(1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of
the clothesline, it sags a distance of 2 meters. What is the magnitude of the tension on the ends of the clothesline
1 answer:
Answer:
Ft = 50.25 N
Explanation:
As you can see in the attached diagram, the angle θ is given by
tanθ = opp/adj
The opposite is the sag of distance 2 m and the adjacent is the half of the horizontal clothesline that is 20 m.
tanθ = 2/10
θ = tan⁻¹(2/10)
θ = 11.30°
There are two forces along the y axis, one is the F = mg acting downwards and the other is the Ftsinθ component of tension force in the string.
The sum of forces along y-axis is
5Ftsin(11.30) - mg = 0
Ft = mg/5sin(11.30)
Ft = 5*9.8/5sin(11.30)
Ft = 9.8/0.195
Ft = 50.25 N
Therefore, the magnitude of the tension on the ends of the clothesline is 50.25 N.
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Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms =
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms = = <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
=
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
=
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Answer:
The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s
= 16.125 Pa
Explanation:
The Bernoulli's equation is essentially a law of conservation of energy.
It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.
For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.
We also assume that the initial velocity of wind is 0 m/s.
This calculation is presented in the attached images to this solution.
Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.
The density is obtained to be 1.29 kg/m³.
Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.
We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.
Hope this Helps!!!
