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For this problem, use the formula:
s = Vo * t + (at^2) / 2
Since the initial velocity is zero, the formula simplifies like this:
s = (at^2) / 2
Clear a:
2s = at^2
(2s) / t^2 = a
a = (2s) / t^2
Data:
s = Distance = 518 m
t = Time = 7,48 s
a = Aceleration = ¿?
Replace according formula:
a = (2*518 m) / (7,48 s)^2
Resolving:
a = 1036 m / 55,95 s^2
a = 23,34 m/s^2
The aceleration must be <u>23,34 meters per second squared</u>
Answer:
Explanation:
a) using the energy conservation equation
mgh = 0.5mv^2 + 0.5Iω^2
I(moment of inertia) (basket ball) = (2/3)mr^2
mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)
gh = 1/2v^2 + 1/3v^2
gh = v^2( 5/6)
v =
putting the values we get

solving for h( height)
h = 3.704 m apprx
b) velocity of solid cylinder
mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)
g*h = 1/2v^2 + 1/4v^2
g*h = 3/4v^2
putting the value of h and g we get
v= = 6.957 m/s apprx
Answer:
potential.
Explanation:
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Answer:
m/s
Explanation:
Assumptions: 0° is true North, and 90° is east (along the x-axis).
To solve this problem we must use the expression:

Where
is the velocity in the y-direction (East),
is the total velocity in the direction which the aircraft is travelling, and
is the direction the aircraft is travelling (angle from the y-axis).
Using the equation above, we obtain the y-component of velocity
m/s which is rounded to 227 m/s (due to the number of significant figures in the question).