The group of test subject that are NOT given the experimental treatment is called what?

What property makes ocean water a good conductor of electricity

Lostsunrise [7]

Answer:

The salt molecules (Sodium ions and Chlorine ions) in the ocean water makes it a good conductor of electricity.

Explanation:

Conductivity refers to the ability of an object to transmit electricity. Transmitting electricity means allowing electricity to pass through the material. The best conductor of electricity is "silver." Other elements which are also good conductors are: gold, aluminum, copper, etc. Those which are poor in conducting electricity are called insulators. These include glass, plastic, rubber, paper, styrofoam, etc.

Ocean water is considered a good conductor of electricity because it contains Sodium ions and Chlorine ions. Remember that ions carry electrical charges, which could either be positive or negative. The Sodium ion here is positively charged, while the Chlorine ion is negatively charged. Being mixed with water makes it an electrolyte solution. When electricity travels, it needs to have two sources: a positive terminal and a negative terminal. This means that the Sodium ions will be attracted to the electricity's negative terminal, while the Chlorine ions will be attracted to the positive terminal. This will allow the electrons to flow through the medium, thus creating a circuit for electricity to pass through.

This is the reason why ocean water is a good conductor of electricity.

5 0

3 years ago

Consider a traveling wave described by the formula

Harrizon [31]

Answer:

The wave is traveling in the +x direction.

Explanation:

The equation of a wave is given by the formula as :

$y(x,t)=A\ sin(kx-\omega t)$

Here,

A is the amplitude of wave

$(kx-\omega t)$ is the phase of wave

$\omega$ is the angular frequency of the wave

We need to find the correct statement out of given options. The given equation can be rewritten as :

$y(x,t)=A\ sin(\omega t-kx)$

Here, the propagation constant is negative. So, the wave is moving in +x direction. Hence, the correct option is (a).

4 0

3 years ago

A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch p

kirill115 [55]

Answer:

Change in mechanical energy, $\Delta E=283.2\ J$

Explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2$

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

$P=12\ kg\times 9.8\ m/s^2\times 18\ m$

P =2116.8 J

The change in mechanical energy is given by :

$\Delta E=K-P$

$\Delta E=2400-2116.8$

$\Delta E=283.2\ J$

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.

8 0

2 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago

Can You Please Give Me Some Examples of Density-Independent and Density-Dependent Limiting Factors?

ivolga24 [154]

Density-Dependent:
1. competition.
2. overcrowding.
3. predators.

(These are a few from a test I took, hopefully they help you a bit >.<)

7 0

3 years ago