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oksano4ka [1.4K]
3 years ago
13

While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density

of aluminum is 2.70 g/cm3. What was the student's percent error?
Physics
1 answer:
bezimeni [28]3 years ago
3 0
The equation for percent error is

% Error = 100*|Experimental-Theoretical|/Theoretical

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%
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Answer:

q₁ = + 1.25 nC

Explanation:

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Known data

q₃=5 nC

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Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

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\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

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q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

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