Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
A window is the most transparent object from these, so that is the answer.
Answer:
(a) The range of the projectile is 31,813.18 m
(b) The maximum height of the projectile is 4,591.84 m
(c) The speed with which the projectile hits the ground is 670.82 m/s.
Explanation:
Given;
initial speed of the projectile, u = 600 m/s
angle of projection, θ = 30⁰
acceleration due to gravity, g = 9.8 m/s²
(a) The range of the projectile in meters;

(b) The maximum height of the projectile in meters;

(c) The speed with which the projectile hits the ground is;

The electrical force acting on a charge q immersed in an electric field is equal to

where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
Answer: see the graph attached (straight line, passing through the origin and positive slope).
Justification:1)
Kinetic energy and temperature are in direct proportion. That means:
i) Being kinetic energy y and temperature x:
y α xii) That implies:
y = kx,where k is the constant of proportionality.
iii) The graph is a
line that passes through the origin and has positive slope k (k = y / x).2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the
Boltzman law, which states:
Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.