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3 years ago

If a cars velocity is slowing down is it considered a positive or negative acceleration?

Physics

1 answer:

Natali [406]3 years ago

4 0

Answer:

Negative

Explanation:

Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

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When one member of a binary star system is a black hole, and astronomers detect flickering x-rays coming from the system, where

puteri [66]

When one member of a binary star system is a black hole, and astronomers detect flickering x-rays coming from the system, these x-rays usually coming from a disk of material around the black hole (material that has been pulled from the companion star and is falling toward the black hole).

A binary black hole (BBH) is a system consisting of two black holes in close orbit around each other. Like black holes themselves, binary black holes are often divided into stellar binary black holes, formed either as remnants of high-mass binary star systems or by dynamic processes and mutual capture; and binary supermassive black holes, believed to be a result of galactic mergers.

For many years, proving the existence of binary black holes was made difficult because of the nature of black holes themselves and the limited means of detection available.

However, in the event that a pair of black holes were to merge, an immense amount of energy should be given off as gravitational waves, with distinctive waveforms that can be calculated using general relativity.

Therefore, during the late 20th and early 21st century, binary black holes became of great interest scientifically as a potential source of such waves and a means by which gravitational waves could be proven to exist. Binary black hole mergers would be one of the strongest known sources of gravitational waves in the universe, and thus offer a good chance of directly detecting such waves.

As the orbiting black holes give off these waves, the orbit decays, and the orbital period decreases. This stage is called binary black hole inspiral. The black holes will merge once they are close enough. Once merged, the single hole settles down to a stable form, via a stage called ringdown, where any distortion in the shape is dissipated as more gravitational waves. In the final fraction of a second the black holes can reach extremely high velocity, and the gravitational wave amplitude reaches its peak.

Learn more about binary black hole here : brainly.com/question/16199119

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8 0

2 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula

astraxan [27]

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I = 8.53 kg*m²

b) The rotational kinetic energy of the rigid body composed by the rod and the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

7 0

3 years ago

Can sumbody help me wit dis

stepladder [879]

C, exothermic reaction. These types of reaction releases heat so that you can heat up your ready-to-eat meals.

3 0

3 years ago

is it possibly to make special swords made out of light that can cute through anything like in the movie star wars?

g100num [7]

No because light travels in waves and can not stop 3 feet away an having power supplied like that

6 0

3 years ago