Question

Both the La Plata river dolphin (Pontoporia blainvillei) and

the spermwhale (Physeter macrocephalus) belong to the suborder
Odontoceti (individuals that have teeth). A La Plata river dolphin
weighs between 30 and 50 kg, whereas a sperm whale weighs
between 35,000 and 40,000 kg. A sperm whale is order(s)
of magnitude heavier than a La Plata river dolphin.
39. Compare a ball of radius 1 cm against a ball of radius 10 cm.
The radius of the larger ball is order(s) of magnitude bigger
than the radius of the smaller ball. The volume of the larger ball is
order(s) of magnitude bigger than the volume of the smaller
ball.

Answer 1

A sperm whale is orders of magnitude heavier than a La Plata river dolphin. The radius of a larger ball is orders of magnitude bigger than the radius of a smaller ball. The volume of a larger ball is orders of magnitude bigger than the volume of a smaller ball.

Step-by-step explanation:

The question compares the size and weight of a La Plata river dolphin and a sperm whale. It is mentioned that a sperm whale is orders of magnitude heavier than a dolphin. This means that the weight of a sperm whale is at least 10 times (or more) heavier than that of a dolphin. The weight comparison between the two species demonstrates the significant size difference between them.

Similarly, when comparing the size of two balls with different radii, a ball with a larger radius is orders of magnitude bigger than a ball with a smaller radius. The volume of a ball is directly proportional to the cube of its radius, so if the radius increases by a factor of 10, the volume increases by a factor of 10^3 or 1000. This illustrates the exponential increase in volume with an increase in radius.

Answer 2

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

`\\ N = a * 10^(b)` (1)

Where

`\\ (1)/(√(10)) \leq a < √(10)` (2)

or

`\\ 1 \leq a < 10` (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

First case

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

`\\ 30kg \leq Dolphin_(weight) \leq 50kg`

`\\ 3*10^(1)kg \leq Dolphin_(weight) \leq 5*10^(1)kg`

`\\ 35000kg \leq Whale_(weight) \leq 40000kg`

`\\ 3.5*10^(4)kg \leq Whale_(weight) \leq 4.0*10^(4)kg`

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

`\\ Dolphin_(weight) = 10^(1)\;(order\;of\;magnitude=1)`

`\\ Whale_(weight) = 10^(4)\;(order\;of\;magnitude=4)`

Then

`\\ (Whale_(weight) = 10^(4))/(Dolphin_(weight) = 10^(1))`

`\\ (Whale_(weight))/(Dolphin_(weight)) = (10^(4))/(10^(1))`

`\\ (Whale_(weight))/(Dolphin_(weight)) = 10^(4-1)`

`\\ (Whale_(weight))/(Dolphin_(weight)) = 10^(3)`

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

Second case

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

`\\ (Larger\;ball_(radius))/(Smaller\;ball_(radius)) = (10^(1))/(10^(0))`

`\\ (Larger\;ball_(radius))/(Smaller\;ball_(radius)) = 10^(1-0) = 10^(1)`

Third case

For this case, we need to calculate the volume of a sphere for both radii (1cm and 10cm).

The volume of a sphere is

`\\ V_(sphere) = (4)/(3)*\pi*R^(3)`

Then, the volume of the ball of radius 1cm is:

`\\ V_(radius=1) = (4)/(3)*\pi*(1cm)^(3)`

`\\ V_(radius=1) \approx 4.19*10^(0)cm^(3)`

And, the volume of the ball of radius 10cm is:

`\\ V_(radius=10) = (4)/(3)*\pi*(10cm)^(3)`

`\\ V_(radius=10) \approx 4.19*10^(3)cm^(3)`

Thus

`\\ (10^(3))/(10^(0)) = 10^(3)`

As a result, the volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.