Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
Now, the distance covered by the player in this time will be:
<u>s₁ = 0.022 m</u>
Answer:
490.5 N
Explanation:
Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction
F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.
F=kmg=0.5*100*9.81=490.5 N
Answer:
The actual elevation angle is 12.87 degrees
Explanation:
In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that 
.
Using Snell's Law we can write:
,
Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.
We thus have:
![\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13](https://tex.z-dn.net/?f=%5Cimplies%5Ctheta_1%3D%5Carcsin%5Bn_2%5Csin%28%5Ctheta_2%29%5D%3D%5Carcsin%5B1.333%5Csin%2847%29%5D%5Capprox%2077.13)
. Calling 
the actual angle of elevation, we get from the picture that 
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
