Answer:
a) Magnitude = 1.03 m/s², Direction: south
b) ![V_{f} = 8.16 m/s](https://tex.z-dn.net/?f=V_%7Bf%7D%20%3D%208.16%20m%2Fs%20)
Explanation:
a) The magnitude and direction of the acceleration can be calculated using the following equation:
(1)
Where:
: is the final speed = 9.40 m/s
: is the initial speed = 13.0 m/s
t: is the time = 3.50 s
Solving equation (1) for a, we have:
![a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7BV_%7Bf%7D%20-%20V_%7B0%7D%7D%7Bt%7D%20%3D%20%5Cfrac%7B9.40%20m%2Fs%20-%2013.0%20m%2Fs%7D%7B3.50%20s%7D%20%3D%20-1.03%20m%2Fs%5E%7B2%7D%20)
Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating.
b) The final velocity of the bird can be found using the same equation 1:
![V_{f} = V_{0} + at](https://tex.z-dn.net/?f=%20V_%7Bf%7D%20%3D%20V_%7B0%7D%20%2B%20at%20)
![V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s](https://tex.z-dn.net/?f=%20V_%7Bf%7D%20%3D%2013.0%20m%2Fs%20%2B%20%28-1.03%20m%2Fs%5E%7B2%7D%29%2A%283.50%20s%20%2B%201.20%20s%29%20%3D%208.16%20m%2Fs%20)
Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.
I hope it helps you!
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
Answer:
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.