Standard and Compound Units: past paper questions & worked solutions

So \[ \text{volume}=\frac{\text{mass}}{\text{density}} =\frac{48.3}{10.5}=4.6 \]

So \[ 1\text{ m}^3=100^3\text{ cm}^3=1,000,000\text{ cm}^3 \]

There are 3600 seconds in one hour.

There are 1000 metres in one kilometre.

\(198\text{ km}=198000\text{ m}\) and \(1\text{ hour}=3600\text{ s}\).

Convert both the distance and the time in the rate.

Metres per second means metres divided by seconds.

Use the cylinder volume formula.

Mass equals density times volume; keep full precision until the end.

Round the final mass as requested.

The contact area is one square face:

To convert m/s to km/h, multiply by \(3.6\).

The base area is \(6x\), so

To convert m/s to km/h, multiply by \(3.6\).

The volume of a cylinder is \[ \text{volume}=\text{area of base}\times \text{height} \]

So the area in contact with the table is \[ 1575\div 21=75\text{ cm}^2 \]

Time \(=3\) hours \(18\) minutes:

To convert km/h to m/s, divide by \(3.6\).

For the cuboid,

Average speed: \[ 3980\div 5.4=737.037\ldots \]

To convert km/h to m/s, divide by \(3.6\).

Average speed: \[ 429\div 6.65=64.511\ldots \]

Average speed: \[ 433.5\div 3.4=127.5 \]

Average speed: \[ 2470\div 3.6=686.111\ldots \]

To convert m/s to km/h, multiply by \(3.6\).

Convert hours to seconds: \[ \frac{100}{28440}\times 3600=12.658\ldots \]

To the nearest second, \[ 12.658\ldots \approx 13 \]

The area of the square face is \[ 1.5\times 1.5=2.25\text{ m}^2 \]

So \[ \text{force}=\text{pressure}\times \text{area} \]

Cycling time: \(60\div24=2.5\) hours. Running time: \(1+15/60=1.25\) hours.

Time equals distance divided by speed.

Running distance: \(12\times1.25=15\) km. Total distance: \(60+15=75\) km. Total time: \(2.5+1.25=3.75\) hours.

Combine the actual distances and times.

Average speed is total distance divided by total time, not the mean of the two speeds.

For the cylinder, \[ V=\pi r^2h \]

So, to the nearest whole number, \[ h=6 \]

For the mass, use the given volume: \[ \text{mass}=\text{density}\times \text{volume} \]

The base area of the cylinder is \(51.428\ldots\text{ cm}^2\).

First part: \[ \text{distance}=245\text{ km}, \qquad \text{time}=2.5\text{ h} \]

Second part: \[ \text{time}=\frac{220}{80}=2.75\text{ h} \]

Third part: \[ 50\text{ minutes}=\frac56\text{ h} \]

Total distance: \[ 245+220+60=525\text{ km} \]

Total time: \[ 2.5+2.75+\frac56=6.0833\ldots\text{ h} \]

Average speed: \[ \frac{525}{6.0833\ldots}=86.301\ldots \]

The cube rests on one square face, so \[ w^2=400 \]

Pedro's time was \[ 2\text{ h }42\text{ min}=2.7\text{ h} \]

Distance from Toulouse to Montpellier: \[ 90\times 2.7=243\text{ km} \]

Janine took 3 hours, so her average speed was \[ 243\div 3=81 \]

So the area of the base is

Correct to 3 significant figures,

Use Pythagoras to find the other side of the rectangle.

The mass is \(5.4\text{ kg}\), and the density is \(0.0027\text{ kg/cm}^3\).

For the cylinder, \[ V=\pi r^2h \]

Convert the mass to grams: \[ 5.73\text{ kg}=5730\text{ g} \]

Using \[ \text{density}=\frac{\text{mass}}{\text{volume}} \]

Volume of one cuboid block is \[ 1.5\times2\times8=24\text{ cm}^3 \]

Pablo must have a whole number of blocks, so he needs at least 13 blocks.

Volume of cone:

Volume of hemisphere:

Total volume:

Cone volume:

Mass of copper cone:

Mass of hemisphere:

Hemisphere volume:

Density of the hemisphere:

The outside surface area is:

curved surface area of cylinder:

bottom of cylinder:

curved surface area of hemisphere:

So

Volume of the cylinder:

Volume of the hemisphere:

Total volume:

Density:

This matches aluminium, whose density is \(2.7\text{ g/cm}^3\).