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4 years ago

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Physics

1 answer:

mojhsa [17]4 years ago

5 0

B. the distance the star is from Earth

Explanation:

The apparent magnitude of star is a function of its distance from the earth. It is one of the physical properties that is used to study a star.

The apparent magnitude of a star or other astronomical bodies is a measure of their brightness as seen from a location on the earth.

The apparent magnitude of a star depends on:

Distance of the star from the location on earth.
luminosity of the star
the particles along the part of the star and earth that cuts off the light the earth receives.

learn more:

Star luminosity brainly.com/question/9084808

#learnwithBrainly

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A car is traveling down the highway at speed 5 m/s when the driver slams on the brakes and skids to a stop in a distance 80 m. A

Murljashka [212]

The car is very incredibly fast

8 0

3 years ago

Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Di

qwelly [4]

Answer:

__________________

I think the answer is A.

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3 0

3 years ago

A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin

ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f = v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so $V_{max1}$ = A × ω

$V_{max1}$ = 2.2×10⁻³ × 2722.69 m/s

$V_{max1}$ = 5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin( 0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

$V_{max2}$ = A' × ω

$V_{max2}$ = 1.555×10⁻³ × 2722.69

$V_{max2}$ = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

Hydrogen bonds create unusual properties in water. What are they?

xeze [42]

Answer:

High boiling and melting points: Hydrogen bonds increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.

High specific heat: Hydrogen bonds increase the amount of energy required for molecules to increase in speed, thereby raising the specific heat.

Lower density as a solid than as a liquid: Hydrogen bonds increase the volume of the solid by holding molecules apart, thereby decreasing the density

High surface tension: Hydrogen bonds produce strong intermolecular attractions, which increase surface tension

Explanation:

7 0

3 years ago

Read 2 more answers