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Dennis_Churaev [7]

3 years ago

6

A length of copper wire has a resistance 29 Ω. The wire is cut into three pieces of equal length, which are then connected as pa

rallel lengths between points A and B. What resistance will this new "wire" of length L0 3 have between points A and B? Answer in units of Ω.

1 answer:

erik [133]3 years ago

6 0

Answer:

3.222 ohms

Explanation:

If the total wire had a resistance of 29 ohms, when cut in three, each piece will have a resistance of 9.666 ohms.

As these three pieces (R1, R2 and R3) are now connected in parallel, the equivalent resistance R can be calculated using this equation:

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/9.666 + 1/9.666 + 1/9.666

1/R = 3/9.666

R = 9.666/3 = 3.222 ohms

The resistance between A and B will be 3.222 ohms

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3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

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$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

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3 years ago

What are the products of linear electron flow during the light reactions of photosynthesis?

Katena32 [7]

Answer:

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The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

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3 years ago

A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$

Explanation:

From the question we are told that:

Electric field $B=1.5*10N/C$

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At negative plate

Generally the equation for Velocity is mathematically given by

$V^2=2as$

Therefore

$V^2=\frac{2*e_0E*d}{m}$

$V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}$

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5 0

3 years ago