Ask question

Ask question

All categories

3 years ago

15

Scientific evidence documents the pattern of evolution. The evidence exists in a variety of categories. What are these categorie

s?

1 answer:

Sauron [17]3 years ago

3 0

Answer:

Explanation:

Scientific evidence takes note of the pattern of evolution. The evidence exists in a variety of categories, including direct observation of evolutionary change, the fossil record, homology, and biogeography.

Examples of the categories with their respective examples are:

Direct observation of evolutionary change: Development of drug resistant bacteria

Fossil record: Discovery of transitional forms of horses, Discovery of shells of extinct species

Homology: Similarities in mammalian forelimbs, Same genetic code in fireflies and tobacco plants, Vestigial pelvis in right whales

Biogeography: Similarity of endemic island species to nearby mainland species, The high concentration of marsupial species in Australia

You might be interested in

Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

Ganezh [65]

As we know that the formula of kinetic energy will be

$KE = \frac{1}{2} mv^2$

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

$KE = \frac{1}{2}(2)(1^2)$

$KE = 1 J$

7 0

2 years ago

Alright eksqijakojqnlqozjzbw.wlisj

saw5 [17]

Answer:

eksqijakojqnozjzbw.wlisjaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Explanation:

6 0

2 years ago

True or False A scientific law only states that an event occurs?

Kamila [148]

True

A scientific law only states that an event occurs.

Hope this helps!

3 0

3 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago