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4 years ago

15

Why is AU (astronomical unit) the unit of distance that we used instead of kilometers?

1 answer:

Novay_Z [31]4 years ago

7 0

Answer:

The solar system is enormous, and interstellar space is even bigger. One astronomical unit is equal to 150 million kilometers. This makes it much easier to count the distances if they're in counts of Astronomic Units instead of having to count everything in millions or billions of kilometers

Explanation:

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The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

astra-53 [7]

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

4 0

3 years ago

A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

4 0

3 years ago

What is the average (mean) of these numbers: 13,43, 12, 3,66

algol13

Answer:

27.4

Explanation:

(13+43+12+3+66)/5

137/5

27.4

3 0

3 years ago

Read 2 more answers

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Verizon [17]

Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

So

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$

7 0

3 years ago

9. A plane starts at rest & accelerates along the ground before takeoff. It

Phoenix [80]

Answer:

9.877 m/s^2

Explanation:

The acceleration can be computed from ...

d = (1/2)at^2

(1600 m) = (1/2)a(18 s)^2

a = (1600/162) m/s^2 ≈ 9.877 m/s^2

6 0

3 years ago