Answer:
E0
Explanation:
Yes. The "0" indicates that it is spherical
Answer:
v_{4}= 80.92[m/s] (Heading south)
Explanation:
In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.
ΣPbefore = ΣPafter
where:
P = linear momentum [kg*m/s]
Let's take the southward movement as negative and the northward movement as positive.

where:
m₁ = mass of car 1 = 14650 [kg]
v₁ = velocity of car 1 = 18 [m/s]
m₂ = mass of car 2 = 3825 [kg]
v₂ = velocity of car 2 = 11 [m/s]
v₃ = velocity of car 1 after the collison = 6 [m/s]
v₄ = velocity of car 2 after the collision [m/s]
![-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]](https://tex.z-dn.net/?f=-%2814650%2A18%29%2B%283825%2A11%29%3D%2814650%2A6%29-%283825%2Av_%7B4%7D%29%5C%5Cv_%7B4%7D%3D80.92%5Bm%2Fs%5D)
Answer:
The rate of change of distance between the two ships is 18.63 km/h
Explanation:
Given;
distance between the two ships, d = 140 km
speed of ship A = 30 km/h
speed of ship B = 25 km/h
between noon (12 pm) to 4 pm = 4 hours
The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =
140 km - 120 km = 20 km
(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)
The displacement of ship B at 4pm = 25 km/h x 4h = 100 km
Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;
r² = a² + b²
r² = 20² + 100²
r = √10,400
r = 101.98 km
The rate of change of this distance is calculated as;
r² = a² + b²
r = 101.98 km, a = 20 km, b = 100 km

Answer:
(i) the resistive force exerted by the sand on the bullet is - 1 x 10⁵ N
(ii) the time taken by the bullet to come to rest is 1 x 10⁻⁴ s
Explanation:
Given;
mass of bullet, m = 10 g = 0.01 kg
speed of the bullet, v = 10³ m/s
distance traveled by the bullet, d = 5 cm = 0.05 m
(i) the resistive force exerted by the sand on the bullet
the acceleration of the bullet is given by;
v² = u² + 2as
0 = (10³)² + a(2 x 0.05)
-0.1a = (10⁶)
-a = (10⁶) / 0.1
a = -1 x 10⁷ m/s²
The resistive force is given by;
F = ma
F = (0.01)(-1 x 10⁷)
F = - 1 x 10⁵ N
(ii) the time taken by the bullet to come to rest.
v = u + at
0 = 10³ + (-1 x 10⁷ t)
1 x 10⁷ t = 10³
t = 10³ / 10⁷
t = 1 x 10⁻⁴ s