Which is an example of current electricity?

Why do astronomers hypothesize that a massive black hole lies at the center of M87? Historical records show that a supermassive

antiseptic1488 [7]

<h2>Answer: A very small region at the center of M87 releases an enormous amount of energy.</h2>

According to Einstein's theory of relativity, a<u> black hole is a "singularity"</u> that consists of a region of the space in which the density of matter tends to infinity. In consequence, this huge massive body has a gravitational pull so strong that not even light can escape from it.

In addition, "the surface" of a black hole is called the event horizon, which is the border of space-time in which the events on one side of it can not affect an observer on the other side.

In other words, at this border also called "point of no return", nothing can escape (not even light) and no event that occurs within it can be seen from outside.

In this sense, and according to the relativity, <u>it is possible to determine where a black hole is if it is "observed" an enormous amount of energy released.</u> So, in accordance to this, galaxies like ours must have a black hole in its center.

On the other hand, the elliptical galaxy Mesier 87 (also called <u>Virgo A,</u> but from now on M87) was showing the above described behaviour, with enormous jets of high-energy particles shooting away from its vicinity . This was imaged by the Hubble Space Telescope years ago; that is why astronemers were hypothesizing about the existence of a massive black hole there.

Well now, on April, 10th 2019 this was demonstrated with the publication of the image, for the first time, of the event horizon of the black hole in M87. This is the first time in human history a picture of a black hole is taken.

This was done by the huge effort of diverse scientist and by the syncronization of eight radio telescopes scattered across the Earth (located at: Hawaii, Spain, Chile, Mexico, Arizona and the South Pole), which took the same point of the sky at the same time.

7 0

3 years ago

3. A projectile is fired from a vertical cliff edge (height = 100 m) with a horizontal velocity of 20

Klio2033 [76]

Refer to the diagram shown below.

u = 20 m/s, the horiontal launch velocity
v = 0, initial vertical velocity

Assume that wind resistance may be ignored.

The time, t, of the flight is given by
(1/2)*(9.8 m/s²)*t² = (100 m)
t² = 100/4.9 = 20.4082
t = 4.5175 s

Th horizontal distance traveled is
d = (20 m/s)*(4.5175 s) = 90.35 m

Answer: B
The projectile hits the ground at approximately 90 m from the base of the cliff.

6 0

3 years ago

The angular position of objects as a function of time is given, where a, b, and care constants. In which of these cases is the a

GalinKa [24]

Answer:

Explanation:

The options is not well presented

This are the options

A. θ = at³ + b

B. θ = at² + bt + c

C. θ = at² — b

D. θ = Sin(at)

So, we want to prove which of the following option have a constant angular acceleration I.e. does not depend on time

Now,

Angular acceleration can be determine using.

α = d²θ / dt²

α = θ''(t)

So, second deferential of each θ(t) will give the angular acceleration

A. θ = at³ + b

dθ/dt = 3at² + 0 = 3at²

d²θ/dt² = 6at

α = d²θ/dt² = 6at

The angular acceleration here still depend on time

B. θ = at² + bt + c

dθ/dt = 2at + b + 0 = 2at + b

d²θ/dt² = 2a + 0 = 2a

α = d²θ/dt² = 2a

Then, the angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

C. θ = at² —b

dθ/dt = 2at — 0 = 2at

d²θ/dt² = 2a

α = d²θ/dt² = 2a

Same as above in B. The angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

D. θ = Sin(at)

dθ/dt = aCos(at)

d²θ/dt² = —a²Sin(at) = —a²θ

α = d²θ/dt² = -a²θ

Since θ is not a constant, then, the angular acceleration is dependent on time and angular displacement

So,

The answer is B and C

4 0

3 years ago

At the nose of a missile in flight, the pressure and temperature are 5.6 atm and 850°R, respectively. Calculate the density and

Contact [7]

To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

$1atm = 2116lb/ft^2$

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = Amount of substance (at this case the mass)

Then

$\frac{n}{V} = \frac{P}{RT}$

The amount of substance per volume is the density, then

$\rho = \frac{P}{RT}$

Replacing with our values,

$\rho = \frac{5.6*2116}{1716*850}$

$\rho = 0.00812slug/ft^3$

Finally the specific volume would be

$v = \frac{1}{\rho}$

$v = 123ft^3/slug$

6 0

3 years ago

Which of the following statements is true for ideal gases, but is not always true for real gases?

xxMikexx [17]

Answer:

C. Replacing one gas by another under the same conditions, has no effect on pressure.

Explanation:

Ideal gas:

A gas is treated as an ideal gas if temperature is high and pressure is low.

Kinetic energy for ideal gas given as

$K.E.=\dfrac{3}{2}KT$

So when temperature of gas is increases then Average molecular kinetic energy will also increases.

The size of molecule is negligible as compare to the dimension of container. It mean that volume occupied by molecule is less as compare to the volume of container.

The between molecules is perfectly elastic.

Ideal gas equation

P V = m R T

So the option C is not always true.

4 0

3 years ago