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4 years ago

Why is the absolute magnitude of some stars greater than their apparent magnitude?

Physics

1 answer:

Sedbober [7]4 years ago

7 0

<h3>Apparent magnitude of star:</h3>

Apparent magnitude of a star is how bright it appears on earth.

<h3>Absolute magnitude of star:</h3>

Absolute magnitude of star is how bright it appear from a distance of 32.6 light years from earth.

Due to farthest distance of star from earth, it appears smaller. But when we look at it from near it will appear greater in size or magnitude. That is why, we can say that absolute magnitude of some stars is greater than their apparent magnitude.

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What are the four fundamental forces of nature?

Nostrana [21]

The Four Fundamental Forces of Nature are Gravitational force, Weak Nuclear force, Electromagnetic force and Strong Nuclear force.

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3 years ago

When developing an experimental design which action could a scientist take to improve the quality of the results

Anni [7]

Answer:

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3 years ago

While standing at the top of an 100 m high observatory, you accidentally dropped your phone through the grates (it is now fallin

Arturiano [62]

Using the idea of motion under gravity, it will take 0.5 seconds more before the phone hits the ground.

<h3>What is the distance covered?</h3>

The distance covered is obtained form the equations of kinematics under gravity.

Now;

v = u + gt

Recall that it was dropped from a height hence u = 0 m/s

v = gt

v = 9.8 m/s^2 * 4 s

v = 39.2 m/s

Now;

h = ut + 1/2gt^2

but u = 0 m/s

h = 1/2gt^2

h = 1/2 * 9.8 * (4)^2

h = 78.4 m

The phone will not hit the ground within this time

For the phone to hit the ground;

h = 1/2gt^2

if h = 100 m

100 = 1/2 * 9.8 * t^2

2 * 100/9.8 = t^2

t = √2 * 100/9.8

t = 4.5 seconds

It will take about 0.5 seconds more before the phone will hit the ground.

Learn more about motion under gravity:brainly.com/question/15120445

#SPJ1

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2 years ago

MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle

labwork [276]

Answer:

(a) $y_{max}=0.423m$

(b) $\alpha =64.3^{o}$

Explanation:

Given data

$v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle$

Solution

For Part (a)

As the velocity component in direction of y is given by:

$v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s$

The maximum displacement is given by:

$v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m$

For Part (b)

To reach y=46cm =0.46m apply:

$0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}$