draw a diagram and solve: masses m1 amd m2 at the 20cm amd 65cm marks respectively of a uniform meter rule freely suspended at its centre of gravity. If the meter rule balances horizontally, determine the ratio m2:m1.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

because if the tool that you are using is not balanced it there would be false mass.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

amplitude

hope this