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3 years ago

How can an animal regeneration produce two results?

Physics

2 answers:

GenaCL600 [577]3 years ago

8 0

In biology, regeneration<span> refers to the process by which plants and </span>animals<span> replace lost or damaged parts by growing them anew. Some </span>animals can regenerate<span> their limbs, tails, or even parts of internal organs, such as the liver. In plant </span>regeneration<span>, neighboring cells replace missing tissue.</span>

igomit [66]3 years ago

3 0

Answer: Animals can regenerate in two ways: A complete organism or a part of the organism.

Explanation:

The regeneration can be defined as the process by which a part of the organism is produced or a whole new organism is produced.

The regeneration when takes place in the lower organism where whole whole organism is regenerated. Example: Planaria, hydra et cetera.

In case of higher organism where the whole organism is made of different types of cell, it is not possible to synthesize the whole organism.

In this case a small part of the organism regenerates. Such as the injury can be recovered in higher animals.

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In a longitudinal wave, amplitude can be measured

nignag [31]

Answer:

Compressions

Explanation:

A longitudinal wave is a wave in which the direction of propagation of the wave and the direction of the displacement of the particles of the wave are parallel.

As the wave propagates, the particles do not move, but, oscillate back and forth about their equilibrium position, and there are thus regions of high pressure called compressions and regions of low pressure called rarefactions.

The amplitude is thus the distance between successive compressions.

4 0

3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot

nordsb [41]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

6 0

3 years ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .