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Kitty [74]
3 years ago
5

The stars in the sky are organized into groups of stars called constellations which appear near each other in the sky but are no

t necessarily close together in space. How many constellations are currently accepted by the IAU?
Physics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

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I will mark you brainlist please help
denis-greek [22]

Answer:15th

Explanation:

I believe it will fall on the 15th stair because 10 times 15 is 150 divided by 10 m/s equals 15

8 0
3 years ago
Any of two or more forms of the same element which differ only in the number of neutrons their atoms contain is called an
telo118 [61]
C. isotope

Isotopes have the same atomic number but different mass number because of the difference in the number of neutrons in the nucleus of the atom.
8 0
3 years ago
A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second
JulijaS [17]

Answer:

i_2 = 978750 A

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

F = \frac{\mu_0 i_1 i_2}{2 \pi d}

here we know that

F = 7.83 \times 10 N/m

d = 7.0 cm = 0.07 m

i_1 = 28 A

now we will have

F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}

7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}

i_2 = 978750 A

Since the force between wires is attraction type of force so current must be flowing in upward direction

3 0
3 years ago
Can work be done on a system if there is no motion?
Minchanka [31]

Answer:

D) No, because of the way work is defined

Explanation:

The work done on an object is given by:

W=Fd cos \theta

where

F is the force applied on the object

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement

From the formula, we observe that if there is no motion involved, then the displacement of the object is zero:

d = 0

As a result, the product in the formula is also zero, therefore the work done will be zero as well.

5 0
3 years ago
A 0.537-kg basketball is dropped out of a window that is 5.88 m above the ground. The ball is caught by a person whose hands are
Monica [59]

Answer:

a) 20.54 J

b) 30.97 J

c) 10.43 J

d) -20.54 J

Explanation:

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

h = Height

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88-1.98)\\\Rightarrow W=0.537\times 9.81\times 3.9\\\Rightarrow W=20.54\ J

Work done by the ball's weight is 20.54 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88)\\\Rightarrow W=30.97\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J

W=mgh\\\Rightarrow W=0.537\times 9.81\times (1.98)\\\Rightarrow W=10.43\ J

Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J

Change in gravitational potential energy

\Delta U=10.43-30.97=-20.54\ J

Change in gravitational potential energy is given by -20.54 J

4 0
4 years ago
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