All categories

1 year ago

Which is brighter in our sky, a star with apparent magnitude 2 or a star with apparent magnitude 7?

Physics

1 answer:

Lina20 [59]1 year ago

7 0

The star with apparent magnitude 2 is more brighter than 7.

To find the answer, we have to know about apparent magnitude.

<h3>What is apparent magnitude?</h3>

100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
The apparent magnitude of bigger stars is always smaller.
The brightest star in the night sky is Sirius.
The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.

Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.

Learn more about the apparent magnitude here:

brainly.com/question/350008

#SPJ4

You might be interested in

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

NeTakaya

5.0*10^3 this is the answer. I know for a fact.

7 0

3 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

A metal pot feels hot to the touch, but the plastic handle does not. Which type of material is the plastic handle? A. A thermal

gtnhenbr [62]

D is the answer..........

3 0

3 years ago

Read 2 more answers

Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi

insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

s−15

return trip is then :

s+15

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0

3 years ago

Read 2 more answers

When two objects with electrical charges interact, which affect the strength of that interaction?

noname [10]

The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span>

6 0

3 years ago