8.5 Adding and subtracting rational expressions (Page 2/2)

Elementary algebra Page 2 / 2

$\begin{array}{l} \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{x^{2} + 4 x - 12} - \frac{x^{2} - 4 x - 6}{x^{2} + 4 x - 12} \\ Factor the denominators to determine if they're the same . \\ \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{(x + 6) (x - 2)} - \frac{x^{2} - 4 x - 6}{(x + 6) (x - 2)} \\ The denominators are the same . Combine the numerators being careful to note the negative sign . \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - (x^{2} - 4 x + 6)}{(x + 6) (x - 2)} \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - x^{2} + 4 x + 6}{(x + 6) (x - 2)} \\ \frac{4 x^{2} + 9 x + 17}{(x + 6) (x - 2)} \end{array}$

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Practice set a

Add or Subtract the following rational expressions.

$\frac{4}{9} + \frac{2}{9}$

$\frac{2}{3}$

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$\frac{3}{b} + \frac{2}{b}$

$\frac{5}{b}$

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$\frac{5 x}{2 y^{2}} - \frac{3 x}{2 y^{2}}$

$\frac{x}{y^{2}}$

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$\frac{x + y}{x - y} + \frac{2 x + 3 y}{x - y}$

$\frac{3 x + 4 y}{x - y}$

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$\frac{4 x^{2} - x + 4}{3 x + 10} - \frac{x^{2} + 2 x + 5}{3 x + 10}$

$\frac{3 x^{2} - 3 x - 1}{3 x + 10}$

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$\frac{x (x + 1)}{x (2 x + 3)} + \frac{3 x^{2} - x + 7}{2 x^{2} + 3 x}$

$\frac{4 x^{2} + 7}{x (2 x + 3)}$

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$\frac{4 x + 3}{x^{2} - x - 6} - \frac{8 x - 4}{(x + 2) (x - 3)}$

$\frac{- 4 x + 7}{(x + 2) (x - 3)}$

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$\frac{5 a^{2} + a - 4}{2 a (a - 6)} + \frac{2 a^{2} + 3 a + 4}{2 a^{2} - 12 a} + \frac{a^{2} + 2}{2 a^{2} - 12 a}$

$\frac{4 a^{2} + 2 a + 1}{a (a - 6)}$

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$\frac{8 x^{2} + x - 1}{x^{2} - 6 x + 8} + \frac{2 x^{2} + 3 x}{x^{2} - 6 x + 8} - \frac{5 x^{2} + 3 x - 4}{(x - 4) (x - 2)}$

$\frac{5 x^{2} + x + 3}{(x - 4) (x - 2)}$

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Fractions with different denominators

Sample set b

Add or Subtract the following rational expressions.

$\begin{array}{l} \frac{4 a}{3 y} + \frac{2 a}{9 y^{2}} . & The denominators are n o t the same . Find the LCD . By inspection, the LCD is 9 y^{2} . \\ \frac{}{9 y^{2}} + \frac{2 a}{9 y^{2}} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 3 y, \\ so the numerator must be multiplied by 3 y . \\ 4 a \cdot 3 y = 12 a y \end{array} \\ \frac{12 a y}{9 y^{2}} + \frac{2 a}{9 y^{2}} & The denominators are now the same . Add the numerators . \\ \frac{12 a y + 2 a}{9 y^{2}} \end{array}$

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$\begin{array}{l} \frac{3 b}{b + 2} + \frac{5 b}{b - 3} . & \begin{array}{l} The denominators are n o t the same . The LCD is (b + 2) (b - 3) . \end{array} \\ \frac{}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by b - 3, \\ so the numerator must be multiplied by b - 3. 3 b (b - 3) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by b + 2, \\ so the numerator must be multiplied by b + 2. 5 b (b + 2) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{5 b (b + 2)}{(b + 2) (b - 3)} & \begin{array}{l} The denominators are now the same . Add the numerators . \end{array} \\ \frac{3 b (b - 3) + 5 b (b + 2)}{(b - 3) (b + 2)} & = & \frac{3 b^{2} - 9 b + 5 b^{2} + 10 b}{(b - 3) (b + 2)} \\ = & \frac{8 b^{2} + b}{(b - 3) (b - 2)} \end{array}$

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$\begin{array}{l} \frac{x + 3}{x - 1} + \frac{x - 2}{4 x + 4} . & \begin{array}{l} The denominators are n o t the same . \\ Find the LCD . \end{array} \\ \frac{x + 3}{x - 1} + \frac{x - 2}{4 (x + 1)} & The LCD is (x + 1) (x - 1) \\ \frac{}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 4 (x + 1) so \\ the numerator must be multiplied by 4 (x + 1) . 4 (x + 3) (x + 1) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by x - 1 \\ so the numerator must be multiplied by x - 1. (x - 1) (x - 2) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{(x - 1) (x - 2)}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominators are now the same. \\ Add the numerators. \end{array} \\ \frac{4 (x + 3) (x + 1) + (x - 1) (x - 2)}{4 (x + 1) (x - 1)} \\ \frac{4 (x^{2} + 4 x + 3) + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} \\ \frac{4 x^{2} + 16 x + 12 + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} & = & \frac{5 x^{2} + 13 x + 14}{4 (x + 1) (x - 1)} \end{array}$

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$\begin{array}{l} \frac{x + 5}{x^{2} - 7 x + 12} + \frac{3 x - 1}{x^{2} - 2 x - 3} & Determine the LCD . \\ \frac{x + 5}{(x - 4) (x - 3)} + \frac{3 x - 1}{(x - 3) (x + 1)} & The LCD is (x - 4) (x - 3) (x + 1) . \\ \frac{}{(x - 4) (x - 3) (x + 1)} + \frac{}{(x - 4) (x - 3) (x + 1)} & The first numerator must be multiplied by x + 1 and the second by x - 4. \\ \frac{(x + 5) (x + 1)}{(x - 4) (x - 3) (x + 1)} + \frac{(3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} & The denominators are now the same . Add the numerators . \\ \frac{(x + 5) (x + 1) + (3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} \\ \frac{x^{2} + 6 x + 5 + 3 x^{2} - 13 x + 4}{(x - 4) (x - 3) (x + 1)} \\ \frac{4 x^{2} - 7 x + 9}{(x - 4) (x - 3) (x + 1)} \end{array}$

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$\begin{array}{l} \frac{a + 4}{a^{2} + 5 a + 6} - \frac{a - 4}{a^{2} - 5 a - 24} & Determine the LCD. \\ \frac{a + 4}{(a + 3) (a + 2)} - \frac{a - 4}{(a + 3) (a - 8)} & The LCD is (a + 3) (a + 2) (a - 8) . \\ \frac{}{(a + 3) (a + 2) (a - 8)} - \frac{}{(a + 3) (a + 2) (a - 8)} & The first numerator must be multiplied by a - 8 and the second by a + 2. \\ \frac{(a + 4) (a - 8)}{(a + 3) (a + 2) (a - 8)} - \frac{(a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} & The denominators are now the same . Subtract the numerators . \\ \frac{(a + 4) (a - 8) - (a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - (a^{2} - 2 a - 8)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - a^{2} + 2 a + 8}{(a + 3) (a + 2) (a - 8)} \\ \frac{- 2 a - 24}{(a + 3) (a + 2) (a - 8)} & Factor - 2 from the numerator . \\ \frac{- 2 (a + 12)}{(a + 3) (a + 2) (a - 8)} \end{array}$

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$\begin{array}{l} \frac{3 x}{7 - x} + \frac{5 x}{x - 7} . & \begin{array}{l} The denominators are n e a r l y the same. They differ only in sign. \\ Our technique is to factor - 1 from one of them. \end{array} \\ \frac{3 x}{7 - x} = \frac{3 x}{- (x - 7)} & = & \frac{- 3 x}{x - 7} & Factor - 1 from the first term . \\ \frac{3 x}{7 - x} + \frac{5 x}{x - 7} & = & \frac{- 3 x}{x - 7} + \frac{5 x}{x - 7} \\ = & \frac{- 3 x + 5 x}{x - 7} \\ = & \frac{2 x}{x - 7} \end{array}$

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Practice set b

Add or Subtract the following rational expressions.

$\frac{3 x}{4 a^{2}} + \frac{5 x}{12 a^{3}}$

$\frac{9 a x + 5 x}{12 a^{3}}$

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$\frac{5 b}{b + 1} + \frac{3 b}{b - 2}$

$\frac{8 b^{2} - 7 b}{(b + 1) (b - 2)}$

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$\frac{a - 7}{a + 2} + \frac{a - 2}{a + 3}$

$\frac{2 a^{2} - 4 a - 25}{(a + 2) (a + 3)}$

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$\frac{4 x + 1}{x + 3} - \frac{x + 5}{x - 3}$

$\frac{3 x^{2} - 19 x - 18}{(x + 3) (x - 3)}$

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$\frac{2 y - 3}{y} + \frac{3 y + 1}{y + 4}$

$\frac{5 y^{2} + 6 y - 12}{y (y + 4)}$

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$\frac{a - 7}{a^{2} - 3 a + 2} + \frac{a + 2}{a^{2} - 6 a + 8}$

$\frac{2 a^{2} - 10 a + 26}{(a - 2) (a - 1) (a - 4)}$

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$\frac{6}{b^{2} + 6 b + 9} - \frac{2}{b^{2} + 4 b + 4}$

$\frac{4 b^{2} + 12 b + 6}{{(b + 3)}^{2} {(b + 2)}^{2}}$

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$\frac{x}{x + 4} - \frac{x - 2}{3 x - 3}$

$\frac{2 x^{2} - 5 x + 8}{3 (x + 4) (x - 1)}$

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$\frac{5 x}{4 - x} + \frac{7 x}{x - 4}$

$\frac{2 x}{x - 4}$

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Sample set c

Combine the following rational expressions.

$\begin{array}{l} 3 + \frac{7}{x - 1} . & Rewrite the expression. \\ \frac{3}{1} + \frac{7}{x - 1} & The LCD is x - 1. \\ \frac{3 (x - 1)}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3 + 7}{x - 1} \\ = & \frac{3 x + 4}{x - 1} \end{array}$

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$\begin{array}{l} 3 y + 4 - \frac{y^{2} - y + 3}{y - 6} . & Rewrite the expression. \\ \frac{3 y + 4}{1} - \frac{y^{2} - y + 3}{y - 6} & The LCD is y - 6. \\ \frac{(3 y + 4) (y - 6)}{y - 6} - \frac{y^{2} - y + 3}{y - 6} & = & \frac{(3 y + 4) (y - 6) - (y^{2} - y + 3)}{y - 6} \\ = & \frac{3 y^{2} - 14 y - 24 - y^{2} + y - 3}{y - 6} \\ = & \frac{2 y^{2} - 13 y - 27}{y - 6} \end{array}$

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Practice set c

Simplify $8 + \frac{3}{x - 6} .$

$\frac{8 x - 45}{x - 6}$

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Simplify $2 a - 5 - \frac{a^{2} + 2 a - 1}{a + 3} .$

$\frac{a^{2} - a - 14}{a + 3}$

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Exercises

For the following problems, add or subtract the rational expressions.

$\frac{3}{8} + \frac{1}{8}$

$\frac{1}{2}$

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$\frac{1}{9} + \frac{4}{9}$

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$\frac{7}{10} - \frac{2}{5}$

$\frac{3}{10}$

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$\frac{3}{4} - \frac{5}{12}$

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$\frac{3}{4 x} + \frac{5}{4 x}$

$\frac{2}{x}$

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$\frac{2}{7 y} + \frac{3}{7 y}$

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$\frac{6 y}{5 x} + \frac{8 y}{5 x}$

$\frac{14 y}{5 x}$

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$\frac{9 a}{7 b} + \frac{3 a}{7 b}$

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$\frac{15 n}{2 m} - \frac{6 n}{2 m}$

$\frac{9 n}{2 m}$

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$\frac{8 p}{11 q} - \frac{3 p}{11 q}$

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$\frac{y + 4}{y - 6} + \frac{y + 8}{y - 6}$

$\frac{2 y + 12}{y - 6}$

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$\frac{y - 1}{y + 4} + \frac{y + 7}{y + 4}$

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$\frac{a + 6}{a - 1} + \frac{3 a + 5}{a - 1}$

$\frac{4 a + 11}{a - 1}$

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$\frac{5 a + 1}{a + 7} + \frac{2 a - 6}{a + 7}$

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$\frac{x + 1}{5 x} + \frac{x + 3}{5 x}$

$\frac{2 x + 4}{5 x}$

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$\frac{a - 6}{a + 2} + \frac{a - 2}{a + 2}$

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$\frac{b + 1}{b - 3} + \frac{b + 2}{b - 3}$

$\frac{2 b + 3}{b - 3}$

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$\frac{a + 2}{a - 5} - \frac{a + 3}{a - 5}$

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$\frac{b + 7}{b - 6} - \frac{b - 1}{b - 6}$

$\frac{8}{b - 6}$

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$\frac{2 b + 3}{b + 1} - \frac{b - 4}{b + 1}$

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$\frac{3 y + 4}{y + 8} - \frac{2 y - 5}{y + 8}$

$\frac{y + 9}{y + 8}$

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$\frac{2 a - 7}{a - 9} + \frac{3 a + 5}{a - 9}$

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$\frac{8 x - 1}{x + 2} - \frac{15 x + 7}{x + 2}$

$\frac{- 7 x - 8}{x + 2}$

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$\frac{7}{2 x^{2}} + \frac{1}{6 x^{3}}$

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$\frac{2}{3 x} + \frac{4}{6 x^{2}}$

$\frac{2 (x + 1)}{3 x^{2}}$

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$\frac{5}{6 y^{3}} - \frac{2}{18 y^{5}}$

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$\frac{2}{5 a^{2}} - \frac{1}{10 a^{3}}$

$\frac{4 a - 1}{10 a^{3}}$

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$\frac{3}{x + 1} + \frac{5}{x - 2}$

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$\frac{4}{x - 6} + \frac{1}{x - 1}$

$\frac{5 (x - 2)}{(x - 6) (x - 1)}$

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$\frac{2 a}{a + 1} - \frac{3 a}{a + 4}$

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$\frac{6 y}{y + 4} + \frac{2 y}{y + 3}$

$\frac{2 y (4 y + 13)}{(y + 4) (y + 3)}$

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$\frac{x - 1}{x - 3} + \frac{x + 4}{x - 4}$

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$\frac{x + 2}{x - 5} + \frac{x - 1}{x + 2}$

$\frac{2 x^{2} - 2 x + 9}{(x - 5) (x + 2)}$

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$\frac{a + 3}{a - 3} - \frac{a + 2}{a - 2}$

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$\frac{y + 1}{y - 1} - \frac{y + 4}{y - 4}$

$\frac{- 6 y}{(y - 1) (y - 4)}$

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$\frac{x - 1}{(x + 2) (x - 3)} + \frac{x + 4}{x - 3}$

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$\frac{y + 2}{(y + 1) (y + 6)} + \frac{y - 2}{y + 6}$

$\frac{y^{2}}{(y + 1) (y + 6)}$

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$\frac{2 a + 1}{(a + 3) (a - 3)} - \frac{a + 2}{a + 3}$

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$\frac{3 a + 5}{(a + 4) (a - 1)} - \frac{2 a - 1}{a - 1}$

$\frac{- 2 a^{2} - 4 a + 9}{(a + 4) (a - 1)}$

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$\frac{2 x}{x^{2} - 3 x + 2} + \frac{3}{x - 2}$

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$\frac{4 a}{a^{2} - 2 a - 3} + \frac{3}{a + 1}$

$\frac{7 a - 9}{(a + 1) (a - 3)}$

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$\frac{3 y}{y^{2} - 7 y + 12} - \frac{y^{2}}{y - 3}$

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$\frac{x - 1}{x^{2} + 6 x + 8} + \frac{x + 3}{x^{2} + 2 x - 8}$

$\frac{2 (x^{2} + x + 4)}{(x + 2) (x - 2) (x + 4)}$

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$\frac{a - 4}{a^{2} + 2 a - 3} + \frac{a + 2}{a^{2} + 3 a - 4}$

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$\frac{b - 3}{b^{2} + 9 b + 20} + \frac{b + 4}{b^{2} + b - 12}$

$\frac{2 b^{2} + 3 b + 29}{(b - 3) (b + 4) (b + 5)}$

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$\frac{y - 1}{y^{2} + 4 y - 12} - \frac{y + 3}{y^{2} + 6 y - 16}$

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$\frac{x + 3}{x^{2} + 9 x + 14} - \frac{x - 5}{x^{2} - 4}$

$\frac{- x + 29}{(x - 2) (x + 2) (x + 7)}$

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$\frac{x - 1}{x^{2} - 4 x + 3} + \frac{x + 3}{x^{2} - 5 x + 6} + \frac{2 x}{x^{2} - 3 x + 2}$

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$\frac{4 x}{x^{2} + 6 x + 8} + \frac{3}{x^{2} + x - 6} + \frac{x - 1}{x^{2} + x - 12}$

$\frac{5 x^{4} - 3 x^{3} - 34 x^{2} + 34 x - 60}{(x - 2) (x + 2) (x - 3) (x + 3) (x + 4)}$

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$\frac{y + 2}{y^{2} - 1} + \frac{y - 3}{y^{2} - 3 y - 4} - \frac{y + 3}{y^{2} - 5 y + 4}$

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$\frac{a - 2}{a^{2} - 9 a + 18} + \frac{a - 2}{a^{2} - 4 a - 12} - \frac{a - 2}{a^{2} - a - 6}$

$\frac{(a + 5) (a - 2)}{(a + 2) (a - 3) (a - 6)}$

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$\frac{y - 2}{y^{2} + 6 y} + \frac{y + 4}{y^{2} + 5 y - 6}$

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$\frac{a + 1}{a^{3} + 3 a^{2}} - \frac{a + 6}{a^{2} - a}$

$\frac{- a^{3} - 8 a^{2} - 18 a - 1}{a^{2} (a + 3) (a - 1)}$

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$\frac{4}{3 b^{2} - 12 b} - \frac{2}{6 b^{2} - 6 b}$

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$\frac{3}{2 x^{5} - 4 x^{4}} + \frac{- 2}{8 x^{3} + 24 x^{2}}$

$\frac{- x^{3} + 2 x^{2} + 6 x + 18}{4 x^{4} (x - 2) (x + 3)}$

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$\frac{x + 2}{12 x^{3}} + \frac{x + 1}{4 x^{2} + 8 x - 12} - \frac{x + 3}{16 x^{2} - 32 x + 16}$

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$\frac{2 x}{x^{2} - 9} - \frac{x + 1}{4 x^{2} - 12 x} - \frac{x - 4}{8 x^{3}}$

$\frac{14 x^{4} - 9 x^{3} - 2 x^{2} + 9 x - 36}{8 x^{3} (x + 3) (x - 3)}$

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$4 + \frac{3}{x + 2}$

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$8 + \frac{2}{x + 6}$

$\frac{8 x + 50}{x + 6}$

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$1 + \frac{4}{x - 7}$

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$3 + \frac{5}{x - 6}$

$\frac{3 x - 13}{x - 6}$

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$- 2 + \frac{4 x}{x + 5}$

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$- 1 + \frac{3 a}{a - 1}$

$\frac{2 a + 1}{a - 1}$

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$6 - \frac{4 y}{y + 2}$

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$2 x + \frac{x^{2} - 4}{x + 1}$

$\frac{3 x^{2} + 2 x - 4}{x + 1}$

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$- 3 y + \frac{4 y^{2} + 2 y - 5}{y + 3}$

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$x + 2 + \frac{x^{2} + 4}{x - 1}$

$\frac{2 x^{2} + x + 2}{x - 1}$

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$b + 6 + \frac{2 b + 5}{b - 2}$

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$\frac{3 x - 1}{x - 4} - 8$

$\frac{- 5 x + 31}{x - 4}$

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$\frac{4 y + 5}{y + 1} - 9$

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$\frac{2 y^{2} + 11 y - 1}{y + 4} - 3 y$

$\frac{- (y^{2} + y + 1)}{y + 4}$

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$\frac{5 y^{2} - 2 y + 1}{y^{2} + y - 6} - 2$

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$\frac{4 a^{3} + 2 a^{2} + a - 1}{a^{2} + 11 a + 28} + 3 a$

$\frac{7 a^{3} + 35 a^{2} + 85 a - 1}{(a + 7) (a + 4)}$

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$\frac{2 x}{1 - x} + \frac{6 x}{x - 1}$

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$\frac{5 m}{6 - m} + \frac{3 m}{m - 6}$

$\frac{- 2 m}{m - 6}$

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$\frac{- a + 7}{8 - 3 a} + \frac{2 a + 1}{3 a - 8}$

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$\frac{- 2 y + 4}{4 - 5 y} - \frac{9}{5 y - 4}$

$\frac{2 y - 13}{5 y - 4}$

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$\frac{m - 1}{1 - m} - \frac{2}{m - 1}$

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Exercises for review

( [link] ) Simplify ${(x^{3} y^{2} z^{5})}^{6} {(x^{2} y z)}^{2} .$

$x^{22} y^{14} z^{32}$

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( [link] ) Write $6 a^{- 3} b^{4} c^{- 2} a^{- 1} b^{- 5} c^{3}$ so that only positive exponents appear.

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( [link] ) Construct the graph of $y = - 2 x + 4.$
An xy coordinate plane with gridlines, labeled negative five to five on both axes.

A graph of a line passing through three points with coordinates zero, four; one, two; and two, zero.

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( [link] ) Find the product: $\frac{x^{2} - 3 x - 4}{x^{2} + 6 x + 5} \cdot \frac{x^{2} + 5 x + 6}{x^{2} - 2 x - 8} .$

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( [link] ) Replace N with the proper quantity: $\frac{x + 3}{x - 5} = \frac{N}{x^{2} - 7 x + 10} .$

$(x + 3) (x - 2)$

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.

Kate Reply

To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.

Muhammed

What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?

Kate Reply

what is the change in momentum of a body?

Eunice Reply

what is a capacitor?

Raymond Reply

Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.

Gautam

A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate

Maria Reply

please solve

Sharon

8m/s²

Aishat

What is Thermodynamics

Muordit

velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.

Mehmet

A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat

Saheed Reply

50 m/s due south east

Someone

which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer

Ramon Reply

I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.

Someone

Scratch that

Someone

temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....

Someone

about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle

Someone

definitely of physics

Haryormhidey Reply

how many start and codon

Esrael Reply

what is field

Felix Reply

physics, biology and chemistry this is my Field

ALIYU

field is a region of space under the influence of some physical properties

Collete

what is ogarnic chemistry

WISDOM Reply

determine the slope giving that 3y+ 2x-14=0

WISDOM

Another formula for Acceleration

Belty Reply

a=v/t. a=f/m a

IHUMA

innocent

Adah

pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?

Nassze Reply

how do lnternal energy measures

Esrael

Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

JALLAH Reply

No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.

Dlovan

Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?

Robert

like charges repel while unlike charges atttact

Raymond

What is specific heat capacity

Destiny Reply

Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).

AI-Robot

specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin

ROKEEB

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Source: OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3

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