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3 years ago

What is a small body that follows a highly elliptical orbit around the Sun

2 answers:

6 0

That could be a comet, or any one of the billions of meteoroids
moving in a cloud that's actually the remains of a shattered comet.

Nikitich [7]3 years ago

5 0

A small body of ice, rock, and dirt that follows a extremely elliptical orbit round the sun. a spherical region that surrounds the scheme and extends virtually halfway to the closest star. A small, on an irregular basis formed, rocky object that orbits the sun.

A estraterrestrial body is an icy tiny scheme body that, once passing on the brink of the Sun, warms and begins to unharness gases, a method known as outgassing.

This produces an evident atmosphere or coma, and typically additionally a tail.

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A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

4 years ago

What would happen to the function or purpose of flower if the stigma wasn't sticky?

sashaice [31]

Answer:

The plant would not reproduce because the flower uses the stigma to catch the pollen

8 0

3 years ago

Read 2 more answers

What wa can friction do to things like brake pads

andrey2020 [161]

Ware them down, its like rubbing two pieces of chalk together.

4 0

3 years ago

Read 2 more answers

At a drag race, the light turns green and 0.00125 hours later, a dragster is traveling 300 miles per hour. Calculate the acceler

Talja [164]

Answer:

240000 mph² or miles/hour²

Explanation:

<em>Use the formula</em>

<h3>acceleration = change in velocity ÷ time</h3>

change in velocity = 300 mph - 0 mph (final velocity - intial velocity)

time = 0.00125 hours

<em>Substitute the values into the formula:</em>

acceleration = 300 ÷ 0.00125 = 240000 mph²

4 0

3 years ago

In a home stereo system, low sound frequencies are handled by large "woofer" speakers, and high frequencies by smaller "tweeter"

Hitman42 [59]

Answer:

C = 2.9 10⁻⁵ F = 29 μF

Explanation:

In this exercise we must use that the voltage is

V = i X

i = V/X

where X is the impedance of the system

in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is

X = $\sqrt{R^2 + ( wL - \frac{1}{wC})^2 }$

tells us to take inductance L = 0.

The angular velocity is

w = 2π f

the current is required to be half the current at high frequency.

Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small

$\frac{1}{wC}$ →0 when w → ∞

therefore in this frequency regime

X₀ = $\sqrt{R^2 + ( \frac{1}{2\pi 2 10^4 C} )^2 } = R \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC} }$

the very small fraction for which we can despise it

X₀ = R

to halve the current at f = 200 H, from equation 1 we obtain

X = 2X₀

let's write the two equations of inductance

X₀ = R w → ∞

X= 2X₀ = $\sqrt{R^2 +( \frac{1}{wC} )^2 }$ w = 2π 200

we solve the system

2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }

4 R² = R² + 1 / (wC) ²

1 / (wC) ² = 3 R²

w C = $\frac{1}{\sqrt{3} } \ \frac{1}{R}$

C = $\frac{1}{\sqrt{3} } \ \frac{1}{wR}$

let's calculate

C = $\frac{1}{\sqrt{3} } \ \frac{1}{2\pi \ 200 \ 9}$

C = 2.9 10⁻⁵ F

C = 29 μF

7 0

3 years ago