A sphere of mass 2.9x 10^-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 41° with the vertical
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To find the tension in the cord, we can analyze the forces acting on the sphere:
1. The weight of the sphere (mg): The weight is given by the formula mg, where m is the mass of the sphere and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Weight = mg = (2.9 x 10^-4 kg) * (9.8 m/s^2)
2. The horizontal component of the tension (Tcosθ): The tension in the cord can be divided into two components, one acting horizontally and the other vertically. The horizontal component can be found using the formula Tcosθ, where T is the tension in the cord and θ is the angle made by the cord with the vertical.
Horizontal Component = Tcosθ
3. The vertical component of the tension (Tsinθ): Similar to the horizontal component, the vertical component of the tension can be found using the formula Tsinθ.
Vertical Component = Tsinθ
In equilibrium, the vertical component of the tension balances the weight of the sphere, so:
Vertical Component = Weight
Tsinθ = (2.9 x 10^-4 kg) * (9.8 m/s^2)
Now that we have the equation for the vertical component, we can substitute it into the equation for the horizontal component:
Horizontal Component = Tcosθ
Tcosθ = T * (cosθ)
= T * (sqrt(1 - sin^2θ))
Equating the horizontal component to the weight:
T * (sqrt(1 - sin^2θ)) = (2.9 x 10^-4 kg) * (9.8 m/s^2)
Now, divide both sides by sqrt(1 - sin^2θ) to solve for T:
T = [(2.9 x 10^-4 kg) * (9.8 m/s^2)] / sqrt(1 - sin^2θ)
Substitute the value of θ (41°) into the equation to find the tension in the cord.