In deep space( far from other objects) a 3000 of asteroid travels at 500 m/s in the xy plane in the east north east direction. if that’s = 30 degrees and the asteroid passed through the x axis at x= 200 m, what is the angular momentum measured around the origin (1.5x10^8)k
(3x10^8)k
-(3x10^8)k
0 kg m^2/s
-(1.5x10^8)k

angular momentum = 1.5 × k

option 1st is correct

Explanation:

Given data

mass = 3000

velocity = 500 m/s

angle = 30 degree

x = 200 m

to find out

angular momentum

solution

we know angular momentum  about origin that is

angular momentum = x × p

angular momentum = x × m × v ×sinθ

put here all these value

angular momentum = 200 × 3000 × 500 ×sin30

angular momentum = 300000000 × 0.5

angular momentum = 150000000 k

angular momentum = 1.5 × k

so option 1st is correct

wavelegth as in spectrum red oygbiv

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