All categories

3 years ago

Many organisms on Earth exhibit similar ____________.

Physics

1 answer:

valentinak56 [21]3 years ago

7 0

Answer:

The correct answer is - Characteristics.

Explanation:

On Earth, there are many organisms that shared similar characteristics with other organisms in various ways. These similarities of the characteristics could result from similar habitat, common ancestor, similar function, genetics, and many other reasons.

The example of such shared characteristics are different kinds of birds that have wings and lay eggs, while mammals give birth to babies and many other traits and characteristics. On the basis of the traits and characteristics organisms shared they are grouped and classified.

You might be interested in

Corrosion is the tendency of a metal to revert back to its natural state.

algol13

Corrosion is the irreversible damage or destruction of living tissue or material due to a chemical or electrochemical reaction.

3 0

3 years ago

As the wavelength increases, the frequency (2 points) Select one:

LiRa [457]

A.
According to the equation: v=λ*f
λ stands for wavelength. wavelength increases, the frequency will decrease. lower frequency deserves low energy.

7 0

4 years ago

States that there is an exchange of materials when two objects come into contact with each other

pishuonlain [190]

"Edmond Locard" states that there is an exchange of materials when two objects come into contact with each other.

<u>Explanation:</u>

A French criminologist who was popular as the "Sherlock Holmes of France," the pioneer in forensic science named as Dr. Edmond Locard. He articulated forensic science's fundamental principle "Each touch leaves a trace." This became known as Locard's philosophy of exchange. A Locard hypothesized that each and every time you touch another person, place or object, the result would be an exchange of materials. Burglars, for instance, will leave evidence of their existence behind and take traces with them too.

3 0

3 years ago

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

Temka [501]

The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

8 0

3 years ago

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

REY [17]

Answer: $85.46\ kJ$

Explanation:

Given

Volume of air $V=105\ m^3$

Temperature of air $T=305\ K$

Increase in temperature $\Delta T=0.7^{\circ}C$

Specific heat for diatomic gas is $C_p=\dfrac{7R}{2}$

Energy required to increase the temperature is

$\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]$

Insert the values

$\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ$

6 0

3 years ago