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3 years ago

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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

the spring elongated a total length 17 cm. What was the relaxed length of the spring? Answer in units of m.

1 answer:

Liula [17]3 years ago

6 0

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

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Neporo4naja [7]

Answer:

Explanation:

Given

Load $W=20000\ N$

height to which load is raised $h=250\ m$

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Second Crane requires 6 times more power than the slow crane

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2 years ago

Calculate the mass (in g) of 0.8 cm³ of steel. The density of steel is 7.8 g/cm³. Give your answer to 2 decimal places.

zysi [14]

mass= density times volume so 7.8 g/cm3 by0.8 cm3 answer is 6.24 g and it's already in two decimal places

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3 years ago

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Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a

Tasya [4]

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length, $\rm L_f = 1.45 m = 1450 \ mm$

Initial length, $\rm L_i = 7.5 mm$

$\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N$

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

To learn more about the force refer to the link;

brainly.com/question/26115859

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1 year ago

the guage pressure in a car tire is 30.0 psi when the air temperature is 0 C as the day warms up and brighten sun shines What is

Viktor [21]

Answer:

$P_2 = 33.297\ psi$

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give,

Gauge pressure of car, P₁ = 30 psi

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$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

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$P_2 = 33.297\ psi$

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miss Akunina [59]

Answer:

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3 years ago

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