To calculate the horizontal velocity component, use the cosine identity. where cos A = a / h
where A is the angle the golfer hits the ball
a is the horizontal component velocity
h is the speed the golfer hits the ball
so a = h cos A
a = 31 cos 35
a = 25.4 m/s
They are experiencing the pull of leaving the atmosphere
<span>The weight lifted by a machine to the applied force on a machine is called mechanical advantage.
This is written as Mechanical advantage, M. A, = load(weight)/effort.
So for 1) M.A = 2 and load = 2, 000lb = 8896.446N.
So 2 = 8896.446/ effort
Effort = 8896.446/2 = 4448.48
Similarly for M.A of 2, 000, 000 we have
Effort = 8896.446/ 2, 000, 000 = 0.004448</span>
True hope this helps, <span>Nictheuse !</span>
Answer: 6m/s
Explanation:
Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.
After collision, the two objects will move at the same velocity (v).
Let mA and mB be the mass of the two objects
uA and uB be their velocities before collision.
v be their velocity after collision
Since the two objects has the same mass, mA= mB= m
Also since object A is at rest, its velocity = 0m/s
Velocity of object B = 12m/s
Mathematically,
mAuA + mBuB = (mA+mB )v
m(0) + m(12) = (m+m)v
0+12m = (2m)v
12m = 2mv
12 = 2v
v = 6m/s
Therefore the speed of the composite body (A B) after the collision is 6m/s
