The SI unit for speed is
2 answers:
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Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α = =
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²
Answer:
a) 98.1 Joules
b) 49.05 N × sin(θ)
c) 9.81 × sin(θ)
d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s
e) 98.1 Joules
Explanation:
The given parameters of the block are;
The mass of the block, m = 5.0 kg
The distance down the plane the block slides, h = 2.0 m
The friction between the block and the surface = 0
Let θ represent the angle of inclination oof the plane
a) The gravitational potential energy, P.E. = m·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. ≈ 98.1 Joules
b) The component of the weight of the block parallel to the plane, , is given as follows;
= w × sin(θ) = m·g·sin(θ)
∴ ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N
The component of the weight of the block parallel to the plane, ≈ 49.05 N × sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
= m·g·sin(θ) = m·a
Where <em>a</em> represents the acceleration of the block along the plane
Therefore, by comparison, we have;
g·sin(θ) = a
∴ a ≈ 9.81 × sin(θ)
d) Given that the motion of the block is 2.0 m downwards, we have;
The velocity of the block at the bottom of the plane, v² = 2·g·h
Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²
v = √(39.24 m²/s²) ≈ 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²
∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J