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-BARSIC- [3]
3 years ago
6

A 1.35 V potential difference is maintained across a 1.1 m length of tungsten wire that has a cross-sectional area of 0.72 mm2 .

What is the current in the wire? The resistivity of the tungsten is 5.6 × 10−8 Ω · m . Answer in units of A.
Physics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

Therefore,

The current in the wire is 15.77 Ampere.

Explanation:

Given:

Length = l = 1.1 meter

V = 1.35 Volt

Area = 0.72\ mm^{2}=\dfrac{0.72}{1000000}=0.72\times 10^{-6}\ m^{3}

To Find:

Current, I =?

Solution:

Resistance for 1.1-m long tungsten wire with a cross sectional area, if it is connected across a 1.35 V potential difference given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section

\rho=Resistivity=5.6\times 10^{-8}\ ohm-meter

Substituting the values we get

R=\dfrac{5.6\times 10^{-8}\times 1.1}{0.72\times 10^{-6}}=8.56\times 10^{-2}\ ohm

Now by Ohm's Law,

I=\dfrac{V}{R}

Substituting the values we get

I=\dfrac{1.35}{8.56\times 10^{-2}}=15.77\ Ampere

Therefore,

The current in the wire is 15.77 Ampere.

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Which of the following does not make use of total internal reflection.

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2 years ago
N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

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from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

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v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

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x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

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Ga) No. Stars such as Cosmo shine for 3 Million years.
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Answer:

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Explanation:

If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.

We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.

With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that <em>beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.</em>

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3 years ago
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