Answer:
Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Explanation:
<em>a)Which of the two has uniform acceleration?</em>
Acceleration is the second derivative of position. The acceleration of the first particle is:
x = 4t² − 2t
v = 8t − 2
a = 8
The acceleration of the second particle is:
x = 6t³ + 8t
v = 18t² + 8
a = 36t
The first particle has uniform acceleration.
<em>b)Which one is likely to come to rest at some time during its motion?</em>
The particles come to rest when v = 0. The first particle's velocity has a real zero at t = 4. The second particle's velocity has only imaginary zeros, meaning v is never 0.
Answer:
<em>1. c. Same in both</em>
<em>2. a. Case 1</em>
<em></em>
Explanation:
1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.
2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.
Answer:
The amplitude of the wing tip's motion is 1.6 mm.
Explanation:
Given that,
Beat = 250 /s
Speed = 2.5 m/s
We need to calculate the amplitude of the wing tip's motion
Using the equation for the maximum velocity
Where,
v = speed
f = frequency
A = amplitude
Put the value into the formula
Hence, The amplitude of the wing tip's motion is 1.6 mm.
