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NikAS [45]
3 years ago
6

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

n the stars at the center bulge and the stars in the arms?
Physics
1 answer:
nadya68 [22]3 years ago
5 0

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

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Why is nut-cracker 2nd class lever?​
Dmitry_Shevchenko [17]

2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......

4 0
3 years ago
What is the definition of an unbalanced force?
Shtirlitz [24]
<span>an unbalanced force is Forces that are not equal</span>
3 0
3 years ago
"In the far future, a visiting tourist from another planetary system asks to see the most massive object in our solar system. Wh
Andrei [34K]

Answer:

Obviously Our Yellow Star: The Sun....

Explanation:

With a staggering mass of 1.989 × 10^30 kg as well as a gravitation pull of about 274 ms^-1. I think, no other object in our solar system at least have those properties. Not to mention Sun makes <em>99.86% of our solar system </em>combined.

7 0
3 years ago
A BODY STARTING FROM REST MOVES WITH CONSTANT ACCELERATON. what is the ratio of distance covered by the body during the fifth se
Art [367]

Answer:

9/25

Explanation:

Distance covered in the first 5 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₅ = (0) (5) + ½ a (5)²

Δx₀₋₅ = 25a/2

Distance covered in the first 4 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₄ = (0) (4) + ½ a (4)²

Δx₀₋₄ = 8a

So the distance covered during the 5th second is:

Δx₅ = 25a/2 − 8a

Δx₅ = 9a/2

So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:

Δx₅ / Δx₀₋₅

(9a/2) / (25a/2)

9/25

8 0
3 years ago
A thin, square metal plate measures 15 cm on each side and has emissivity of 0.80. The plate is heated to a temperature of 605°C
Sholpan [36]
L = 15cm = 0.15m
A = 0.15 × 0.15 = 0.0225
Sradiation is transfered from bottom
σ = 5.67 × 10⁻⁸ W/m²k
T = 605 c = 878k
Rate of energy transfer
Q = 
σ × A × E × T⁴
5.67 × 10⁻⁸ × 0.0225 × 2 × 0.8 × 878⁴
1792.17 W
6 0
3 years ago
Read 2 more answers
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