The answer is: " 208 g " .
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Explanation:
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The formula/ equation for density is:
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D = m / V ; That is, "mass divided by volume" ;
Density is expressed as:
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"mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
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{Note the exact equivalent: 1 cm³ = 1 mL }.
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→ The formula is: " D = m / V " ;
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in which:
"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);
"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
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We want to find the mass, "m" ; so we take the original equation/formula for the density:
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D = m / V ;
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And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
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Multiply each side of the equation by "V" ;
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V * { D = m / V } ; to get:
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V * D = m ; ↔ m = V * D ;
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Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
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m = V * D ;
m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;
→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
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By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:
This is very hard bit I think 6.3 my, I'm not shure.
