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posledela
1 year ago
9

State two factors that affect the spring constant of a spring made using a wire of a certain material and given thickness

Physics
1 answer:
MaRussiya [10]1 year ago
8 0

The spring's rigidity is quantified by the spring constant. The spring constant increases as the wire's thickness increases. The spring constant lowers as the coil's diameter increases.

Material: Steel and copper are two materials that have differing constants when used to make identical springs.

Describe the spring constant ?

The restoring force applied to the spring for a given length of deflection is known as the spring constant. It is often referred to as the amount of force needed to extend or compress a spring one unit of length.

The level of resistance to spring deformation is expressed by the spring constant. In addition to being known as spring stiffness or spring rate, the spring constant and It is represented by the letter "K."

The following variables affect the coil spring's spring constant:-

a) Modulus of rigidity (G): As the modulus of rigidity of the spring material increases, so does the coil spring's spring constant.

b) Wire diameter (d): As the diameter of the spring wire grows, so does the spring constant.

c) Coil diameter (D): As the diameter of the spring coil increases, so does the spring's spring constant.

d) Number of coils (n): As the number of spring coils increases, the spring constant decreases.

To know more about spring constant :

brainly.com/question/14159361

#SPJ4

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A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about
Misha Larkins [42]
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
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             1 / (1/A + 1/B + 1/C + 1/D + .....) .

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In this problem, we have a humongous one and a tiny one.
Let's call them  1000  and  1 .
Then the series combination is

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5 0
2 years ago
A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmoni
SVETLANKA909090 [29]

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

E=\dfrac{1}{2}kA^2

Put all the values,

E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J

So, the value of total mechanical energy is equal to 0.01 J.

3 0
2 years ago
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Bas_tet [7]

Answer:

The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.

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