All categories

2 years ago

1: In a longitudinal wave, the particle displacement is __________ the direction of wave movement

Physics

1 answer:

jenyasd209 [6]2 years ago

5 0

The answer is parallel

You might be interested in

What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?

Andru [333]

First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
$N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26}$ protons

Then, the total charge is the number of protons times the charge of a single proton:
$Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C$

8 0

3 years ago

Read 2 more answers

Which of the following is not a galilean moon?

REY [17]

The answer is B) titan

7 0

3 years ago

Read 2 more answers

What is the resultant force on the human cannonball in the

Svetllana [295]

Answer:

heart

Explanation:

4 0

2 years ago

Which best describes the law of conservation of mass?

Sergio039 [100]

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

7 0

2 years ago

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago