A model airplane of mass 0.750 kg flies in a horizontal circle at the end of a 60.0 m control wire, with a speed of 35.0 m/s. Compute the tension in the wire if it makes a constant angle of 20.0° with the horizontal. The forces exerted on the airplane are the pull of the control wire, the gravitational force, and aerodynamic lift, which acts at 20.0° inward from the vertical.

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Nathaniel Kramer · Accepted Answer

T+mgsin⁡(θ)=ma  T+mgsin⁡(θ)=mv2r  T=mv2r−mgsin⁡(θ)  T=(0.75)(35)260−(0.75)(9.8)sin⁡(20)  T=12.8 N

madeleinejames20 · Accepted Answer

Let&#039;s break down the forces acting on the airplane along the horizontal and vertical directions.In the horizontal direction, the only force acting on the airplane is the tension in the control wire (T). Since the airplane is moving in a horizontal circle, the net force in the horizontal direction must provide the centripetal force. The centripetal force is given by Fc=mv2r, where m is the mass of the airplane, v is its speed, and r is the radius of the circle. In this case, the radius is equal to the length of the control wire, r=60.0m. Therefore, we can write the equation for the horizontal forces as:T=mv2r(equation&amp;nbsp;1)In the vertical direction, we have two forces acting on the airplane: the gravitational force (mg) and the aerodynamic lift (L). The vertical component of the tension force cancels out the gravitational force, while the vertical component of the aerodynamic lift balances the centripetal force.Let&#039;s consider the forces in the vertical direction. The vertical component of the tension force is given by Tsinθ, where θ is the angle the wire makes with the horizontal (20.0° in this case). The gravitational force is given by mg, where g is the acceleration due to gravity. The vertical component of the aerodynamic lift is given by Lcosθ. Therefore, we can write the equation for the vertical forces as:Tsinθ−mg=Lcosθ(equation&amp;nbsp;2)Now, we can solve the system of equations (equations 1 and 2) to find the tension in the wire (T). Let&#039;s substitute the values given in the problem into the equations:T=mv2r=(0.750kg)(35.0m/s)260.0m≈11.46NTsinθ−mg=LcosθSubstituting the known values:(11.46N)sin(20.0°)−(0.750kg)(9.8m/s2)=Lcos(20.0°)Simplifying:Lcos(20.0°)≈4.56NTherefore, the tension in the wire is approximately 11.46N and the aerodynamic lift is approximately 4.56N.

Eliza Beth13 · Accepted Answer

Answer:45.22 NExplanation:The tension in the wire (T) can be determined by analyzing the forces in the vertical and horizontal directions.In the vertical direction, we have the gravitational force (mg) acting downward and the vertical component of the aerodynamic lift (Lsinθ) acting upward. Since the airplane is in equilibrium vertically, these two forces cancel each other out:mg=LsinθIn the horizontal direction, the tension in the wire provides the centripetal force required to keep the airplane moving in a circle. The horizontal component of the aerodynamic lift (Lcosθ) also contributes to the centripetal force. Therefore, the sum of these two forces equals the centripetal force:T+Lcosθ=mv2rSubstituting the given values into these equations, we can solve for T.Given:m=0.750kg (mass of the airplane)v=35.0m/s (speed of the airplane)r=60.0m (radius of the circle)θ=20.0∘ (angle with the horizontal)From the equation mg=Lsinθ, we can solve for L:L=mgsinθSubstituting this value into the equation T+Lcosθ=mv2r, we can solve for T:T+mgsinθcosθ=mv2rSimplifying further:T=mv2r−mgtanθNow we can substitute the given values and calculate T.T=(0.750kg)(35.0m/s)260.0m−(0.750kg)(9.8m/s2)tan(20.0∘)Calculating the expression:T=0.750×35.0260.0−0.750×9.8tan(20.0∘)Using a calculator, we can find:T≈45.22NTherefore, the tension in the wire is approximately 45.22 N.

Nick Camelot · Accepted Answer

Step 1. Gravitational Force (Fg): This force is acting vertically downward with a magnitude equal to the weight of the airplane. We can calculate it using the equation:Fg=m·g where m is the mass of the airplane and g is the acceleration due to gravity.Step 2. Aerodynamic Lift (Flift): This force acts at an angle of 20.0° inward from the vertical. We need to find the vertical component of this force, which we&#039;ll denote as Flifty.Flifty=Flift·sin(20.0∘)Step 3. Tension in the Control Wire (T): This is the force exerted by the control wire on the airplane. It acts horizontally and provides the centripetal force required for the airplane to move in a circle.Now, let&#039;s calculate the tension in the control wire:The centripetal force is given by:Fcentripetal=m·v2r where v is the speed of the airplane and r is the radius of the circular path (length of the control wire).In our case, r=60.0 m and v=35.0 m/s.Step 4: Next, we can find the horizontal component of the tension force using:Tx=FcentripetalFinally, we can find the total tension in the wire (T) by considering the vertical component of the aerodynamic lift:T=Tx2+Flifty2Now, let&#039;s calculate the values:Gravitational force (Fg):Fg=(0.750kg)·(9.8m/s2)=7.35NAerodynamic lift (Flifty):Flifty=Flift·sin(20.0∘)Tension in the control wire (T):T=Tx2+Flifty2We can substitute the known values into these equations and solve for the tension in the wire (T).

A model airplane of mass 0.750 kg flies in a horizontal circle at the end of a 60.0 m control wire, with a speed of 35.0 m/s. Compute the tension in t

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