Energy Conservation Theory,
<h3>What is law of energy conservation?</h3>
The principle of energy conservation states that energy is neither created nor destroyed. It may change from one sort to another. Just like the mass conservation rule, the legitimacy of the preservation of energy depends on experimental perceptions; hence, it is an experimental law. The law of preservation of energy, too known as the primary law of thermodynamics
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Answer:
accelerating
Explanation:
If we consider(v > u) Acceleration:
final velocity(v)= 14m/s
initial velocity(u)=10m/s
time taken(t)= 2 seconds
a= 
=2m/s²
If we consider (v<u) Deceleration:
final velocity(v)= 3m/s
initial velocity(u)=9m/s
time taken(t)=2 seconds
a= 
= -3m/s²
Answer:
76.78 km/h To calculate the average velocity for the total trip, you need to first determine the total distance traveled and the total time taken. First, let's calculate the total distance traveled. The trip consists of 2 legs. The 1st leg is 280 km and the 2nd leg is 210 km. So the total distance is 280 km + 210 km = 490 km. Now you need to calculate the total time taken. For this problem, there are 3 intervals that need to be accounted for. The travel time for the 1st leg, the duration of the rest stop in the middle, and the travel time for the 2nd leg. The travel time for both legs is calculated by dividing the distance traveled by the average speed. So for the first leg we have 280 km / (88 km / h) = 3.181818 h The 2nd leg is 210 km / (75 km/h) = 2.8 h The rest stop in hours is 24 min / (60 min/h) = 0.4 h The total time is 3.181818 h + 2.8 h + 0.4 h = 6.381818 h The average velocity is the distance divided by the time, giving: 490 km / (6.381818 h) = 76.78 km/h
Explanation:
Hope this helps!!
Answer:
v= 0.2 m/s
Explanation:
Given that
m₁ = 50 kg
m₂ = 100 g = 0.1 kg
u =10 0 m/s
If there is no any external force on the system then the total linear momentum of the system will be conserve.
Initial linear momentum = Final momentum
m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁
50 x 0 + 0.1 x 100 = 50 v + 0
0+ 10 = 50 v
v= 0.2 m/s
Therefore the recoil speed will be 0.2 m/s.
