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babymother [125]
3 years ago
8

A girl sitting on a merry-go-round moves counterclockwise through an arc length of 3.15 m.

Physics
1 answer:
Delicious77 [7]3 years ago
4 0
Well, you get the movement as the arc in radians times the radius, which we need to find. Doing \frac{3.15}{1.82}=1.73, which is about \sqrt{3} meters.
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Which statement represents a difference between horizontal and vertical relationships?
marishachu [46]
D.) Vertical relationships involve unequal status, while horizontal relationships represent equal status.
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6 0
3 years ago
Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa
Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

v_r = Speed of river

v_c = Speed of canoe

v_r+v_c=2.8\ m/s

v_r-v_c=-1.9\ m/s

Adding the equations we get

2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s

0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
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