Answer:
230.4 s
Explanation:
The speed of car A is
![v_A = 35 mi/h](https://tex.z-dn.net/?f=v_A%20%3D%2035%20mi%2Fh)
and the distance travelled is
![d = 10 mi](https://tex.z-dn.net/?f=d%20%3D%2010%20mi)
so the time taken for car A is
![t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h](https://tex.z-dn.net/?f=t_A%20%3D%20%5Cfrac%7Bd%7D%7Bv_A%7D%3D%5Cfrac%7B10%20mi%7D%7B35%20mi%2Fh%7D%3D0.286%20h)
The speed of car B is
![v_B = 45 mi/h](https://tex.z-dn.net/?f=v_B%20%3D%2045%20mi%2Fh)
and the distance travelled is
![d = 10 mi](https://tex.z-dn.net/?f=d%20%3D%2010%20mi)
so the time taken for car B is
![t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h](https://tex.z-dn.net/?f=t_B%20%3D%20%5Cfrac%7Bd%7D%7Bv_B%7D%3D%5Cfrac%7B10%20mi%7D%7B45%20mi%2Fh%7D%3D0.222%20h)
So the difference in time is
![\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20t_A%20-%20t_B%20%3D%200.286%20h%20-0.222%20h%3D0.064%20h)
Which corresponds to
![\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%200.064%20h%20%5Ccdot%203600%20s%2Fh%20%3D%20230.4%20s)
so car B arrived 230.4 s before car A.
Answer:
0 to 145 degrees
Explanation:
The normal range of flexion and extension is from 0 to 145 degrees.
Answer:
20 m
Explanation:
Given:
v₀ = 15 m/s
v = -25 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy
Δy = 20 m