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Vlad1618 [11]
3 years ago
7

Which rf characteristic best determines the range of a 2.4 ghz ism signal?

Physics
1 answer:
Rainbow [258]3 years ago
3 0
<span>The radio frequency characteristic that best determines the range of a 2.4 GHz ism signal is the wavelength.

This frequency can be used in WiFi and can reach up to 46 meters when indoors and about 92 meters when outdoors.
</span><span>
</span>
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Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
A generator converts mechanical energy into _____________________ energy.
erica [24]
Electrical energy is your answer.
6 0
3 years ago
Read 2 more answers
What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

4 0
3 years ago
What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport
ad-work [718]

Answer: 1.13(10)^{3} Pa

Explanation:

This problem can be solved by the following equation:

\Delta P=\frac{8 \eta L Q}{\pi r^{4}}

Where:

\Delta P is the pressure difference between the two ends of the pipe

\eta=0.20 Pa.s is the viscosity of oil

L=2.6 km=2600 m is the length of the pipe

Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s} is the Rate of flow of the fluid

d=36 cm=0.36 m is the diameter of the pipe

r=\frac{d}{2}=0.18 m is the radius of the pipe

Soving for \Delta P:

\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}

Finally:

\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa

7 0
3 years ago
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