I think its b im not sure im in 8th grade
Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)
The distance covered by an object accelerating from rest is
D = (1/2) · (acceleration) · (time)² .
In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.
D = (1/2) · (9.8 m/s²) · (1.67 s)²
D = (4.9 m/s²) · (2.789 s²)
D = 13.67 meters
Answer:
54.6°
Explanation:
From law of reflection i=r.
So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.
Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.
If you consider triangle AOB, one angle is ∠AOB=90°
and ∠OAB is 54.6°
From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°
So, the second incident angle will be 54.6°
Hence, the second reflected angle will be 54.6 degrees.
