Answer:
D
Explanation: this is why they have so much more energy released
Answer:
For these reasons at 98 mph the path is straighter
Explanation:
To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels
X = Vox t
Y = Yo + Voy t - ½ g t²
They tell us that the only parameter that changes is the speed, so the distance to the plate is known
t = Vox / x
We replace
Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2
Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²
Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)
The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.
Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter
Answer:
Explanation:
We know that
Δr = r₁ - r₀
r₀ = 0 i + 0 j
r₁ = (162+137*Cos(31º)+137*Cos(-48º)) i + (0+137*Sin(31º)+137*Sin(-48º)) j = (371.1028 i - 31.2506 j) ft
Δr = r₁ - r₀ = (371.1028 i - 31.2506 j) - (0 i + 0 j) = (371.1028 i - 31.2506 j) ft
Magnitude:
Δr = √((371.1028)²+(-31.2506)²) = 372.4163 ft
Angle:
tan θ = (- 31.2506 / 371.1028) = -0.0839 ⇒ θ = tan⁻¹(-0.0839) = - 4.8135º
(below the horizontal).
Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)
Explanation: Please see the attachments below
