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17 days ago

Why didn't the astronauts land on the moon 3.17 answers punchline?

2 answers:

3 0

There are numerous responses to punchline 3.17 explaining why astronauts didn’t touch down on the moon. Some participants believe it was due to the full moon, while others assert they launched during daylight when the moon wasn’t seen. There's also mention that 'astro' signifies star, not satellite, and that 'naut' equates to zero or naught.

kicyunya [2.2K]17 days ago

3 0

Answer:

Responses to the 3.17 punchline varied among many individuals, with some suggesting that it was a "full" moon day which prevented the astronauts from landing.

Others claimed that the astronauts took off during daylight hours when the moon was not visible. There were also comments that indicated that 'astro' refers to stars rather than satellites, explaining why they did not land.

A few even noted that 'astro naut' sounds like 'naught,' meaning zero (0), as a possible reason for their failure to land.

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

Yuliya22 [2420]

Answer:

The box experiences an acceleration with a magnitude of 2 m/s².

Explanation:

Details provided:

Mass of the box: $m=5.0$ kg

Force acting towards the east: $F=27$ N

Frictional force acting towards the west: $f=17$ N

Let’s denote the acceleration as $a$ m/s².

Thus, the net force acting on the box in the east direction is stated as:

$F_{net}=F-f=27-17=10\textrm{ N}$

According to Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Thus, the magnitude of the box's acceleration is determined to be 2 m/s².

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16 days ago

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

Keith_Richards [2256]

Answer:

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

Explanation:

The question specifies 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 the electron is $m_e = 9.109*10^{-31} \ kg$ .

The energy of the incoming light is typically depicted mathematically as

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

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

h stands for Planck's constant with a value of $h = 6.62607015 * 10^{-34 } J\cdot s$ .

Thus,

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

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

Typically, kinetic energy is represented as

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

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

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

The ionization energy is generally expressed mathematically as

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

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

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27 days ago

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

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Answer:

$8616.7468 \ kg/m^3$

Explanation:

The measurement of pressure is indicated as $p=\rho gh$ where p denotes the pressure, $\rho$ signifies density, and h represents height

Given values include pressure $p=9.891\times 10^4\ Pa$ , gravity's acceleration $g=9.9870\ m/sec^2$ , and height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

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8 days ago

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If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the

Yuliya22 [2420]

Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

Input power = 6 Watts

To find:

What is the output power?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss corresponds to a ratio of 0.5

Output power can be calculated as follows:

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watts

Therefore, the output power of the circuit is 3 Watts.

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kicyunya [2264]

B. red................

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17 days ago

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