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choli [55]
3 years ago
10

Are Constellations based on the magnitude of stars.

Physics
2 answers:
salantis [7]3 years ago
8 0

no they are not they are mostly based off roman and Greek Mithos

Nikolay [14]3 years ago
4 0
Constellations are based off of many Greek and Roman fables. Many of their gods and beliefs are pictured in the stars, which is where we get most of our constellations. Hope this helps!
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If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi
saul85 [17]
The gravitational potential energy:
E p = m x g x h
where m is the mass and h is the height;
m = 50 kg,  g = 9.81 m/s² ,  h = 40 m
E p = 50 kg  x 9.81 m/s²  x  40 m
Answer:
E p = 19,620 J = 19.62 kJ
8 0
3 years ago
Read 2 more answers
The following questions consist of a statement and a reason .Answer these questions selecting the appropriate options given belo
Mamont248 [21]

Answer:

<em><u>1)A)</u></em>

<em><u>1)A)2)A)</u></em>

<h3><em><u>Hope it helps you </u></em><em><u>♡</u></em><em><u>♡</u></em></h3>

4 0
3 years ago
If distance between two charges increased 5 times then force between them
Marrrta [24]

decreased 5 times

Explanation: if the force increases 5 times between them would decrease 5 times

4 0
3 years ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the
meriva

Answer:

a.2.86 m/s^2

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

\mu_s=0.14

g=9.8 m/s^2

By Newton's second law

8F-f=(8M+m)a

a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}

a=2.86 m/s^2

b.By Newton's second law

T=ma+\mu_s mg

Substitute the values

T=250\times 2.86+0.14(250)(9.8)=1058 N

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0
3 years ago
What will be the value of both charges if they are 5 cm apart and suffer a
Mrac [35]

Answer:

\boxed{q = 1.2 \times  {10}^{ - 6} C}

Explanation:

f_e =   \frac{{q}^{2}k }{ {r}^{2} }  \\ q =  \sqrt{ \frac{f_e( {r}^{2} )}{k} }  =  \sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  \\ q =\sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  =  \sqrt{ \frac{0.013}{9 \times  {10}^{9} } }  \\ q = 1.2 \times  {10}^{ - 6}

3 0
3 years ago
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