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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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Degger [83]

The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

<h3>What is induced emf?</h3>

Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.

When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T

V(velocity)=125 M/SEC

L(length)=25 cm=0.25 m

The maximum emf is found as;

E=VBLsin90°

E=125 × 5.0 × 10⁻⁵ ×0.25

E=1.56 ×10 ⁻³ V

Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V

To learn more about the induced emf, refer to the link;

brainly.com/question/16764848

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7 0
2 years ago
Sound waves travel fastest through a A) gas. B) liquid. C) solid. D) vacuum.
earnstyle [38]

Sound waves travel faster through <em>solids</em> than they do through gases or liquids.  <em>(C)  </em>They don't travel through vacuum at all.

Example:

Speed of sound in normal air . . . around 340 m/s

Speed of sound in water . . . around 1,480 m/s

Speed of sound in iron . . . around 5,120 m/s

3 0
3 years ago
Read 2 more answers
A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U
Lubov Fominskaja [6]

Answer:

2u

Explanation:

2u

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

8 0
3 years ago
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lord [1]
I would say the plastic grip because glass, wood, and plastic are all good conductors of electricity
5 0
3 years ago
the measure of each exterior angle of a regular pentagon is ___ the measure of each exterior angle of a regular nonagon
Cerrena [4.2K]

Answer:

(a) 72°

(b) 40°

Explanation:

PENTAGON

First, we calculate the total angles in a Pentagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 5.

Hence, the total angle in a polygon is

180(5 - 2) = 180 * 3 = 540°

Therefore, each angle will be:

540°/5 = 108°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular pentagon will be:

180 - 108 = 72°

The exterior angle of a regular Pentagon is 72°

NONAGON

First, we calculate the total angles in a Nonagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 9.

Hence, the total angle in a polygon is

180(9 - 2) = 180 * 7 = 1260°

Therefore, each angle will be:

1260°/9 = 140°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular nonagon will be:

180 - 140 = 40°

The exterior angle of a regular Nonagon is 40°

4 0
3 years ago
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