Explanation:
The linear analog of angle is angle itself.
The linear analog of angular velocity is linear velocity.
ω is angular velocity, therefore linear velocity is given by v
∴ for linear velocity, 
for angular velocity, 
The linear analog of angular acceleration is acceleration.
α is angular acceleration whereas as a is linear acceleration.
∴ for linear acceleration, v = u + a.t
for angular acceleration, 
The linear analog of moment of inertia is mass.
I is moment of inertia and m is mass,
∴ for linear analog, F = m.a
for angular analog, τ - I.α
The y-component of the acceleration is 
Explanation:
The y-component of the acceleration is given by
where
is the y-component of the final velocity
is the y-component of the initial velocity
t is the time elapsed
For the ice skater in this problem, we have:
u = 2.25 m/s is the initial velocity, in a direction 
v = 4.65 m/s is the final velocity, in a direction 
t = 8.33 s is the time elapsed
The y-components of the initial and final velocity are:
So the y-component of the acceleration is
Learn more about acceleration:
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According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass
. The acceleration of the insect is
, therefore we can calculate the force exerted by the car on the insect:
How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:
11 m/s
Explanation:
Draw a free body diagram. There are two forces acting on the car:
Weigh force mg pulling down
Normal force N pushing perpendicular to the incline
Sum the forces in the +y direction:
∑F = ma
N cos θ − mg = 0
N = mg / cos θ
Sum the forces in the radial (+x) direction:
∑F = ma
N sin θ = m v² / r
Substitute and solve for v:
(mg / cos θ) sin θ = m v² / r
g tan θ = v² / r
v = √(gr tan θ)
Plug in values:
v = √(9.8 m/s² × 48 m × tan 15°)
v = 11.2 m/s
Rounded to 2 significant figures, the maximum speed is 11 m/s.
Answer:
move the decimal 6 places to the left.
Explanation:
um I assume you meant to say area m^3
