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OverLord2011 [107]
3 years ago
7

how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?

Physics
1 answer:
HACTEHA [7]3 years ago
3 0

Answer:

mass

Explanation:

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A coconut falls from a height of 18m to the ground. What is the velocity at which the coconut hits the ground?
Luden [163]

Answer:

v=18.97 m/s

Explanation:

the equation for the above question is v²=u²+2as

7 0
4 years ago
10 PTS.
gayaneshka [121]

1.2 x (2.2 x 10⁵) = 264,000 Ω

0.8 x (2.2 x 10⁵) = 176,000 Ω

With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .

8 0
3 years ago
How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a po
miss Akunina [59]

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m  from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² .  How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

E=\frac{3.90*10^{-12}C/m^{2}  }{2(8.85*10^{-12}C^{2} /N.m^{2} )}\\ E=0.22V/m

Now we need to find force on electron

F=eE\\F=(1.6*10^{-19}C )(0.22V/m)\\F=3.525*10^{-20}N

Now for Work done on the electron

W=F*d\\W=(3.525*10^{-20} N)(3.00*10^{-2}m)\\W=1.06*10^{-21}J

4 0
4 years ago
6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces
Vesna [10]

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket V=15\ gal\approx 56.78\ L

time to fill it t=50\ s

Volume flow rate

\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s

The inner diameter of the hose D=1.5\ cm

diameter of the nozzle exit d=0.8\ cm

we can volume flow rate as

\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s

Mass flow rate

\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s

5 0
3 years ago
Tectonic plates move due to _____ forces. A. internal B. external C. destructive D. constructive
Oliga [24]
Internal forcessssssssssss
6 0
3 years ago
Read 2 more answers
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