We can use the textbook results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall embedded in the Earth. Suppose a professional baseball pitcher hurls a baseball (m = 155 grams = 0.155 kg) with a speed of 103 miles per hour (vball = 45.3 m/s) at a wall, and the ball bounces back with little loss of kinetic energy.(a) What is the recoil speed of the Earth (M = 6multiply1024 kg)?

Recoil velocity; v2 = 1.17 x 10^(-24) m/s

Explanation:

We are given;

Mass of baseball; m1 = 0.155 kg

Velocity of baseball; v1 = 45.3 m/s

Mass of earth; m2 = 6 x 10^(24) kg

Lets the recoil velocity be v2.

From conservation of momentum, initial momentum is equal to final momentum.

Thus,

m1•v1 = m2v2

Thus, v2 = m1•v1/m2

Plugging in the relevant values to get ;

v2 = 0.155 x 45.3/(6 x 10^(24))

v2 = 1.17 x 10^(-24) m/s

no

explanation:

no i dont know how to solve this you should probably try to google the answer

the answer you are looking for is d. there are no holes for blood to enter or leave the system. your welcome