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PIT_PIT [208]
3 years ago
5

What is the frequency of radiation whose wavelength is 0.80 nm ? express the frequency in inverse seconds to three significant d

igits?
Physics
1 answer:
Mashcka [7]3 years ago
5 0
The relationship between the frequency (f) of an electromagnetic wave, the speed of the wave (which is the speed of light, c) and its wavelength \lambda is given by
f= \frac{c}{\lambda}

Since the wavelength of the radiation in the problem is \lambda= 0.80 nm = 0.80 \cdot 10^{-9} m, by substituting numbers in the formula we  can get the frequency:
f= \frac{3 \cdot 10^8 m/s}{0.80 \cdot 10^{-9} m} =3.75 \cdot 10^{17} s^{-1}
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What do all members of a biological species have in common?
Vadim26 [7]
A)they can interbreed and produce fertile offsprings
6 0
4 years ago
Read 2 more answers
Neutral hydrogen can be modeled as a positive point charge +1.6×10^−19C surrounded by a distribution of negative charge with vol
Archy [21]

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

4 0
3 years ago
For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0
4 years ago
The distance a toy car travels over time is shown in the graph.
Korvikt [17]

Answer:

a toy car speed is about 2.5 to 3.5 mph

6 0
3 years ago
A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.
puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0
3 years ago
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