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Dima020 [189]
2 years ago
7

(a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Physics
1 answer:
sertanlavr [38]2 years ago
6 0

a)The tension in the vertical strand will be 0.00049 N.

b))The tension in the horizontal strand will be 0.001284 N.

c)The ratio of tension in the horizontal strand to the tension in the vertical strand will be  2.62.

<h3>What is tension force?</h3>

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

Given data;

Mass of spiderweb = 5.00 ✕ 10⁻⁵ kg

The tension in a horizontal strand of a spiderweb, Tₓ

The tension in a  vertical strand of a spiderweb, Tₐ

The angle at which the strand sags, Θ = 10.0°

R is the ratio of  tension in the horizontal strand to the tension in the vertical strand

a)

From the FBD,

The tension  in the vertical strand is equal to the weight of the spider;

Tₐ = W

Tₐ = mg

Tₐ= 5 x 10⁻⁵ kg x 9.8 m / s²

Tₐ= 0.00049 N.

(b)

The super's total vertical forces are zero since it is not moving the value of the tension in the horizontal strand;

Tₓ = mg/( 2sinΘ)

Tₓ = 5 x 10⁻⁵ kg x 9.8 m /s² / ( 2 sin 11° )

Tₓ= 0.001284 N.

c)

R is the tension in the vertical strand divided by the tension in the horizontal strand.

R= Tₐ / Tₓ

R= 0.001284 N / 0.00049 N

R= 2.62

Hence, the tension in a vertical strand of a spiderweb, the tension in a horizontal strand of a spiderweb, and the ratio of tension in the horizontal strand to the tension in the vertical strand will be 0.00049 N, 0.001284 N, and 2.62 respectively.

To learn more about the tension force refer to the link;

brainly.com/question/2287912

#SPJ1

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A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho
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Answer:

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Explanation:

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Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

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When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

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⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass (m) and angle (\theta) .

⇒ F=190\times 9.8\times sin(18)

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b.

When the force is parallel to the floor.

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⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

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So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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