Answer:
5 Days to Seconds = 432000
Explanation:
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is 
Therefore acceleration during these 5 s


therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s


(b)total distance traveled before stoppage


s=136.89 m
Answer:

Explanation:
Given that
The speed of the airplane ,v= 142 m/s
The speed of the air ,u = 30 m/s
Lets take angle make by airplane from east direction towards north direction is θ .
Now by using diagram ,we can say that

Now by putting the values in the above equation we get



Therefore the angle will be 12.19° .
Answer:
1. 610,000 lb ft
2. 490 J
Explanation:
1. First, convert mi/hr to ft/s:
100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s
Now find the kinetic energy:
KE = ½ mv²
KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²
KE = 610,000 lb ft
2. KE = ½ mv²
KE = ½ (5 kg) (14 m/s)²
KE = 490 J
I think you almost got it.
At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.
With the value of the initial speed (28 m/s, which is the total speed), you can set
v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7
or theta = 64.62 deg, it is D. Think about it. I hope you see it.