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worty [1.4K]
3 years ago
9

Which coordinate system would be most useful for two observers (one in Michigan and one in Florida) who wants to observe the sam

e astronomical object and discuss their observations over the phone as they’re making them?
a. celestial or azimuth-altitude


b. celestial


c. horizontal


d. azimuth-altitude
Physics
2 answers:
morpeh [17]3 years ago
4 0

Answer:

The answer is actually A: celestial

Explanation:

The answer is not azimuth-altitude because using that system means that the coordinates change with your geographic location. With celestial, you can coordinate the object's location using right ascension and declination no matter what latitude you are standing at.

Mkey [24]3 years ago
3 0

Answer : Celestial or azimuth - altitude

Explanation :  

Celestial : The celestial coordinates that are analogous to longitude and latitude are called RA and Dec.

RA = Right Ascension

Dec = Declination

RA is the measured in unit of time and Dec is measured in degree.

The equatorial coordinate system is the projection of the latitude and longitude coordinate system on the celestial sphere.

Azimuth - altitude :  Azimuth - altitude define the location of an object in the sky.

The altitude is the distance of an object appears to be above the horizon.

The Azimuth of the object is the angular distance  along the horizon to the location of the object.

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What are three key aerodynamics principles?
mafiozo [28]

Answer:

A

Explanation:

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8 0
3 years ago
If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

5 0
3 years ago
Which two formulas are used to calculate potential and kinetic energy
goblinko [34]

Answer:

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6 0
2 years ago
Read 2 more answers
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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