Temperature and Scale Conversion Quiz - SAT Physics Practice Questions

First, find the increase in the X scale: \( \\frac{4}{3} \\times 15 = 20 \). Now, use this increase as input for function Y: \( Y(20 + 200) = \\frac{5}{4}(220 - 200) + 250 = \\frac{5}{4} \\times 20 + 250 = 25 + 250 = 275 \).

First, calculate the increase in the new scale: \( \\frac{7}{3} \\times 12 = 28 \). After decreasing by 5 units, the net change in the new scale is \( 28 - 5 = 23 \). To find the net change in the original scale, reverse the conversion: \( \\frac{3}{7} \\times 23 = 9.857 \).

First, convert the Kelvin increase to Fahrenheit: \( \\frac{9}{5} \\times 5.20 = 9.36 \). Then, convert this to Celsius: \( (9.36 - 32) \\times \\frac{5}{9} = -17.58 \). Finally, convert the Celsius decrease to the custom scale Z: \( \\frac{3}{4} \\times (-17.58) = -13.185 \).

First, find the increase in the second scale: \( \\frac{5}{4} \\times 16 = 20 \). Then, convert this increase to the third scale: \( \\frac{3}{2} \\times 20 = 30 \).

First, convert the Kelvin increase to Fahrenheit using \( \\frac{9}{5} \\times 2.10 = 3.78 \). Then, convert this Fahrenheit increase to Celsius using \( (3.78 - 32) \\times \\frac{5}{9} = 2.10 \).

The change in the new scale is determined by multiplying the increase in value V by \( \\frac{7}{5} \), since the \( v \) term in the function is being multiplied by this factor. Therefore, the increase in the new scale is \( \\frac{7}{5} \\times 25 = 35 \).

The change in the new system is determined by multiplying the increase in measurement U by \( \\frac{2}{3} \), since the \( u \) term in the function is being multiplied by this factor. Therefore, the increase in the new system is \( \\frac{2}{3} \\times 30 = 20 \).

The change in the new unit is determined by multiplying the increase in signal S by \( \\frac{5}{3} \), since the \( s \) term in the function is being multiplied by this factor. Therefore, the increase in the new unit is \( \\frac{5}{3} \\times 18 = 30 \).

The change in the new scale is determined by multiplying the increase in reading R by \( \\frac{3}{2} \), since the \( r \) term in the function is being multiplied by this factor. Therefore, the increase in the new scale is \( \\frac{3}{2} \\times 20 = 30 \).

The change in the new system is determined by multiplying the increase in measure M by \( \\frac{4}{5} \), since the \( m \) term in the function is being multiplied by this factor. Therefore, the increase in the new system is \( \\frac{4}{5} \\times 25 = 20 \).

The change in the new unit is determined by multiplying the increase in weight W by \( \\frac{7}{3} \), since the \( w \) term in the function is being multiplied by this factor. Therefore, the increase in the new unit is \( \\frac{7}{3} \\times 3 = 7 \).

The change in scale Q is determined by multiplying the increase in pressure P by \( \\frac{3}{7} \), since the \( p \) term in the function is being multiplied by this factor. Therefore, the increase in scale Q is \( \\frac{3}{7} \\times 14 = 6 \).

The change in scale Y is determined by multiplying the increase in scale Z by \( \\frac{5}{2} \), since the \( z \) term in the function is being multiplied by this factor. Therefore, the increase in scale Y is \( \\frac{5}{2} \\times 12 = 30 \).

The change in the converted scale is determined by multiplying the temperature increase in the custom scale by \( \\frac{9}{4} \), since the \( t \) term in the function is being multiplied by this factor. Therefore, the increase in the converted scale is \( \\frac{9}{4} \\times 8 = 18 \).

The change in kelvins is determined by dividing the temperature decrease in Fahrenheit by \( \\frac{9}{5} \), since the \( y \) term in the function is being divided by this factor. Therefore, the decrease in kelvins is \( \\frac{5}{9} \\times 10 = 5.56 \).