If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick

whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that
part 1
mass = ρ x V
mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg
PE (potential energy)= mgh
PE = 6608.2 kg x 9.81 x 403
PE = 2.61 x 10⁷ J
part 2
megaton of TNT (Mt) =4.2 x 10¹⁵ J
convert PE to Mt:
2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt
Answer:
The final angular velocity is 20rad/s
Explanation:
We are given;
mass, m = 12 kg
radius, r = 0.25 m
Work done;W = 75 J
Moment of inertia of cylinder, I = (1/2) mr²
Thus,
I = (1/2) x 12 x 0.25² = 0.375 kg.m²
Now, from work energy theorem,
Work done = Change in kinetic energy
So, W = KE_f - KE_i
Now, Initial Kinetic Energy (KE_i) = 0
Final Kinetic Energy; KE_f = (1/2)Iω²
So, KE_f = (1/2) x 0.375 x ω²
KE_f = 0.1875 ω²
Now, W = 75 J
Thus,
From, W = KE_f - KE_i, we have;
75 = 0.1875 ω² - 0
75 = 0.1875 ω²
ω² = 75/0.1875
ω² = 400
ω = √400
ω = 20 rad/s
Volume of the original block (V) of length (l), width (w) and height (h) is given by the following equation

hence volume of original block, V = (6)(4)(10) = 
Now say we cut this block in two equal parts, than the length of new block will be
; where as its width and height will be same as of original block i.e. 4 cm and 10 cm respectively.
So, Volume of half-block is
; or we can say that the volume of half-block is half that of orignal block.
Answer:
The cannon has an initial speed of 13.25 m/s.
Explanation:
The launched cannonball is an example of a projectile. Thus, its launch speed can be determined by the application of the formula;
R = u
Where: R is the range of the projectile, u is its initial speed, H is the height of the cliff and g is the gravitaty.
R = 26.3 m, H = 19.3 m, g = 9.8 m/
.
So that:
26.3 = u
=
x 
691.69 =
x 
= 
= 
= 175.6104
⇒ u = 
= 13.2518
u = 13.25 m/s
The initial speed of the cannon is 13.25 m/s.