A hawk flies in a horizontal arc of radius 8.3 m
at a constant speed of 4.4 m/s.
Find its centripetal acceleration.
Answer in units of m/s
It continues to fly along the same horizontal arc but increases its speed at the rate of
1.11 m/s
2
.
Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s
at a constant speed of 4.4 m/s.
Find its centripetal acceleration.
Answer in units of m/s
It continues to fly along the same horizontal arc but increases its speed at the rate of
1.11 m/s
2
.
Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s
-
centripetal acceleration = v^2 /r
= 4.4^2 / 8.3
= 2.33 m/s^2
answer
centripetal acceleration = v^2 /r
= 11^2 / 8.3
= 14.58 m/s^2
answer
= 4.4^2 / 8.3
= 2.33 m/s^2
answer
centripetal acceleration = v^2 /r
= 11^2 / 8.3
= 14.58 m/s^2
answer
-
Cenrtipetal force = v^2/r towards the centre of the circle (centripetal means centre seeking)
v^2 = 4.4*4.4 = 19.36 m^2/s^2
r = 8.3 m
a=19.36/8.3 = 2.333 m/s^2 towards the centre
The hawk accelerates at 1.11m/s^2, so dv/dt = 1.11 m^s and dv = 1.11dt m/s
The velocity becomes v + dv = (4.4 ms+ 1.11dt m/s) and acceleration = (4.4 +1.11dt)/8.3 m/s^2
v^2 = 4.4*4.4 = 19.36 m^2/s^2
r = 8.3 m
a=19.36/8.3 = 2.333 m/s^2 towards the centre
The hawk accelerates at 1.11m/s^2, so dv/dt = 1.11 m^s and dv = 1.11dt m/s
The velocity becomes v + dv = (4.4 ms+ 1.11dt m/s) and acceleration = (4.4 +1.11dt)/8.3 m/s^2