The volume of air in a model car is #40cm^3# . A scale of 1:50 was used to create the model. What is the volume of air in the actual car?

I got the answer #2000cm^3# .
Is that right?

Answer 1

#V_("air in the actual car") = 5,000,000 cm^3#

Given: #V_("air in the model") = 40 cm^3#. Scale of #1:50# model to actual.
The scale #1:50# is a linear scale. This means for each centimeter in length, width and height of the model, there is #50# centimeters in length, width and height in the actual model.
The volume scale is then #50^3 #
#V_("air in the actual car") = 40 cm^3 xx 50^3 = 5,000,000 cm^3#
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Answer 2

#"5,000,000"cm^3=5m^3#

Let's consider the computation as follows:

For example, if we have a line segment of length 1, and we want to scale it using the 1:50 ratio, the resultant scaled line segment will be 50. This is a very simple calculation to perform.

Let's now look at a square. It's got length of 1 and width of 1 and an area of 1 unit squared. Let's now apply the scale. What happens? The length is now 50 and the width is also 50. The area of the square is #50xx50=2500# units squared.
The same thing happens when we move to three dimensions: our cube with length 1, width 1, height 1, and volume 1 unit cubed, when scaled, now has width 50, length 50, height 50 and volume #50xx50xx50="125,000"# units cubed.
Let's now move to our question. We have a car that has a volume of 40 cm cubed. What factor do we use for scaling to the actual size of the car? #"125,000"#. So we end up with the original car having:
#40xx"125,000"="5,000,000"cm^3=5m^3# of air.
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Answer 3

See below

If #1/50# was a linear ratio, meaning that it was applied to each of the three dimensions of the car and not just the volume then:
#(40 cm^3)/x= (1)^3/(50)^3#
#x= 40cm^3*50^3=5000000" cm"^3"#

If you would like to convert to meters cubed:

#5000000 "cm"^3xx(1 m)^3/(100 cm)^3= 5 m^3#
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Answer 4

#5m^3# is the volume of the real car.

Recognizing that the model and the actual car have comparable numbers is crucial.

They have exactly the same shape, but different sizes. The scale is #1 : 50# which means that all lengths on the real car are #50# times bigger than on the model.

Their volumes in comparable figures match the lengths' cube in ratio.

#1^3/50^3 = (40cm^3)/x" "(larr "model")/(larr "real car")#
#x = 50^3 xx 40 cm^3#
#x = 5,000,000 cm^3#
This is not a practical unit, convert to #m^3#
#(5,000,000)/(1000xx1000)#
#= 5m^3#
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Answer 5

To find the volume of air in the actual car, we need to scale up the volume of the model car based on the given scale of 1:50.

Since the scale is 1:50, it means that every unit in the model represents 50 units in the actual car. Therefore, to find the volume of air in the actual car, we need to multiply the volume of air in the model car by 503 50^3 (since volume is measured in three dimensions).

Volume of air in actual car=Volume of air in model car×(503)\text{Volume of air in actual car} = \text{Volume of air in model car} \times (50^3)

Volume of air in actual car=40cm3×(503)\text{Volume of air in actual car} = 40 \, \text{cm}^3 \times (50^3)

Volume of air in actual car=40cm3×125,000\text{Volume of air in actual car} = 40 \, \text{cm}^3 \times 125,000

Volume of air in actual car=5,000,000cm3\text{Volume of air in actual car} = 5,000,000 \, \text{cm}^3

So, the volume of air in the actual car is 5,000,000cm3 5,000,000 \, \text{cm}^3 .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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