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s2008m [1.1K]
2 years ago
8

HELPPP BESTIESSS An astronomer is studying two stars that have about the same size and brightness. From the Earth, one of the st

ars appears very bright in the night sky. The other star appears very faint and dim.
Why does one star appear to be brighter than the other?
A. The Earth is moving toward the brighter star.
B. The Earth is moving toward the dimmer star.
C. The Earth is farther from the brighter star.
D. The Earth is closer to the brighter star.

Physics
2 answers:
otez555 [7]2 years ago
7 0

Answer:

If the stars are about the same  size and brightness then the earth must be nearer to the stat that appears brighter, The stars movement, nearer or farther,  would only effect the frequency received (redshift).

Fittoniya [83]2 years ago
6 0

Answer:

D- The earth is closer to the bright star

Explanation:

If two stars have SAME brightness and size then that means the only point of difference is distance . Earth should be closer to the brighter star for the star to be bright.

I hope im right!!

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
4 years ago
Which optical devices spread light apart?
Harman [31]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Optical Devices.

Now basically here the devices that can spread light over apart are the CONCAVE LENS and Convex mirror.

so the Answer is

C. B and D

4 0
3 years ago
Read 2 more answers
The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud
motikmotik
<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = \frac{Gm_1m_2}{r^2}

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = \sqrt{\frac{Gm_1m_2}{F} }

If we substitute the values in the above equation

r = \sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }

= 2.46 m

3 0
3 years ago
Read 2 more answers
What is the formula of finding displacement​
xenn [34]
Displacement is the final position of the object minus the initial position of the object.
Xf - Xi. Displacement is not the distance of the object. If you go to the right 10m and to the left another 10m, your displacement is 0m. But your distance is 20m
8 0
4 years ago
A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod
AleksAgata [21]

Answer:

5 N acting in the same direction as the 10 N force

Explanation:

10+5=15

15=15

7 0
3 years ago
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