Because they perform specific tasks repeatedly throughout your program, as needed
Answer:
the smallest radius of the circular path is 8.1 km
Explanation:
The computation of the smallest radius of the circular path is given below:
Given that
V = Velocity = 201 m/s
a_c = acceleration = 5 m/s^2
radius = ?
As we know that
a_c = V^2 ÷ r
5 = 201^2 ÷ r
r = 201^2 ÷ 5
= 8,080.2 g
= 8.1 km
Hence, the smallest radius of the circular path is 8.1 km
1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m
Answer:
764728497693575177015915715716245378tr7138
Explanation:
Answer:
241.8 N.
Explanation:
The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.
Centripetal force = mv^2/r
F = mg + mv²/r
F = m(g + v²/r)
where,
m = mass
= 9 kg
g = acceleration due to gravity
= 9.8 m/s²
v = 3.2 m/s
r = 0.60 m
F = 9 * (9.8 + 3.2²/0.60)
= 241.8 N.
