The final velocity of the train at the end of the given distance is 7.81 m/s.
The given parameters;
- initial velocity of the train, u = 6.4 m/s
- acceleration of the train, a = 0.1 m/s²
- distance traveled, s = 100 m
The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;
v² = u² + 2as
v² = (6.4)² + (2 x 0.1 x 100)
v² = 60.96
v = √60.96
v = 7.81 m/s
Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.
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Answer:

Explanation:
We know that acceleration is change in velocity over time.


v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.

Add u to both sides.

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.
Let X be the output voltage of power supply
X∼N 
A
The lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively





Hence probability that a power supply selected at random will conform to the specifications on voltage is 0.9876
Answer:
F₁ = 4,120.2 N
F₂ = 3,924N
Explanation:
1) Balance of angular momentum around the end where F₁ is applied.
F₂ × 0.5m - F₁ × 0 = mass × g × 1m
⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m
F₂ = 196.2 Nm / 0.5m = 3,924 N
2) Balance of forces
F₁ - F₂ = mg
F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N