All categories

3 years ago

A bowling ball moves 18 meters every 2 seconds down the lane at a bowling alley. What is the speed of the bowling ball?

Physics

1 answer:

Strike441 [17]3 years ago

3 0

Answer:

9m/s

Explanation:

18/2 = 9m/s

You might be interested in

Smooth surfaces produce a __________________ reflection.

Galina-37 [17]

Answer:

I really hope this is right I think this is Diffuse I'm sorry if its worng

6 0

3 years ago

Hi i need help. thNKS

Ede4ka [16]

Answer:

Explanation:

Groups go down, periods go across :)

3 0

3 years ago

Please subscribe wonder melodies please on you tube channel

Anuta_ua [19.1K]

Sure , I'm subscribing!!!

8 0

2 years ago

Read 2 more answers

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

On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his

PilotLPTM [1.2K]

Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
33x/2808=11
33x= 30888
x=936miles
U can substitue back to check
at speed of 72mph, he would need 936/72=13hrs
at speed of 39mph, he would need 936/39=24hr
the difference is 24-13=11

5 0

3 years ago