Place least complex to most complex:

Physics

1 answer:

Reika [66]3 years ago

7 0

Cell, tissue, organ, organ sytem, organism, population, community, ecosystem, biome.

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Anthony walks to the pizza place for lunch. He walk 1km east, then 1km south and then 1km east again. What distance did he cover

Komok [63]

Answer:

Pizza pizzza pizza

Explanation:

little ceasrs

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3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch

Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules