A writer is creating a guide to help children learn how to take care of their first pet.

Question

What items will the writer most likely use to develop the topic? Select the two correct answers.
(1 point)

a. statistics from scientific studies

d. quotations from experts

c. examples that support claims

d. simple explanations of basic ideas

e. specialized or academic language

Answer 1

The writer will most likely use the following two items to develop the topic:

c. examples that support claims
d. simple explanations of basic ideas

These choices are suitable for a guide aimed at children, as they would help make the information accessible and relatable.

Answer 2

What is the frequency of a wave carrying 7.35 x 10^-18 J of energy?

Answer 3

To find the frequency of a wave carrying a specific amount of energy, you can use the equation that relates energy (E) to frequency (f):

\[ E = h \cdot f \]

where:

• \( E \) is the energy in joules,
• \( h \) is Planck's constant (\( 6.626 \times 10^{-34} , \text{J}\cdot\text{s} \)),
• \( f \) is the frequency in hertz (Hz).

Rearranging the formula to solve for frequency, we get:

\[ f = \frac{E}{h} \]

Now we can substitute the given energy and Planck's constant into the equation:

\[ f = \frac{7.35 \times 10^{-18} , \text{J}}{6.626 \times 10^{-34} , \text{J}\cdot\text{s}} \]

Calculating this:

\[ f \approx \frac{7.35 \times 10^{-18}}{6.626 \times 10^{-34}} \]

\[ f \approx 1.11 \times 10^{16} , \text{Hz} \]

So, the frequency of the wave is approximately \( 1.11 \times 10^{16} , \text{Hz} \).

Answer 4

What is the frequency of a 2.78 x 10^-15 J wave?

Answer 5

To find the frequency of a wave carrying a specific amount of energy, you can use the equation that relates energy (E) to frequency (f):

\[ E = h \cdot f \]

where:

• \( E \) is the energy in joules,
• \( h \) is Planck's constant (\( 6.626 \times 10^{-34} , \text{J}\cdot\text{s} \)),
• \( f \) is the frequency in hertz (Hz).

Rearranging the formula to solve for frequency, you get:

\[ f = \frac{E}{h} \]

Now, substituting the given energy \( 2.78 \times 10^{-15} , \text{J} \) into the equation:

\[ f = \frac{2.78 \times 10^{-15} , \text{J}}{6.626 \times 10^{-34} , \text{J}\cdot\text{s}} \]

Calculating this:

\[ f \approx \frac{2.78 \times 10^{-15}}{6.626 \times 10^{-34}} \]

\[ f \approx 4.19 \times 10^{18} , \text{Hz} \]

So, the frequency of the wave is approximately \( 4.19 \times 10^{18} , \text{Hz} \).

Answer 6

Provide family for Ne

Answer 7

Neon (Ne) is a chemical element that belongs to the group of noble gases on the periodic table. Its family, or group, consists of the following elements:

1. Helium (He)
2. Neon (Ne)
3. Argon (Ar)
4. Krypton (Kr)
5. Xenon (Xe)
6. Radon (Rn)
7. Oganesson (Og) (though it is a synthetic element and its properties are not fully understood)

These elements are characterized by their lack of reactivity under standard conditions due to having full outer electron shells. Neon itself is known for its use in neon signs and as a refrigerant.