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2 years ago

5

g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor

is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.

1 answer:

Mrrafil [7]2 years ago

6 0

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1, and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

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Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = $\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]$

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = $\frac{\rho dx}{A}$

$dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}$

Taking the integral of both sides;we have:

$\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx$

$R = 3.185 [x + \frac {x^3}{3}}]^2__0$

$R = 3.185 [2 + \frac {2^3}{3}}]$

R = 14.863 Ω

Since V = IR

$I = \frac{V}{R}$

$I = \frac{17}{14.863}$

I = 1.144 A

∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0

2 years ago

What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

serg [7]

Answer: $6.268(10)^{-16}J$

Explanation:

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$ :

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)

Where:

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$

$\lambda_{e}=3.168(10)^{-10}m$ (5)

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)

Where $c=3(10)^{8}m/s$ is the spped of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$

Finally:

$E_{p}=6.268(10)^{-16}J$

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3 years ago

2. The components of vector A are given as follows:

Stella [2.4K]

Answer:

50 degree.

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Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7

The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

Tan Ø = Ay/Ax

Substitute Ay and Ax into the formula above.

Tan Ø = -4.7 / 5.6

Tan Ø = -0.839

Ø = tan^-1(-0. 839)

Ø = - 40 degree

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Answer:

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r = 0.00523 m

or

$r=5.23\times 10^{-3}\ m$

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

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2 years ago

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shusha [124]

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