To develop this problem we will proceed to convert all units previously given to the international system for which we have to:



PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:


Through the Newtonian relationship of the Force we have to:




PART B) For the maximum magnitude of the acceleration downward we have that:


Through the Newtonian relationship of the Force we have to:





Answer:
1) A. 0.44 m/s East + 0.33 m/s North
2) A. 0 m/s²
3) A. a scalar calculated as distance divided by time.
4) B. 31 km per hour
Explanation:
1) Velocity is DISPLACEMENT over time.
at 1 m/s, total time of walking is 9000 seconds
displacement is 3000 m north and 5000 - 1000 = 4000 m east
4000 m/ 9000 s = 0.44 m/s E
3000 m/ 9000s = 0.33 m/s N
2) constant speed means no acceleration
3) A. a scalar calculated as distance divided by time.
4) displacement 50 km N and 80 km W
v = √(50² + 80²) / (1 + 2) = 31.446603... km/hr
Answer:
The bomb will remain in air for <u>17.5 s</u> before hitting the ground.
Explanation:
Given:
Initial vertical height is, 
Initial horizontal velocity is, 
Initial vertical velocity is, 
Let the time taken by the bomb to reach the ground be 't'.
So, consider the equation of motion of the bomb in the vertical direction.
The displacement of the bomb vertically is 
Acceleration in the vertical direction is due to gravity, 
Therefore, the displacement of the bomb is given as:

So, the bomb will remain in air for 17.5 s before hitting the ground.
When the ball has left your hand and is flying on its own, its kinetic energy is
KE = (1/2) (mass) (speed²)
KE = (1/2) (0.145 kg) (25 m/s)²
KE = (0.0725 kg) (625 m²/s²)
<em>KE = 45.3 Joules</em>
If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from: It MUST have come from the hand that threw the ball. The hand would have needed to do <em>45.3 J</em> of work on the ball before releasing it.
(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>