Respuesta :
Firstly, we will find the distance covered by the object in the 20 seconds:
[tex]\text{angle}(in~radians)=\dfrac{\text{arch}}{\text{radius}}\Longrightarrow \dfrac{1}{3}=\dfrac{d}{10~cm}\iff d=\dfrac{10}{3}~cm[/tex]
We must know that [tex]v=\dfrac{d}{\Delta t}[/tex]. So:
[tex]v=\dfrac{d}{\Delta t}\Longrightarrow v=\dfrac{\dfrac{10}{3}~cm}{20~s}\iff \boxed{v=\dfrac{1}{6}~cm/s}[/tex]
[tex]\text{angle}(in~radians)=\dfrac{\text{arch}}{\text{radius}}\Longrightarrow \dfrac{1}{3}=\dfrac{d}{10~cm}\iff d=\dfrac{10}{3}~cm[/tex]
We must know that [tex]v=\dfrac{d}{\Delta t}[/tex]. So:
[tex]v=\dfrac{d}{\Delta t}\Longrightarrow v=\dfrac{\dfrac{10}{3}~cm}{20~s}\iff \boxed{v=\dfrac{1}{6}~cm/s}[/tex]
The linear speed of the object is [tex] \frac{1}{6} [/tex]
- The Radius, r = 10 cm
- The Fraction of Radius swept out = [tex] \frac{1}{3} [/tex]
- The distance = [tex] 10 \div \frac{1}{3} [/tex]
- The distance covered = [tex] \frac{10}{3} [/tex]
The linear speed :
- Radius swept out ÷ time taken
- [tex] \frac{10}{3} \times \frac{1}{20} = \frac{1}{6}[/tex]
The linear speed of the object is [tex] \frac{1}{6} [/tex]
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