A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away

0.0241 m

Explanation:

mass of the hockey player m1 = 90 kg

mass of puck m2 = 0.150 kg

puck velocity v1= 45 m/s

distance traveled by puck to reach the goal =15.0 m.

now accoding to momentum conservation law

90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]

therefore, v2= -0.0725 m/s.

Now time taken by the puck to reach the goal

t= 15/45 = 1/3 sec.

therefore, how far does the player recoil in the time

=0.0725×1/3= 0.0241 m.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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