Physics, 04.07.2020 14:01 naomicervero

For safety in climbing, a mountaineer uses a nylon rope that is 55 m long and 1.3 cm in diameter. When supporting a(n) 83-kg climber, the rope elongates 1.4 m. Find its Young's modulus.

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Answer from: AliMe52

The Young's modulus of the rope is $1.55 \times 10^{5} \;\rm N/m^{2}$ .

Given data:

The length of rope is, L = 55 m.

The diameter of rope is, d = 1.3 cm = 0.013 m.

The mass of climber is, m = 83 kg.

The elongation of rope is, $\Delta L = 1.4 \;\rm m$ .

The ratio of stress and strain produced in a material is known as Young's Modulus of material . Then,

$E=\dfrac{\sigma }{\epsilon}\\\\E=\dfrac{W/A }{L/ \Delta L}\\\\\\E = \dfrac{W \times \Delta L}{AL}$

Here,

W is the weight of climber and its value is, W = mg.

A is the cross-sectional area of rope and its value is, $A =\dfrac{\pi}{4}d^{2}$ .

Solving as,

$E = \dfrac{mg \times \Delta L}{\dfrac{\pi}{4} \times d^{2} \times L}\\\\\\E = \dfrac{83 \times 9.8 \times 1.4}{\dfrac{\pi}{4} \times (0.013)^{2} \times 55}\\\\\\E=1138.76/7.30 \times 10^{-3}\\\\E = 1.55 \times 10^{5} \;\rm N/m^{2}$

Thus, we can conclude that the Young's modulus of the rope is $1.55 \times 10^{5} \;\rm N/m^{2}$ .

Learn more about the Young's modulus here:

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Answer from: sciencefacts4504

Young's modulus(Y) = 2.4 × 10⁸ (Approx)

Explanation:

Given:

Length of rope (l) = 55 m

Diameter of rope = 1.3 cm

Radius of rope (r) = 1.3 / 2 = 0.65 cm = 0.0065 m

Weight of climber (m) = 83 kg

Rope elongate(Δl) = 1.4 m

Find:

Young's modulus(Y).

Computation:

Force (F) = Weight of climber (m) × g

Force (F) = 83 × 9.8

Force (F) = 813.4 N

Area of rope (A) = πr²

Area of rope (A) = 0.000132665

Young's modulus(Y) = $\frac{(F)(l)}{(\delta l)(A)}$

Young's modulus(Y) = 44,737 / 0.000185731

Young's modulus(Y) = 2.4 × 10⁸ (Approx)

Answer from: Quest

(75°f − 32) × 5/9 = 23.889°c

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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For safety in climbing, a mountaineer uses a nylon rope that is 55 m long and 1.3 cm in diameter. Wh...