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2 years ago

What are the real life applications of Hooke's Law?

Physics

1 answer:

Anarel [89]2 years ago

5 0

Answer:

1) a rubber band

2) the spring of retractable pen

3) a spring loaded toy gun

Explanation:

Hooke's law states that; provided the elastic limit of a material is not exceeded, the force exerted on an elastic material is directly proportional to its extension. This relationship was first captured by Robert Hooke in 1660 when he asserted that 'as the extension, so is the force!'.

Hooke's law generally deals with elastic or stretchable materials. These materials can be deformed, but returned to their original shapes when the deforming force is removed. This deforming force causes an extension in the material which is directly proportional to the deforming force. That is F= Kx where K is the called the force constant, F is the deforming force and x is the magnitude of extension brought about by the force.

Various real life applications of Hooke's law have been listed in the answer. Any material that makes use of a loaded spiral spring or indeed any kind of elastic material obeys Hooke's law.

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The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor

mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

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2 years ago

Read 2 more answers

A simple pendulum of length 2.5 m makes 5.0 complete swings in 16 s. What is the acceleration of gravity at the location?

Norma-Jean [14]

<h2>The acceleration of gravity at the location is 9.64 m/s²</h2>

Explanation:

Length of pendulum = 2.5 m

Time taken for 5 swings = 16 seconds

Time taken for 1 swing = 3.2 seconds

Period of pendulum = 3.2 seconds.

We have equation for period of simple pendulum as

$T=2\pi \sqrt{\frac{l}{g}}$

Where l is the length of pendulum and g is acceleration due to gravity.

Substituting

$T=2\pi \sqrt{\frac{l}{g}}\\\\3.2=2\pi \sqrt{\frac{2.5}{g}}\\\\g=\frac{4\pi^2 \times 2.5}{3.2^2}\\\\g=9.64m/s^2$

The acceleration of gravity at the location is 9.64 m/s²

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1 year ago

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to th

Stella [2.4K]

Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

$\tau =F\times R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha$ ( Moment of Inertia of hoop is MR² )

Putting value of M, R and α in above equation, we get :

$F=5\times 2\times 2.5\ N\\\\F = 25 \ N$

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

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1 year ago

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

tatuchka [14]

By Newton's second law, assuming F is horizontal,

• the net horizontal force on the larger block is

F - µmg = 3mA

where µmg is the magnitude of friction felt by the larger block due to rubbing with the smaller one, µ is the coefficient of static friction between the two blocks, and A is the block's acceleration;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3mg is the weight of the larger block, and mg is the weight of the smaller block;

• the net horizontal force on the smaller block is

µmg = ma

where µmg is again the friction between the two blocks, but notice that this points in the same direction as F. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net vertical force on the smaller block is

mg - mg = 0

where mg is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for A and a :

A = (F - µmg) / (3m)

a = µg

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1 year ago

Question #2

Contact [7]

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

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2 years ago