A certain amount of a monatomic gas is maintained at constant volume as it is cooled by 50K. This feat is accomplished by removi
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Answer:
130 m/s (to two significant figures)
Explanation:
In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.
u represents the initial projectile velocity = 150 m/s
uₓ = u cos θ = 150 cos 30° = 129.9 m/s
uᵧ = u sin θ = 150 sin 30° = 75.0 m/s
A projectile's motion can be viewed as made up of independent vertical and horizontal elements.
The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.
Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.
Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.
At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s
Answer:
a) τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Explanation:
a) The torque can be expressed as
τ = r x F
To tackle this equation, using the determinant approach is the most straightforward method
The resulting expression is
τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) Now let's compute
τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)
τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)
τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) To find angular acceleration, we use
τ = I α
α = τ / I
The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.
α = (-0.189i^ -5.6 j^ + 33.978k^) / 241
α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²