A 45 kg object has a momentum of 255 kg m/s northward. What is the objects velocity

57. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.

(a) -16.7 N s; (b) -167 N

Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s

Find: (a) Impulse, (b) Force

(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)

Impulse = -16.7 kg•m/s = -16.7 N•s

where the "-" indicates that the impulse was opposite the original direction of motion.

(Note that a kg•m/s is equivalent to a N•s)

Explanation:

5.7 m/s?

Explanation:

answer: strength i guess

explanation:

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west