The density of lead is 11.34 g/cm3 and the density of aluminum is 2.69 g/cm3. Find the radius of lead and aluminum spheres each having a mass of 70 kg. (Round your answers to two decimal places.)

Iam lost on everything I got this far
p(4/3)πr^3=m
r-(3m/(4πp))^1/3

Well, 70 kg = 70,000 g

so the volume of the lead ball = 70000/11.34 cm^3 = 6172.8395 cm^3
the volume of the aluminium ball = 70000/2.69 cm^3 = 26022.305 cm^3

Now we'll use our sphere formula:
for the lead,
(4/3)πr^3 = 6172.8395
r^3 = 1473.65...
r = (1473.65...)^(1/3) = appr 11.38 cm

for the aluminum,
(4/3)πr^3 = 26022.305
r^3 = 6212.367...
r = 18.38 cm

To find the radius of the lead and aluminum spheres, we need to use their respective densities and the given mass.

We can start by rearranging the volume formula to solve for the radius (r):

Volume of a sphere (V) = (4/3)πr^3

Since the density (p) is given by mass (m) divided by volume (V):

p = m / V

We can substitute the volume formula into the density formula:

p = m / [(4/3)πr^3]

Rearranging the formula to solve for the radius (r):

r^3 = (3m) / (4πp)

Taking the cube root of both sides:

r = [(3m) / (4πp)]^(1/3)

Now, we can calculate the radius for both lead and aluminum using the given mass and densities:

For lead:
p_lead = 11.34 g/cm^3
m_lead = 70 kg

Converting mass to grams:
m_lead = 70 kg * 1000 g/kg = 70,000 g

Substituting the values into the formula:

r_lead = [(3 * 70,000 g) / (4 * π * 11.34 g/cm^3)]^(1/3)
r_lead = [(210,000 g) / (45.36 g/cm^3)]^(1/3)
r_lead = (4631.35)^(1/3)
r_lead ≈ 16.18 cm

For aluminum:
p_aluminum = 2.69 g/cm^3
m_aluminum = 70 kg

Converting mass to grams:
m_aluminum = 70 kg * 1000 g/kg = 70,000 g

Substituting the values into the formula:

r_aluminum = [(3 * 70,000 g) / (4 * π * 2.69 g/cm^3)]^(1/3)
r_aluminum = [(210,000 g) / (10.724 g/cm^3)]^(1/3)
r_aluminum = (19,539.1)^(1/3)
r_aluminum ≈ 27.93 cm

Therefore, the radius of the lead sphere is approximately 16.18 cm, and the radius of the aluminum sphere is approximately 27.93 cm.