Answer:
α=0.625rad/s^2
v=340m/s
w=10rad/s
θ=320rad
Explanation:
Constant angular acceleration = ∆w/∆t
angular acceleration = 20/32
α=0.625rad/s^2
Linear velocity v=wr
v = 20×17= 340m/s
Average angular velocity
w0+w1/2
w= 0+20/2
w= 20/2
w=10rad/s
What angle did it rotate with
θ=wt
θ= 10×32
=320rad
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Answer:
11
Explanation:
1. You are going to be rounding down.
2. change the metric ton to kg.
1.000 * 10^3 kg = 1000 kg
1000 / 87 = 11.49 = 11 people
