Answer:
t = 6.17 s
Explanation:
For a 1 revolution movement,
Torque,
Moment of Inertia,
If the wheel starts from rest,
The angular displacement of the wheel can be given by the formula:
................(1)
Where is the angular acceleration
To get t, put all necessary parameters into equation (1)
that is an example of negative acceleration because it is slowing down
The answer is Decibels. <span />
Answer:
2.72×10^-7
Explanation:
velocity = frequency × wavelength
2.05×10^8=7.55×10^14 wavelength
wavelength = 2.05×10^8/7.55×10^14
wavelength = 2.72×10^-7
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.