I think Im gonna have to go with C 6.00 T/s but Im not sure
Answer:
θ=108rad
t =10.29seconds
α=-8.17rad/s²
Explanation:
Given that
At t=0, Wo=24rad/sec
Constant angular acceleration =30rad/s²
At t=2, θ=432rad as it try to stop because the circuit break
Angular motion
W=Wo+αt
θ=Wot+1/2αt²
W²=Wo²+2αθ
We need to find θ between 0sec to 2sec when the wheel stop
a. θ=Wot+1/2αt²
θ=24×2+1/2×30×2²
θ=48+60
θ=108rad.
b. W=Wo+αt
W=24+30×2
W=84rad/s
This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.
Wo=84rad/sec
W=0rad/s, because the wheel stop at θ=432rad
Using W²=Wo²+2αθ
0²=84²+2×α×432
-84²=864α
α=-8.17rad/s²
It is negative because it is decelerating
Now, time taken for the wheel to stop
W=Wo+αt
0=84-8.17t
-84=-8.17t
Then t =10.29seconds.
a. θ=108rad
b. t =10.29seconds
c. α=-8.17rad/s²
Answer:
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
Explanation:
Given:
Let west be negative and east be positive x axis.
Let north be the positive y axis.
5.00blocks west = -5.00 i
5.00 blocks north = 5.00 j
7.00 blocks east = 7.00i
Addition of the vector form of hee position is;
P = -5i +7i -5j
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
