Suppose a spherical asteroid has a radius of approximately 3.2 × 10^3 m. Use the formula 4/3 pie r^3 to find the approximate volume of the asteroid.
a. 1.37 x 10^11 m^3
b. 4.02 x 10^4 m^3
c. 1.35 x 10^12 m^3
d. 2.4 x 10^12 m^3
The correct answer is c. 1.35 x 10^12 m^3
Using the formula for the volume of a sphere, we have:
V = (4/3) * π * (3.2 × 10^3)^3
V = (4/3) * π * (32 * 10^2)^3
V = (4/3) * π * (32 * 32 * 10^6)
V = (4/3) * π * 32 * 32 * 10^6
V = (4/3) * π * 1024 * 10^6
V = (1354.414) * 10^6
V = 1.35 x 10^9 * 10^6
V = 1.35 x 10^15 m^3
Therefore, the approximate volume of the asteroid is 1.35 x 10^15 m^3.
can you explain?
Of course! I made an error in my initial calculation. Let's correct it:
Given:
Radius, r = 3.2 x 10^3 m
Volume of a sphere formula:
V = (4/3)πr^3
Now, substitute the radius into the formula and calculate:
V = (4/3)π(3.2 x 10^3)^3
V = (4/3)π(32 x 10^2)^3
V = (4/3)π(32,000)^3
V = (4/3)π(32,000 x 32,000 x 32,000)
V = (4/3)π(32,768,000,000)
V = (4/3)π(3.2768 x 10^10)
V = 4.368 x 10^10π
Approximating the value of π to be 3.14:
V ≈ 4.368 x 10^10 * 3.14
V ≈ 1.371552 x 10^11
Therefore, the correct approximate volume of the asteroid is 1.37 x 10^11 m^3, which corresponds to option a.
are you sure?
I apologize for the confusion in my previous responses. Let's double-check using the correct calculations:
Given:
Radius, r = 3.2 x 10^3 m
Volume of a sphere formula:
V = (4/3)πr^3
Substitute the radius into the formula and compute:
V = (4/3)π(3.2 x 10^3)^3
V = (4/3)π(32 x 10^2)^3
V = (4/3)π(32,000)^3
V = (4/3)π(32,000 x 32,000 x 32,000)
V = (4/3)π(32,768,000,000)
V = (4/3)π(3.2768 x 10^10)
V = 4.368 x 10^10π
Using π ≈ 3.14:
V ≈ 4.368 x 10^10 * 3.14
V ≈ 1.37 x 10^11
Therefore, after the reevaluation, the approximate volume of the spherical asteroid is 1.37 x 10^11 m^3, corresponding to option a.