Answer:
Kelvin
Explanation:
fact as per the guideline given
Hello!
This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).
We can write out the conservation of momentum formula:
m1vi + m2vi = m1vf + m2vf
Let:
m1 = mass of puck
m2 = mass of the goalkeeper
We know that the initial velocity of the goalkeeper is 0, so:
m1vi + m2(0) = m1vf + m2vf
m1vi = m1vf + m2vf
The final velocities will be the same, so:
m1vi = (m1 + m2)vf
Plug in the given values:
(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s
Using the equation for momentum:
p = mv
The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of
p = (120)(0.0533) = 6.396 kgm/s.
Answer: Hello mate!
lets define the north as the y-axis and east as the x-axis.
Using the notation (x,y) we can define the initial position of the car as (0,0)
then the car travells 13 mi east, so now the position is (13,0)
then the car travels Y miles to the north, so the position now is (13, Y)
and we know that the final position is 25° degrees north of east of the initial position. This angle says that the distance traveled to the north is less than 13 mi because this angle is closer to the x-axis (or east in this case).
This angle is measured from east to north, then the adjacent cathetus is on the x-axis, in this case, 13mi
And we want to find the distance Y, so we can use the tangent:
Tan(25°) = Y/13
tan(25°)*13 mi = Y = 6.06 mi.
Answer: 5 units
Let's begin by stating clear that movement is the change of position of a body at a certain time. So, during this movement, the body will have a trajectory and a displacement, being both different:
The trajectory is the path followed by the body (is a scalar magnitude).
The <u>displacement</u> is the distance in a straight line between the initial and final position (is a vector magnitude).
According to this, in the description of the object (figure attached) placed at 0 on a number line and moving some units to the left and some oter units to the right, we are talking about the path followed by the object, hence its trajectory. So, 13 units is its trajectory.
But, if we talk about displacement, we have to draw a straight line between the initial position of the object (point 0) to its final position (point 5).
Now, being this an unidimensional problem, the displacement vector for this object is 5 units.