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denis-greek [22]
3 years ago
15

Grain is pouring into a silo to be stored for later use. Due to the friction between pieces of grain as they rub against each ot

her during the pouring process, one piece of grain picks up a charge of 6.0 E -10 C and another piece of grain picks up a charge of 2.3 E -15 C. What is the electric force between them if the pieces of grain are 2 cm apart?
Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.
Physics
1 answer:
Svetlanka [38]3 years ago
3 0

Answer:

The electric force between them if the pieces of grain are 2 cm apart is 3.15\times 10^{-11}\textrm{ N}.

Explanation:

Given:

Charge on one grain, Q_{1}=6\times 10^{-10}\textrm{ C}

Charge on another grain, Q_{2}=2.3\times 10^{-15}\textrm{ C}

Separation between them, d=2\textrm{ cm}=0.02\textrm{ m}

Electric force acting between two charges Q_{1}\textrm{ and }Q_{2} separated by a distance d is given as:

F_{elec}=\frac{kQ_{1}Q_{2}}{d^2}

Where, k is Coulomb's constant equal to 9\times 10^9\textrm{ }Nm^2/C^2.

Now, plug in all the values and solve for F{elec}.

F{elec}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}

Therefore, the electric force between them if the pieces of grain are 2 cm apart is 3.15\times 10^{-11}\textrm{ N}.

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Answer:

a

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b

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Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

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\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

\delta h  = 0.802 \  m

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h_m =   0.950 -  0.802

=> h_m =   0.148 \  m

Generally from manometry principle we have that

P_G + \rho * g  * d   -  \rho *  g  * [h - (h_m + d)] = 0

Here P_G is the pressure of the gas

P_G +13.6 *10^{3} * 9.8  * 0.039    -  13.6 *10^{3}  *  9.8  * [0.950 - (0.148 + 0.039)] = 0

P_G  =  9.6724 04 *10^{4} \  N/m^2

converting to  psig

P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

P_G  = 14.03 \  psig

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