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4 years ago

Where would aliens be if they really do exist here on Earth?

Physics

2 answers:

AysviL [449]4 years ago

7 0

Answer:

I would guess they'd probably be in an area they feel is habitatible.

irakobra [83]4 years ago

7 0

That's not for me to say.

It's YOUR story. You've already invented fictitious aliens. Now you can go ahead and make up all the fascinating details of why they're here, where they live, what they eat, what they like to do in their spare time, how they like it here, what they do when it's raining, what their plans are, what they think of us, and why nobody has ever seen or heard one.

If you can get it published, I'd buy it.

You might be interested in

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

What is the frequency of radiation whose wavelength is 11.5 a0 ?

irakobra [83]

Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$

Explanation:

Given:

Wavelength $\lambda = 11.5 \times 10^{-10}$ m

Speed of light $c = 3 \times 10^{8}$ $\frac{m}{s}$

For finding the frequency of radiation,

$c = f \lambda$

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }$

$f = 2.61 \times 10^{17}$ $s^{-1}$

Therefore, the frequency of radiation is $2.61 \times 10^{17} s^{-1}$

4 0

3 years ago

A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles

slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²

0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²

0.02912 = 1/2 × 0.035 × v²

1/2 × 0.035 × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s

E = PE + KE

0.032 = 1/2 × 40 × 0.025² + KE

KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

4 0

3 years ago

Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

$Ac =V^2/R$

where Ac = centripetal acceleration = $98 m/s^2$

R = radius of rotation = 15 m

V = speed of astronaut

Hence, $\frac{V^2}{15} =98$

solving this we get, V = 38.34 m/s

3 0

3 years ago

Read 2 more answers

The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a

Andreyy89

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

Hence, L(T) = 8T²/π²

But T = 2.1

Pi= 3.14

8(2.1)²/3.14²

35.28/9.85

= 3.6feet

5 0

3 years ago

Where would aliens be if they really do exist here on Earth?​

Where would aliens be if they really do exist here on Earth?