?
a) the velocity vector is tangent to the circle, and since the arm is moving clockwise, we can tell the direction|dw:1516843259816:dw|
|dw:1516843266460:dw|
b) centripetal acceleration points from the car to the center as usual c) a = v^2/r
a=26.8^2/14.6
good d) 8g just means 8 * g where g is the universal gravity constant
anyway for e) just use the acceleration from d) to find the new velocity
sorry i had to go.... this for c right....a=26.8^2/14.6=49.19
good units for acceleration are m/s/s as usual
D). 8 * 9.8=78.4
good
m/s?
the units are given in the problem
right m/s^2
good
okay now for E
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid anyway for e) just use the acceleration from d) to find the new velocity \(\color{#0cbb34}{\text{End of Quote}}\)
yes, so i would use 78.4
yes
OKay so the equation would be 4?
it's the same centripetal motion problem so a = v^2/r
got it
78.4=v^2/8?
or would it b e equal to 8 maybe?
the radius is the same as the radius stated in c)
got it 78.4=v^2/14.6
good
=33.83
good, 33.83 m/s
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