The distance that would be accumulated during the journey is 2.5 meters
The parameters given in the question are written below;
average speed= 5 km/hr
time = 30 minutes
convert 30 minutes to hours
= 30/60
= 0.5 hours
Distance-= speed × time
= 5 × 0.5
= 2.5 meters
Hence the distance of the entire journey is 2.5 meters
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Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, 
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
e is the charge on electron
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:
2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
Sitting = no movement
KE=0
