Answer:
w = 4.712 10⁻³ rad / s
Explanation:
For this exercise, the time it takes for the bullet to travel the distance of 2R must be equal to the time that the hole must travel half a circle.
Let's start by calculating the time it takes for the bullet, which is going at constant speed.
v = x / t
t = x / v
t = 2R / v
t = 2 0.10 / 300
t = 6.666 10⁻⁴ s
As they ask that a single hole is formed in this time, it must be rotated half a circle, that is, θ =π rad, for which we use the angular scientific relations, where the shell has constant angular velocity
w = θ / t
w = π / 6,666 10⁻⁴
w = 4.712 10⁻³ rad / s
Resultant vector= 24 m west.
Explanation:
resultant vector= r1+ r2 when the two vectors are in same direction
r1=10m west
r2= 14 m west
so resultant displacement= 10 + 14
resultant displacement= 24 m west
10 Km.
S= Speed
D= distance
T= time
S= d/t
but since you are solving for "d" the equation is d=st so you plug in 10 km/h for speed and 2.1 hours for time and just multiply them. The hours cancel out so you are left with 10km.
