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3 years ago

From Smallest to Largest Moon asteroid meteorite meteoroid

Physics

2 answers:

Yanka [14]3 years ago

7 0

Meteorite meteoroid asteroid moon give brainlest please

Korolek [52]3 years ago

7 0

In my view
Meteorite
Meteoroid
Asteroid
Moon

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If we made a model with our solar system the size of an Oreo cookie, what would be the largest object that would fit inside your

matrenka [14]

Answer:The television

Explanation:

6 0

3 years ago

A 13.4-mH inductor carries a current i = <img src="https://tex.z-dn.net/?f=I_%7Bmax%7D" id="TexFormula1" title="I_{max}" alt="I_

Digiron [165]

The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

ω = 2π · 60 = 377 /sec

I(max) = 4.80 A

V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

V = 24.25 cos(377 t)

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

6 0

3 years ago

The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

BartSMP [9]

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

8 0

3 years ago

S Problem Set 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

3 years ago

You are on a boat which is crossing the prime meridian. the altitude of polaris is 50Âº.explain how you know the boat's location

Lerok [7]

The definition for the prime meridian is: The prime meridian with 0 degrees longitude runs through Greenwich. Polaris is the north star at 50 degrees above the horizon, which means 50 degrees latitude.
If you observe Polaris at 50 degrees of altitude, you are at latitude 50 North. You that's why you know that your boat's location is 50Âº north latitude and 0Âº longitude.

7 0

3 years ago