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Alenkasestr [34]
2 years ago
11

In the center of the Milky Way galaxy is a A

Physics
2 answers:
olchik [2.2K]2 years ago
5 0
The answer is the Galactic Center I think
LekaFEV [45]2 years ago
4 0

Answer:

a super massive blackhole

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large truck exerts 7000 n of force on a piston with an area of 0.4m squared and the is ton can only support 16,000 pa of pressur
sasho [114]
7000 N / 0,4m^2  = 17 500 Pa.
6 0
3 years ago
A 25 N force stretches a spring 280 cm. What was the spring constant? ​
boyakko [2]

Answer:

  1. F= 25N
  2. ∆l=280cm
  3. K=?
  4. you apply the formula = K:F/∆l
  5. = 0,089N/cm
4 0
2 years ago
Read 2 more answers
Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?
leonid [27]

Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult  to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of  material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout  the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

6 0
3 years ago
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
3 years ago
How many significant figures should be written in the sum 4.65g +9.032g+580.0078+540.439g?
Ne4ueva [31]

Answer: 1.13 X 10^3 g or 1130 g

Explanation: When you add up everything, you will get 1134.1288 g = 1.11341218 X 10^3. But while adding significant figures, we always look for the least significant figures (here 4.65 has least sig. figures  i.e. 3) and take that as a reference to how many significant figures we should have in the answer. (Pretty hard to explain!) Hope you got it!!

7 0
2 years ago
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