Answer:
<em>P=mgh</em>
<em>P=mghm=55</em>
<em>P=mghm=55g=9.8 or ~10</em>
<em>P=mghm=55g=9.8 or ~10h=27</em>
Explanation:
<em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em><em> and</em><em> Mark</em><em> brainlist</em><em> if</em><em> the</em><em> answer</em><em> is</em><em> correct</em>
<em>
</em>
<em> </em><em>#</em><em>c</em><em>arry </em><em>on </em><em>learning</em>
Answer:
Options A, B, and C are all possible.
Explanation:
We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?
If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.
If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.
If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.
Answer:
105 m/s
Explanation:
Given that the speed of train A, 
= 45 m/s from west to east.
Speed of train B, 
= 60 m/s from east to west.
Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.
So, the speed of train A from east to west= - 45 m/s
The speed of train B w.r.t train A 
m/s
Hence, the speed of train B w.r.t train A is 105 m/s from east to west.
