V\rho_{water} g

(M_{woman}+V \rho_{wood})=V\rho_{water}

For the relation between density and Volume we have that,

mass=density\times volume

Where given that,

Mass of woman, M_{man}=70kg

If the density of the wood is 60% the density of the water we will have to

Density of water, \rho_{water}=1000kg/m^{3}

And density of wood is

\rho_{wood}=600kg/m^{3}

Now the mass of the man can be expressed as

M_{woman}=V\rho_{water} -V\rho_{wood}

V=\frac{M_{woman}}{\rho_{water} -\rho_{wood}}=\frac{70\;kg}{1000\;kg/m^{3}-600\;kg/m^{3}}

V=0.175m^{3}

Mass of wood required is given by,

M_{wood}=V\rho_{wood}=0.175m^{3}\times600kg/m^{3}

\mathbf{\therefore M_{wood}=105kg}

Therefore the minimum mass necessary to support a 70kg Woman is 105Kg.

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If 1.3 grams of air has a volume of 1000cm ^3, what is the density of air?
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Density is equal to mass/volume, so in your case:
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