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4 years ago

When a star collapses to form a black hole, its mass _____.

1 answer:

7 0

When a star collapses to form a black hole, its mass increases, that is letter C. One example of this is the Big Bang Theory.
<span>
It is by no doubt that our universe had had its own starting point but how it really started became a debate for the scientists over the past centuries. From theories such as Ekpyrotic theory, White Holes, Matrix theory, and Quantum theory, the Big Bang theory is the theory mostly accepted by scientists. According to the theory, the universe started as a singularity, that is, from a black hole under extreme gravitational pressure and expanded (instead of exploded) and cooled. Its cooling, according to the theory, is still happening now as of the moment.</span>

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The pictures represent three different states of matter.

Veronika [31]

ONLY if ur asking for the 3 states of matter it is solid liquid and gas

4 0

4 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0

3 years ago

Which of the following is an example of potential energy?

Papessa [141]

Answer:

Explanation:

potential energy is stored energy so the ball has potential energy to bounce or roll which would then have been converted to kinetic but the rest are in motion meaning energy is no longer stored but used in motion and therefore cannot be potential energy

4 0

3 years ago

a circus performer launches himself from a springboard with an initial velocity of 21 m/s at an angle of 75 toward a platform ha

kherson [118]

Answer:

The circus performer falls back down to the ground

Explanation:

The question parameters are;

The initial velocity of the circus performer = 21 m/s

The angle in which the performer launches himself = 75° towards the platform

The height of the platform above the ground = 20 m

The horizontal distance of the platform from the springboard = 15 m

The vertical motion of the circus performer is given by the following projectile motion relation;

y = y₀ + v₀·sinθ₀t-1/2·g·t²

Where;

y = Height reached by the circus performer

y₀ = Initial height of the the circus performer (the springboard) = 0 m

v₀ = Initial velocity of the the circus performer = 21 m/s

θ₀ = The angle with which the circus performer launches himself = 75°

t = The time of ,light of the circus performer

g = The acceleration due to gravity

Therefore, when the height is 20 m, we have;

20 = 21*sin(75)*t - 1/2*9.81*t²

Which gives;

21*sin(75)*t - 1/2*9.81*t² - 20 = 0

Factorizing using a graphing calculator, gives;

t = 1.623 or t = 2.513

Therefore, the circus performer passes the 20 m mark twice, in his motion, where the first time is when he is on his way up while the second time is when he is on his way down

The horizontal motion of the circus performer is given by the following projectile motion relation;

x = x₀ + v₀*cos(θ₀)* t

Where;

x₀ = The initial position of the circus performer in relation to the final position = 0

Plugging in the value of t when y = 20, we get;

x = 21×cos(75)×1.623 = 8.82 m, which is less than the 15 m platform distance from the spring board

Checking the other time value, we have;

x = 21×cos(75)×2.513 = 13.66 m which is also less than the 15 m platform distance from the spring board

Therefore, the circus performer misses the platform and falls back down to the ground.

8 0

3 years ago

Mgh=1/2mv^2solve for v

Dmitry [639]

So since m is on both sides of the equal sign, they cancel each other out. Then multiply both sides by 2, so you get 2gh = v². Then (I'm assuming) g = acceleration due to gravity = 9.80 m/s². If you multiply it by the two, you get 19.6 m/s². I'm assuming h is the height, in meters. If you have the number for h, plug it in now and multiply that. You'll get some number x that has a value of m²/s². Since you're looking for v, and right now you only have v², take the square root of both sides of your equation and that'll give you the answer. If you don't have a value for h, though, just write your answer like √(2gh) = v or whatever format like that your teacher usually wants, and you should be good. I don't know how far the teacher wants you to simplify or solve this, but hopefully this gives you what you need! :)

3 0

3 years ago

Read 2 more answers