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4 years ago

A charge Q is uniformly distributed along the x axis from x = a to x = b. If Q = 45

Physics

1 answer:

luda_lava [24]4 years ago

3 0

Answer:

The potential at the point 8 meters is approximately 48.98 Joules/C

Explanation:

Notice that we need to find the potential at a point aligned to the rod's axis (see attached figure), that is located 6 meters from one end of the rod. Notice as well, that the length (L) of the rod is 5 meters.

Since we have a uniformly distributed charge, the charge density per unit of length is defined as:

$\lambda=\frac{Q}{L} \\\lambda=\frac{45}{5}\,\frac{10^{-9}\,C}{m} \\\lambda=9\,\frac{10^{-9}\,C}{m}$

In order to find the contribution of each little segment dx of charge

$dq=\lambda\,dx$

to the potential at the requested point, we need to perform an integral:

$V=\int\limits^{11}_{6} {} \, dV \\V=\int\limits^{11}_{6} {} \, k\,\frac{dq}{x} \\V=\lambda \,*k\,\int\limits^{11}_{6} {} \, \frac{dx}{x} \\V=8.98*9 \, \,ln(\frac{11}{6}) \,\frac{J}{C} \\V=80.82\,* \,0.606 \,\frac{J}{C}\\V=48.98 \frac{J}{C}$

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Answer:

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3 years ago

n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

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Answer:

a). V = 3.13*10⁶ m/s

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c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

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q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

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1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

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V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

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T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

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3 years ago

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4 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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3 years ago