Answer:
Final velocity of both goalie & puck = 0.018116 m/s
Explanation:
M1U1 + M2U2 = (M1+M2)V
70 x 0 + 0.17 x 33.5 = (70+0.17)V
V = 0.08116m/s
Answer:
0.35
Explanation:
According to Newton's second law;
\sum Fx = ma
Fm - Ff =ma
Fm is the moving force = Wsin theta
Fm = 4(9.8)sin55
Fm = 32.1N
Ff is the frictional force = nmgcos theta
Ff = n(4)(9.8)cos55
Ff = 22.48n
Acceleration a = 6.0m/s²
Substitute the given values into the formula and get the coefficient of friction
32.11-23.48n = 4(6)
32.11-24= 23.48n
8.11 = 23.48
n = 8.11/23.48
n = 0.35
Hence the coefficient of friction is 0.35
Answer:
Staying connected to friends
Explanation:
hope this helps
A negative object so it takes in the charged object and nothing will happen until something else would touch it I think
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.