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Bingel [31]
3 years ago
11

ou are sparring in a tae kwon do class. Your opponent executes a roundhouse kick. The average angular velocity of his kicking le

g and foot is 20 rad/s. The angular displacement from his foot to your head is 5 rad. How much time do you have to move if you want to avoid being kicked in the head?
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

0.25 s

Explanation:

The angular velocity, <em>ω</em>, is related to the angular displacement, <em>θ</em>, by

\omega = \dfrac{\theta}{t}

<em>t</em> is the time.

Making <em>t</em> subject of the formula,

t = \dfrac{\theta}{\omega}

Substituting values in the question,

t = \dfrac{5}{20} = 0.25\text{ s}

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Explanation:

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Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick
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6 0
2 years ago
Explain the difference between the synodic period of a planet and the sidereal period. why is the synodic period often much less
Ilia_Sergeevich [38]
-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
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4 0
4 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
3 years ago
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