(10%) Problem 1: The surface area of a ball is measured to be A = 55 cm2 33% Par

Question

(10%) Problem 1: The surface area of a ball is measured to be A = 55 cm2 33% Part (a) Write an equation for the radius of the ball, r, treating it as a sphere, in terms of its surface area - 33% Part (b) The mass is measured to be M= 120 g. Calculate its density Q in g/cm3. Grade Summary Deductions Potential 100% 0% sinO Submissions cosO cotanO asin) acosO atanO acotan)sinh(0 cosh0 tanh) cotanh0 Degrees Radians tanOt Attempts remaining: S % per attempt) detailed view 0 END CLEAR Submit Hint I give up! Hints: 1 % deduction per hint. Hint s remaining: 3 Feedback: 1% deduction per feedback. - 33% Part (c) what is the density ekgr in kg/m3?

Explanation / Answer

Part a)

surface area of ball A = 4*pi*r^2

4*pi*r^2 = A

r = sqrt(A/4pi)

part b)

radius r = sqrt(55/(4*pi)) = 2.09 cm

density = Mass/volume

volume = (4/3)*pi*r^3 = (4/3)*pi*2.09^3 = 38.35 cm^3

density = 120/38.35 = 3.129 g/cm^3

part c

1 g/cm^3 = 10^3 kg/m^3

density = 3129 kg /m^3

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