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2 years ago

At what distance must a star be to have its apparent magnitude equal to its absolute magnitude?

1 answer:

6 0

10 parsecs<span>Another way to look at this is that the apparent magnitude (brightness from true distance) is brighter than the absolute magnitude (brightness from 10 parsecsaway). So Ross 128 must be closer than 10 parsecs, making star look brighter than it "really" is.</span>

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horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes

lbvjy [14]

Answer:

The tension is $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

The distance between the two poles is $D =14 m$

The mass tied between the two cloth line is $m = 3Kg$

The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium

And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

$= tan^{-1} [\frac{1}{7}]$

$=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

5 0

2 years ago

Read 2 more answers

The origin of the universe remains a question. true false

OleMash [197]

True ..............................

8 0

2 years ago

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

lions [1.4K]

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i² = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

= 0.2 A approx .

2 i = .4 A Ans .

6 0

2 years ago

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle conn

suter [353]

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × $\frac{V^{2} }{2g}$

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

5 0

1 year ago

What has to happen to an uncharged aluminum atom so that it will bond with oxygen?

iren [92.7K]

It has to lose electrons to become positively charged, so it can attract a negatively charged oxygen ion.

7 0

3 years ago