The linear speed of the ladybug is 4.1 m/s
Explanation:
First of all, we need to find the angular speed of the lady bug. This is given by:
where
T is the period of revolution
The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:
T = 1 s
Therefore, the angular speed is
Now we can find the linear speed of the ladybug, which is given by
where:
is the angular speed
r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation
Substituting, we find
Learn more about angular speed:
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The acceleration that the cheeseburger experienced is 20 m/s^2.
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
