A force of pounds makes an angle of with a second force. The resultant of the two forces makes an angle of to the first force. Find the magnitudes of the second force and of the resultant.

This question is incomplete, the complete question is;

A force of 193 pounds makes an angle of  79°14' with a second force. The resultant of the two forces makes an angle of 27°0' to the first force. Find the magnitudes of the second force and of the resultant.

magnitudes of the second force (vector) is 110.84 lb

the resultant force force has a magnitude of 239.85 lb

Explanation:

Given that;

Magnitude of resultant vector R = ?

Direction of resultant vector α = 27°0'

Magnitude of vector p = 193

Magnitude of vector Q = ?

Angle between two vectors ∅ = 79°14'

Using the formula

tan∝ = [ Qsin∅ / P + Qcos∅]

tan27°0' = [ Qsin79°14' / 193 + Qcos79°14' )

we cross multiply

(193tan27°0') + (Qcos79°14'tan27°0' ) = Qsin79°14'

Q = 193tan27°0' / (sin79°14' - cos79°14'tan27°0')

Q = 110.84 lb

Therefore magnitudes of the second force (vector) is 110.84 lb

Now

R = √( p² + Q² + 2PQcos∅ )

R = √( 193² + 110.84² + ( 2 × 193 × 110.84 × cos79°14'))

R = 239.85 lb

Therefore the resultant force force has a magnitude of 239.85 lb

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

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