Answer: 11369.46 m/s
Explanation:
We have the following data:
is the mass of the bowling ball
is the velocity of the bowling ball
is the mass of the ping-pong ball
is the velocity of the ping-pong ball
Now, the momentum of the bowling ball is:
(1)
(2)
And the momentum of the ping-pong ball is:
(3)
If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:
(4)
(5)
Isolating :
(6)
(7)
Finally:
Work done against gravity to climb upwards is always stored in the form of gravitational potential energy
so we can say
here h = vertical height raised
so here we know that
here we have
now from above equation
so work done will be given by above value
Answers:
a) -2.54 m/s
b) -2351.25 J
Explanation:
This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum must be equal to the final momentum :
(1)
Where:
(2)
(3)
is the mass of the first football player
is the velocity of the first football player (to the south)
is the mass of the second football player
is the velocity of the second football player (to the north)
is the final velocity of both football players
With this in mind, let's begin with the answers:
a) Velocity of the players just after the tackle
Substituting (2) and (3) in (1):
(4)
Isolating :
(5)
(6)
(7) The negative sign indicates the direction of the final velocity, to the south
b) Decrease in kinetic energy of the 110kg player
The change in Kinetic energy is defined as:
(8)
Simplifying:
(9)
(10)
Finally:
(10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
I believe the answer is c