Answer:
(a) -202 m/s²
(b) 198 m
Explanation:
Given data
- Initial speed (v₀): 283 m/s
![\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s](https://tex.z-dn.net/?f=%5Cfrac%7B632mi%7D%7Bh%7D%20.%5Cfrac%7B1609.34m%7D%7B1mi%7D%20.%5Cfrac%7B1h%7D%7B3600s%7D%20%3D283m%2Fs)
- Final speed (vf): 0 (rest)
(a) The acceleration (a) is the change in the speed over the time elapsed.
a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²
(b) We can find the distance traveled (d) using the following kinematic expression.
y = v₀ × t + 1/2 × a × t²
y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²
y = 198 m
Answer:
B
Explanation:
nothing to do with black holes creating star or related
<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>