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A tin can has a radius of 0.0414 m and is 0.185 m high. If the air is removed from inside the can, how much force will the outside air (at 1 atm) exert on the can?

Hint: Surface area of a cylinder [tex]= 2 \pi r h + 2 \pi r^2[/tex]

(Unit = N)


Sagot :

To solve this problem, we will follow these steps:

### Step 1: Calculate the surface area of the tin can
The surface area of a cylinder is given by the formula:
[tex]\[ \text{Surface Area} = 2 \pi r h + 2 \pi r^2 \][/tex]

Given values:
- Radius, [tex]\( r \)[/tex] = 0.0414 meters
- Height, [tex]\( h \)[/tex] = 0.185 meters

Plugging these values into the formula, we get:
[tex]\[ \text{Surface Area} = 2 \pi (0.0414) (0.185) + 2 \pi (0.0414)^2 \][/tex]

### Step 2: Calculate the force exerted by the outside air
The force exerted by the air on the surface of the can is calculated using the following formula:
[tex]\[ \text{Force} = \text{Surface Area} \times \text{Pressure} \][/tex]

Given:
- Atmospheric pressure, [tex]\( \text{Pressure} = 1 \)[/tex] atmosphere
- Note: 1 atmosphere is equivalent to 101325 Pascals

Combining the values:
[tex]\[ \text{Force} = \text{Surface Area} \times 101325 \][/tex]

### Step 3: Plug in the values to find the numerical results

Using the given answer:
- Surface area of the can is approximately 0.05889204455678198 square meters
- The force exerted by the outside air on the can is approximately 5967.236414715934 Newtons

### Conclusion
Therefore, the surface area of the tin can is approximately [tex]\(0.0589 \, \text{m}^2\)[/tex], and the force exerted by the outside air on the can is approximately 5967.24 Newtons.