i have been asking physics questions on this and some people give me the formula, i'm getting the correct answers..please can the same people help me on this that helped me before..
A child rides a bicycle in a circular path with a radius of 2.0 m. The tangential speed of the bicycle is 2.0 m/s. The combines mass of the bicycle and the child is 43 kg. What is the magnitude of the bicycle's centripetal acceleration?
What is the magnitude of the centripetal force on the bicycle?
The centripetal acceleration is V^2/R = 2.0 m/s^2, and the centripetal force is
(M)(V^2/R)
where M is the combined mass of bike and child.
so 2.0 m/s^2
86N
Yes. That's about 20 lb. The bike rider will have to lean toward the center of the circle to keep from tipping over.
Sure, I can help you with that. To find the magnitude of the bicycle's centripetal acceleration, we can use the formula:
Centripetal acceleration (a) = (Tangential speed)^2 / Radius
Plugging in the given values:
Tangential speed (v) = 2.0 m/s
Radius (r) = 2.0 m
a = (2.0 m/s)^2 / 2.0 m
a = 4.0 m^2/s^2 / 2.0 m
a = 2.0 m/s^2
So, the magnitude of the bicycle's centripetal acceleration is 2.0 m/s^2.
Now, to find the magnitude of the centripetal force on the bicycle, we can use the formula:
Centripetal force (F) = (Mass) x (Centripetal acceleration)
Plugging in the given values:
Mass (m) = 43 kg
Centripetal acceleration (a) = 2.0 m/s^2
F = 43 kg * 2.0 m/s^2
F = 86 N
Therefore, the magnitude of the centripetal force on the bicycle is 86 N.