Sorry Man If you ask this question no one wants to write an essay and waste their time. You are smart enough to write an essay.
Answer:443.1 s
Explanation:
Given
Engine of a locomotive exerts a force of 
Mass of train
Final speed (v)
F=ma
so 
and acceleration is
<h2>
After 26.28 seconds projectile returns 26.28 seconds.</h2>
Explanation:
Initial velocity = 450 ft/s = 137.16 m/s
Angle, θ = 70°
Consider the vertical motion of projectile,
When the projectile return to the ground we have
Displacement, s = 0 m
Acceleration, a = -9.81 m/s²
Initial velocity, u = 137.16 x sin70 = 128.89 m/s
Substituting in s = ut + 0.5 at²
s = ut + 0.5 at²
0 = 128.89 x t + 0.5 x (-9.81) x t²
t² - 26.28 t = 0
t ( t- 26.28) = 0
t = 0 s or t = 26.28 s
After 26.28 seconds projectile returns 26.28 seconds.
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps
