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NNADVOKAT [17]
3 years ago
15

The diameter of a hydrogen atom is 0.000000000106 m. How can this

Physics
1 answer:
Andreas93 [3]3 years ago
8 0

Answer:

The diameter of a hydrogen atom based on scientific notation is 1.06 x 10^-10 m

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a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra
Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= \frac{v}{f}

λ= \frac{30000}{90500}

λ= 0.311m

5 0
2 years ago
Calculate the speed of a car that travels 250 miles in 4.0 hours. (remember your unit)
MrRa [10]
250/4= 62.5 mph

to find the mph of a car, you need to divide the number of miles traveled by the hours that it took to travel that many miles
6 0
3 years ago
"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s
Tamiku [17]

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

6 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
Which explains whether boron or arsenic is the dopant most likely used in the semiconductor layer marked with +?
liubo4ka [24]

Answer:

D) This layer is the emitter of a PNP transistor, so it could include boron.

   Got it on quiz! Good luck!

   

Explanation:

4 0
3 years ago
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