All categories

4 years ago

Which statement best describes the process that produced the ending populations of moths

Physics

2 answers:

Yuki888 [10]4 years ago

8 0

The answer is B. Dark-colored moths were selected for in the new environment

zzz [600]4 years ago

3 0

C. Light color became an adaptation in the surviving moths? Not fully sure.

You might be interested in

Can water and wind change the shape of a mountain

Art [367]

Answer:

Yes, through erosion.

Explanation:

Water, wind, and ice shape earths surface. Water, wind, & ice move sediment to another area this process is called erosion.

Mark me brainliest, hope this helps

4 0

3 years ago

Many years ago, a scientist discovered an animal P and studied its characteristics. Some of the characteristics of animal P are

tatuchka [14]

Answer:

Explanation:

most mammals are viviparous instead of oviparous. oviparous means laying eggs

5 0

2 years ago

The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. Th

tiny-mole [99]

Answer:

c. There would be a series of spectral lines in hydrogen with the longest wavelength one at 122 nm.

d. The hydrogen atom binds its electron more tightly than the sodium atom does, and would require more energy to remove its electron completely.

Explanation:

The hydrogen atom which changes from the excited state to the lower ground state, it emits light having a wavelength of 122 m. And the sodium atom also gets excited and emits light at 589 nm when it moves from the 1st excited state to the lowest excited state.

Therefore, when the electrons jumps from the 1st excited state to the ground state, only one wavelength is observed as there is only one transition.

The hydrogen atom will bind the electron tightly but the sodium atom does not and would require more energy to remove the electron the electron completely as the binding energy is higher when the electron is closer to the nucleus.

5 0

3 years ago

16 En un juego de béisbol, un lanzador logra

Anuta_ua [19.1K]

Answer:

B transferida.

Explanation:

The kinetic energy of the baseball would be transferred to the bat and to the handler.

The law of conservation of energy states that "energy is neither created nor destroyed but transformed from one form to another".

We know that the baseball posses kinetic energy as it begins to accelerate towards the other player.
Kinetic energy is the energy due to the motion of a body.
As it collides with the bat, it is converted to another form of energy.
The energy form would have been converted to kinetic energy, heat energy and some mechanical energy component.
Therefore, the energy has been transferred from the ball to the bat.

5 0

3 years ago

A 3.00 kg mass is pushed against a spring and released. If the spring constant of the

Phoenix [80]

Answer:

Energy stored in the compressed spring: $37.5\; \rm J$ before the mass is released.

Maximum horizontal speed of the mass: $5.0\; \rm m \cdot s^{-1}$ .

(Assumption: the surface between the mass and the ground is frictionless until the mass separates from the spring.)

(Maximum horizontal acceleration of the mass: $250\;\rm m \cdot s^{-2}$ .)

Work that friction did on the mass: $13.5\; \rm J$ .

Maximum height of the mass: approximately $0.459\; \rm m$ (assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .)

Explanation:

<h3>Energy stored in the spring</h3>

When an ideal spring of spring constant $k$ is compressed by a displacement of $x$ from the equilibrium position, the elastic potential energy stored in the spring is:

$\displaystyle \frac{1}{2}\, k \cdot x^{2}$ .

The question states that for this spring, $k = 7500\; \rm N \cdot m^{-1}$ (Newtons per meters) whereas $x = 10.0\; \rm cm$ (centimeters.) Convert the unit of $x$ to meters to match the unit of $k$ :

$\begin{aligned} x &= 10.0\; \rm cm\\ &= 10.0\; \rm cm \times \frac{1\; \rm m}{100\;\rm cm} = 0.100\; \rm m\end{aligned}$ .

Calculate the energy stored in this spring:

$\begin{aligned}& \text{Elastic Potential Energy} \\ &= \frac{1}{2}\, k \cdot x^{2} \\ &= \frac{1}{2} \times 7500\; \rm N \cdot m^{-1} \times 0.100\; \rm m \\ &=37.5\; \rm J \end{aligned}$ .

<h3>Maximum horizontal speed of the mass</h3>

Assume that the surface under the spring is frictionless. All that $37.5\; \rm J$ of elastic potential energy would be converted to kinetic energy by the time the mass separates from the spring.

The horizontal speed of the mass is largest at that moment. The reason is that immediately after this moment, friction (between the surface and the mass) would start to slow the mass down immediately.

Calculate the speed $v$ of the mass $m$ at that moment.

$\displaystyle \text{Kinetic Energy} = \frac{1}{2}\, m \cdot v^{2}$ .

Therefore:

$\displaystyle v = \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}}$ .

The question states that $m =3.00\; \rm kg$ . At that moment, the kinetic energy of the mass is $37.5\; \rm J$ (or equivalently, $37.5\; \rm kg\cdot m^{2} \cdot s^{-2}$ in base units.)

Calculate the speed of the mass at that moment from the kinetic energy of the mass:

$\begin{aligned}v &= \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}} \\ &= \sqrt{\frac{2 \times 37.5\; \rm kg \cdot m^{2}\cdot s^{2}}{3.00\; \rm kg}} = 5.00\; \rm m \cdot s^{-1}\end{aligned}$ .

<h3>Work friction did on the mass</h3>

Assume that the rough surface is level. Therefore, all the energy loss of the mass should be attributed to friction.

Kinetic energy of the mass before coming onto that rough surface: $37.5 \; \rm J$ .

Kinetic energy of the mass before leaving the rough surface:

$\begin{aligned} &\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 3.00\; \rm kg \times {\left(4.00\; \rm m \cdot s^{-1}\right)}^{2} \\ &= 24.0\; \rm J \end{aligned}$ .

If that rough surface is level, there would be no change to the gravitational potential energy of the mass. Calculate the change to the kinetic energy of the mass:

$37.5\;\rm J - 24.0\; \rm J = 13.5\; \rm J$ .

That should be equal to the size of the work that friction did on the mass.

<h3>Maximum height of the mass</h3>

Assume that the smooth ramp does not change the total energy of the block. Therefore, the total mechanical energy of the block would still be $24.0\; \rm J$ when the height of the block is maximized. However, all that energy would be in the form of gravitational potential energy.

Let $g$ denote the gravitational field strength ( $g \approx 9.8\; \rm m \cdot s^{2}$ near the surface of the earth.) The gravitational potential energy of an object of mass $m$ and height $h$ (relative to the surface of zero gravitational potential energy) would be:

$\begin{aligned}&\text{Gravitational Potential Energy} = m \cdot g \cdot h\end{aligned}$ .

Rewrite this equation to find an expression for $h$ :

$\displaystyle h = \frac{\text{Gravitational Potential Energy}}{m\cdot g}$ .

Assume that $g = 9.8\; \rm m \cdot s^{-2}$ . The question states that $m = 3.00\; \rm kg$ . $\text{Gravitational Potential Energy} = 24.0\; \rm J$ . Therefore:

$\begin{aligned} &h\\ &= \frac{\text{Gravitational Potential Energy}}{m\cdot g}\\ &= \frac{24.0\; \rm kg \cdot m^{2} \cdot s^{-2}}{3.00\; \rm kg \times 9.8\; \rm m \cdot s^{-2}} \approx 0.459\; \rm m\end{aligned}$ .

3 0

3 years ago