An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of sidewalk. If
it has traveled 1 m perpendicular to the edge of sidewalk, how long has it been walking
1 answer:
First draw a diagram (see attached picture).
We know the any traveled 1 m away from the sidewalk, and since we have a 45-45-90 triangle we know that the horizontal distance traveled is also 1 m. Use a^2+ b^2 = c^2 to find the distance the ant traveled. We get sqrt(2).
V=displacement/time
So 0.1 = sqrt(2) / t
Solve for the to get sqrt(2)/0.1 = 14.14 seconds.
We know the any traveled 1 m away from the sidewalk, and since we have a 45-45-90 triangle we know that the horizontal distance traveled is also 1 m. Use a^2+ b^2 = c^2 to find the distance the ant traveled. We get sqrt(2).
V=displacement/time
So 0.1 = sqrt(2) / t
Solve for the to get sqrt(2)/0.1 = 14.14 seconds.
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Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J