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3 years ago

How does a supernova occur?

1 answer:

7 0

Supernova occurs when black holes stretch so much running out of nuclear fuel. When a star collapses to form a black hole, its mass increases. Supernovas have greater masses than black holes. One example application of this is the theory of Big Bang. 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.

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Which person is weightless?

ahrayia [7]

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

7 0

3 years ago

During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, sl

xeze [42]

During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, slowing down and increasing following distance. Adverse weather means that you are driving in difficult and dangerous conditions. Increasing following distance will help you to maintain safe driving and avoid tailgating.

4 0

3 years ago

In a nuclear station if the temperature of the superheated water starts to decrease what should the engineer in charge of the re

Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

then, see if they can fix the problem, im sorry if its wr0ng.

8 0

3 years ago

What causes the pressure to increase as you go deeper below the crust?

Masteriza [31]

The deeper you go, the more rock must be supported so the more force is required and the pressure goes up.

8 0

3 years ago

Read 2 more answers

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

So, the final kinetic energy has lost a 37% of the initial one.

6 0

3 years ago