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Vinil7 [7]
4 years ago
13

What is the most violent star death called

Physics
1 answer:
gogolik [260]4 years ago
6 0
Hi there!

The most violent star death is a Supernova. This is the massive explosion of a supergiant star as it's fuel source runs out and it can no longer fuse iron at its core safely. This causes the star to swell to unstable sizes until it explodes in a Supernova. The amount of energy that is output during a supernova is equivalent to all the energy our own Sun outputs in its whole lifetime. 
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A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to
Varvara68 [4.7K]

Answer: 50J

Explanation:

Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.

Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.

37.5 + 12.5 = 50J

Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.

3 0
3 years ago
While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is
erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0
3 years ago
Uranus has more than 14 times as much mass as earth, yet the gravitational force is less. how can that be?
Anarel [89]
It has more mass, yes. But it has less of a gravitational pull because it is farther away from the sun than the Earth is
3 0
4 years ago
Read 2 more answers
A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is
vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

Learn more about gravity from here

brainly.com/question/4014727

#SPJ1

5 0
1 year ago
A second basemen tosses the ball to the first basemen, who catches it at the same level from which it was thrown. The throw is m
Radda [10]

Answer:

Explanation:

Given

Initial speed of ball u=19\ m/s

Launch angle \theta =34.5^{\circ}

As there is no acceleration in horizontal direction therefore there is no change in velocity

thus horizontal velocity remains unchanged

u_x=u\cos \theta

u_x=19\cdot \cos 34.5

u_x=15.65\ m/s

Time of flight of projectile is

T=\frac{2u\sin \theta }{g}

T=\frac{2\times 19\times \sin (34.5)}{9.8}

T=2.196\ s

Thus time for which it is in air is 2.12 s                        

4 0
4 years ago
Read 2 more answers
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