I pretty sure number 3 is false
Answer:
−x direction decreasing in speed.
Explanation:
I believe it is crude birth rate :)
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km