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BigorU [14]
3 years ago
13

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

the total tension upon the clothesline?

Physics
2 answers:
vredina [299]3 years ago
5 0

Answer:

15√2 N.

Explanation:

Given that the mass of the wet clothes is, m=3kg

Here in the figure ABC the weight is hung from the middle then the sides AC and BC are equal will make their adjacent angle equal and the other angle which is ∠ACB will become 90°.

Now according the figure the tension is apply in two direction than by free body diagram we can solve newton equations.

2Tcos\theta=mg\\T=\frac{mg}{2cos\theta}\\ T=\frac{3\times10}{2cos(45^{\circ})}\\ T=15\sqrt{2}N

Therefore, the total tension upon the clothesline is, T=15\sqrt{2}N

miskamm [114]3 years ago
4 0

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}

Applying values we get

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)

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Answer;

1 second

Explanation;

Two objects moving at the same speed will always stay the same distance apart. If two objects are moving at different speeds, the distance between them must change.

Therefore; if the distance will be the same and the speed is also the same then the time taken will be the same.

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3 years ago
After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter
Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, \Delta L=3.1\ cm

(a) The change in length per unit original length is called strain. So,

\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193

(b) The force acting per unit area is called stress.

\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa

(c) The ratio of stress to the strain is called Young's modulus. So,

Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2

Hence, this is the required solution.

8 0
3 years ago
Speed (mph) 58 72 55 65 70 81 66 What car goes the fastest? Put the car speed in order, from fastest to slowest miles per hour.
RoseWind [281]

Answer:

The answer to your question is letter B

Explanation:

Data

Order from fastest to slowest speed

           58 72 55 65 70 81 66

The fastest is  81, then, 72, 70, then, 66, 65 finally 58 and 55

The correct order

       81 mi/h, 72 mi/h, 70 mi/h, 66 mi/h, 65 mi/h, 58 mi/h, 55 mi/h

The other options are from slowest to fastest or a different order.

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pochemuha

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Not me

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Sorry

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Read 2 more answers
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
gogolik [260]

Answer:

7.54\cdot 10^{-7} C

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{\epsilon_0 A}{d}

where \epsilon_0 is the vacuum permittivity, A is the surface area of the plates, d their separation.

We also know the following relationship

C=\frac{Q}{V}

where Q is the charge stored on the capacitor and V the potential difference between the plates.

Combining the two equations,

\frac{\epsilon_0 A}{d}=\frac{Q}{V}

We also know that for a uniform electric field (such as the one between the plates of a parallel-plate capacitor), we have

V= Ed

where E is the magnitude of the electric field. Substituting into the previous equation and re-arranging it,

\frac{\epsilon_0 A}{d}=\frac{Q}{Ed}\\Q=\frac{\epsilon_0 A E d}{d}=\epsilon_0 A E

For the capacitor in the problem:

A=\pi r^2 = \pi (\frac{d}{2})^2 = \pi (\frac{0.19 m}{2})^2=0.0284 m^2 is the area of the plates

E=3\cdot 10^6 N/C is the maximum electric field before a spark is produced

Solving for Q, we find the maximum charge that can be added before that occurs:

Q=(8.85\cdot 10^{-12})(0.0284)(3\cdot 10^6)=7.54\cdot 10^{-7} C

3 0
3 years ago
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