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3 years ago

What is the scariest thing to established U.S. utilities?

Physics

2 answers:

iogann1982 [59]3 years ago

6 0

<span>They are still structured as monopolies that bundle together disparate energy </span>

murzikaleks [220]3 years ago

4 0

im sorry i dont know

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Which best describes the relationship between the terms “frequency,” “wavelength,” and “hertz”?

Natali5045456 [20]

Answer:

Frequency is the number of wavelengths, which is measured in hertz.

Explanation:

I have my ways ;)

5 0

3 years ago

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In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

4 years ago

How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s

Elis [28]

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

7 0

3 years ago

If there is a difference in air pressure between two locations, what happens?

tino4ka555 [31]

Answer:

High in the atmosphere, air pressure decreases. ... A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.

Explanation:

4 0

3 years ago

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