Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
Answer:
B. Friction with air also affects the fall of the object.
Explanation:
The limitation of this experimental design is that friction with air also affects the fall of the object.
- Therefore, it is difficult to measure effect of gravity on falling objects.
- Air resistance cause friction in the movement of an object falling.
- Frictional force resists the motion of an object subject to free fall.
Therefore, the experiment will be biased due to the influence of the frictional force.
The seatbelt, as it prevents you from flying out of your window.
if this was the appropriate answer make sure to mark as the brainliest!
-procklown
Explanation:
According to the law of conservation of energy, work done by the force is as follows.
= 64.95 J
Now, gain in potential energy is as follows.
P.E = mgh
= 
= 29.4 J
Gain in potential energy will be as follows.
= 
= ![\frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2030%20N%2Fm%20%5Ctimes%20%5B%282.5%20-%201.5%29%5E%7B2%7D%20-%20%282%20-%201.5%29%5E%7B2%7D%5D)
= 11.25
As,
= 64.95 J - 29.4 - 11.25
= 24.3
v = 4.92 m/s
Therefore, we can conclude that relative velocity at point B is 4.92 m/s.
Hot air is less dense than cold air, causing it to displace and rise to the top.
