You plan to take a spaceship to the photon sphere
and hover above the black hole to observe the back
of your head. What sort of acceleration will you
experience as you hover at this point? (Answer
qualitatively, e.g., small, comparable to one g, several times g, much bigger than g, incredibly huge.)

Answer: 2m

Explanation:

x = v(t)

x = (1 m/s)(2s)

x = 2m

Answer:

Thank you for helping use

Answer: Hawking radiation is a result of doing quantum mechanics outside a black hole! In short, a stationary observer outside a black hole, according to quantum mechanics, measures a precisely thermal spectrum of radiation originating from the black hole. What form this radiation takes depends on what quantum fields inhabit the universe, but this crucial result would hold in any case; photons or no photons!

Ultimately I don’t think that there’s really any way to visualise what’s going on. For me I just take it as a consequence of doing quantum mechanics near a black hole *. I suppose that I think (and believe that physicists should perhaps most correctly think) of Hawking radiation as a process, not a “noun”. Some people have cooked up explanations involving virtual particle pairs popping out of the vacuum on the horizon, where one exits and one falls in etc. and the outgoing guy is the Hawking radiation but I don’t really buy this, chiefly because I don’t think that virtual particles exist. Hawking’s original calculation doesn’t care about any of this anyway.

Black holes in nature, naively certainly do appear to violate the conservation of information . This is precisely because of the nature of Hawking radiation: it’s exactly thermal. This means that, no matter what goes in, the same spectrum of radiation emerges. Evidently, it’s then naively impossible to reconstruct the information pertaining to the in-falling stuff from what is, in this sense, just whitenoise. Some argue that if you do the calculation carefully, you can in principle recover this information but as far as I’m aware there’s no consensus as to whether or not this is possible.

As far as things falling into the horizon are concerned (such as your photons), there are two camps. One (I agree with them) maintains that due to Einstein's equivalence principle (loosely speaking, that no local patch of spacetime is in any sense “special”), an observer falling through the event horizon wouldn’t notice anything particularly special (though it’d probably be an interesting light-show!), they’d just go right on in. It’s only when they approached the singularity that things would get tense: the tidal forces near the singularity in the centre would ultimately tear any matter to shreds! The second camp support the firewall argument, which claims that the paradox is solved by an impassible wall of extremely high-energy quanta inhabiting the region just behind the horizon. As a relativist I’m less sympathetic to this view since I’d rather like to hope that solving the information paradox doesn’t require giving up Einstein’s equivalence principle (upon which Einstein’s theory of General Relativity is based). Nonetheless, if this is true, anything entering the black hole horizon would be immediately destroyed by this “firewall”.

*The reason that this happens is rather involved, and, most elegantly, requires understanding the path-integral formulation of quantum mechanical states. At the very least, what one can do is to see that the creation/annihilation operators acting on the vacuum state in Minkowski space correspond to those in a thermal state from the point of view of an observer at a fixed distance outside the black hole (and therefore a uniformally accelerating observer). This latter approach is rather ugly but tractable with some work. This is all a consequence of the fact that, in general, the ontology of quantum mechanical states are dependent on a choice of coordinates. In other words, what one observer calls a “vacuum state”, another observer (with a different coordinate system) might call an “excited state”. Indeed, the example at hand is precisely an instance of this.

Explanation:

1. The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 8% or 0.08.

2. The final kinetic energy of the neutron KE₂ (neutron), is 6.42 x 10⁻¹³ J.

Kinetic energy is the energy possessed by a body by virtue of its motion.

The fraction of the neutron's kinetic energy that is transferred to the atomic nucleus is calculated as follows:

The final velocity of the atom is found using the principle of conservation of linear momentum.

initial momentum of the neutron = final momentum of the atom

m₁u₁  = m₂u₂

where;

m₁ = mass of the neutron

u₁ = initial velocity of the neutron

m₂ = mass of the atomic nucleus

u₂ = final velocity of the atomic nucleus

m₂ = 112.4 m₁

u₂  = m₁u₁ / m₂

u₂  = m₁u₁ / (12.4 m₁)

u₂  = 0.08 u₁

The initial kinetic energy of the neutron is determined as follows:

KE₁ = ¹/₂m₁u₁²

The final kinetic energy of the atomic nucleus is determined as follows:

KE₂ =  ¹/₂m₂u₂²

KE₂ =  ¹/₂(12.4 m₁)(0.08u₁)²

KE₂  = 0.08 (¹/₂m₁u₁²)

KE₂ = 0.08 (KE₁)

The fraction of the neutron's kinetic energy transferred to the atomic nucleus is determined as follows:

Fraction = 0.08 (KE₂) / KE₁

Fraction= 0.08

Fraction = 8 %

The final kinetic energy of the neutron is calculated as follows;

KE₂ (neutron) = (1 - 0.08) x (6.97 x 10⁻¹³ J)

KE₂ (neutron) = 6.42 x 10⁻¹³ J

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The water cycle moves from one pond to another i.e. from the river to the ocean, or from the ocean to the atmosphere.

The water cycle is the path that all water accompanies as it moves around Earth in other states. The warm dry air at this distance then increases evaporation, which leaves the land beneath even drier. There are three basic locations of water storage that occur in the erratic water cycle. Water is stored in the atmosphere; water is deposited on the surface of the earth, and water is stored in the ground. There is no start or end to the water cycle, but for classification purposes, we will start at the sun.

So we can conclude that the water cycle relates to water being exchanged through Earth's land, ocean, and atmosphere.

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The coefficeint of static friction, given that the 12 Kg sled is lying on the hill with an incline of 21 degrees is 0.38

We know that the coefficient of static friction is related to frictional force according to the following formula:

Frictional force (N) = coefficient of friction (μ) × normal reaction (N)

F = μN

μ = F / N

For inclined plane, we have:

F = mgSineθ

N = mgCosθ

Thus,

μ = mgSineθ / mgCosθ

Recall

Sineθ / Cosθ = Tanθ

μ = Tanθ

Where

Now, we shall determine the coefficient of static friction as follow:

μ = Tanθ

μ = Tan21

μ = 0.38

Thus, the coefficient of static friction is 0.38

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The momentum of the sprinter of mass and velocity of 68 kg and 8 m/s respectively is  544 kgm/s

Momentum can be defined as the product of mass and velocity of a body.

To calculate the linear momentum of the sprinter, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitite the values into equation 1

Hence, the momentum of the sprinter is 544 kgm/s.

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The linear momentum of the sprinter with a mass of 68 kg that reaches a speed of 8 m/s during a race is 544kgm/s.

The momentum of a body in motion is the tendency of a body to maintain its inertial motion. It is the product of its mass and velocity, or the vector sum of the products of its masses and velocities.

The linear momentum of a body can be calculated as follows:

p = m × v

According to this question, a sprinter has a mass of 68kg and reaches a speed of 8 m/s during a race. The momentum can be calculated as follows:

p = 68kg × 8m/s

p = 544kgm/s

Therefore, 544kgm/s is the linear momentum of the sprinter.

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Answer:

Explanation:

Given:

F = Fₓ·i + Fy·j

S = Sₓ·i + Sy·j

Fₓ = 3 N

Fy = - 2 N

Sₓ = 5 m

Sy = 2 m

_________

A - ?   - Work

α - ? - Angle

F = 3·i - 2·j

S = 5·i + 2·j

The work is numerically equal to the scalar product of the displacement force

A = (F·S) = 3·5 + (-2)·2 = 15 - 4 = 11 J

Modules:

| F | = √ (3² + (-2)² ) = √ (9 + 4) = √ 13

| S | = √ (5² + 2² ) = √ (25 + 4) = √ 29

Angle:

cos α = (F·S) / ( |F| · |S| ) = 11 / ( √13 · √29) ≈ 0,5665

α ≈ 55.5°

The maximum number of revolutions per minute the fan can run at before she will be flung off is  1,982.24 rev/min.

The maximum angular speed of the fan is calculated by applying the following kinematic equation.

From Newton's second law of motion,

F = ma

F = mv/t

F = m(ωr/t)

where;

ω/t = F/mr

ω/t = (0.0137 N) / (0.00022 kg x 0.3 m)

ω/t = 207.58 rad/s

ω/t = (207.58 rad/s)  x (1 rev / 2π rad) x  (60 s / min)

ω/t = 1,982.24 rev/min

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The statement that is true about energy is as follows;

Kinetic energy is the energy possessed by an object because of its motion, equal (nonrelativistically) to one half the mass of the body times the square of its speed.

Potential energy is the energy possessed by an object because of its position (in a gravitational or electric field), or its condition (as a stretched or compressed spring, as a chemical reactant, or by having rest mass).

The law of conservation of energy is a principle stating that energy may not be created or destroyed but instead can be transformed from one form to another.

This suggests that kinetic energy as a type of energy can be changed into the resting form called potential energy and vice versa.

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The Force of friction on the bicycle of mass 91.4 kg, while moving is 546 N.

Force is the product of mass and acceleration.

To calculate the force of friction while the bicycle moves, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the force of friction on the bicycle is 546 N.

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The gravitational potential energy change during this climb is 83,790 J.

The change in the gravitational potential energy of the bike racer is calculated as follows;

ΔP.E = mgh

where;

The vertical height of the road is calculated as follows;

h = L sinθ

where;

h = 1500 x sin(4.3)

h = 112.5 m

ΔP.E = 76 x 9.8 x 112.5

ΔP.E = 83,790 J

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The energies of the photon in each case are;

a) 2.5 * 10^-5 eV

b) 2.5 eV

c) 2.5 * 10^2 eV

We know that a photon is known to have an energy that can be measured or calculated by the use of the formula;

E = hc/λ

E = energy of the photon

h = Plank's constant

λ = Wavelength

c = Speed of light

Then;

a) E = 6.6 * 10^-34 * 3 * 10^8/5 * 10^-2 m

E = 3.96 * 10^-24 J or 2.5 * 10^-5 eV

b) E = 6.6 * 10^-34 * 3 * 10^8/500 * 10^- 9 m

E = 3.96 * 10^-19 J or 2.5 eV

c) E =  6.6 * 10^-34 * 3 * 10^8/5 * 10^- 9 m

E =  3.96 * 10^-17 or 2.5 * 10^2 eV

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Answer:

Neuroscientists believed that neuroplasticity only manifested itself in childhood, but late 20th-century research suggests that many aspects of the brain can change (or become "plasticized") even in adulthood.

Explanation:

The system acceleration for the pulley having masses m1 = 0.7 kg and m2 = 0.4 kg  Is 35.93 [tex]m/s^2[/tex].

Newton's second law provides a detailed explanation of how a force can alter a body's motion. It is believed that a body's momentum varies over time at a rate equal to the force acting on it in both direction and amplitude. The combined effects of a body's mass and velocity determine its momentum.

Given:

The mass, m₁ = 0.7 kg,

m₂ = 0.4 kg,

Calculate the system acceleration by the following formula,

a = g(m1 − m2)/(m1 + m2)

Here a is the acceleration.

Substitute the values,

a = 9.8 * (0.7 - 0.4 ) / (0.7 + 0.4)

a = 35.93 [tex]m/s^2[/tex]

Therefore, the system acceleration for the pulley is 35.93 [tex]m/s^2[/tex].

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The amount of work done by the 100 W motor in 5 minutes is 30000 J.

Work is said to be done when a force moves a body through a certain distance.

To calculate the amount of work done by the motor, we use  the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the amount of work done by the motor is 30000 J.

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When an object transported from earth to the moon, its mass remains same but weight becomes 1/6th that on earth..

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter.

It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.

The amount of a body's weight indicates how much gravity is pulling on it. Weight is calculated using the method w = mg. Since weight is a force, it has the same SI unit as a force, which is the Newton (N).

So, mass of an object is intrinsic property of the object - it remains same in earth and in moon. But weight is the amount of gravitational  attraction force. Moon has nearly 1/6th gravitational attraction field. So, the weight of the object on moon will be 1/6th that on earth.

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Answer:

mass and size are the main important to any playing person for more the two extremely don't Matcha the person doesn't play any exercise

The pressure of the carbon dioxide is  21.2 kPa.

We know that the pressure of the gas is the force with which the gas does hit the walls of the container. In this case, we are told that a cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25°C.

Thus;

Volume of the cylinder = πd^2/4

Volume = 3.142 * (20 * 10^-2)^2/4

= 0.03142 m^3

From the ideal gas law;

PV = nRT

P = pressure

V = volume

n = Number of moles

R = gas constant

T = temperature

Then;

n = 1.2 * 10^3 g/44 g/mol = 27.3 moles

P = nRT/V

P = 27.3 * 0.082 * 298/0.03142

P = 21.2 kPa

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The amount of work required to move the rock of mass 95 kg up a ramp of 100 m is 93100 J.

Work can be fined as the product of force and distance.

To calculate the amount of work required to get the rock up, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the amount of work required is 93100 J.

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Answer: C

Explanation:

The answers include the following:

This is an institution which provides supervision and care of infants and young children especially so that parents can focus on their jobs.

The director is the head of the daycare and issues such as this should be reported to him/her so as to enable the best possible resolution.

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The genral result of the proton-proton chain is the formation of Helium Nucleus i.e., 4 hydrogen combined to form one helium nucleus.

Proton-proton chains are also called p-p chains, proton-proton cycles, or proton-proton reactions.

Roton-Proton Chain.

⁴H ---> He + energy + other

In the proton-proton chain, four hydrogen nuclei (protons) combine to form a helium nucleus. This results in a loss of the original mass. But some energy escapes in the form of neutrinos. First, two hydrogen nuclei (¹H) combine to release a positive electron (e+, positron) and a neutrino (ν) to form a hydrogen 2 nucleus (²H, deuterium).

¹H + ¹H --> ²H + e + ν

The hydrogen 2 nucleus quickly captures another proton to form a helium 3 (³He) nucleus while emitting a gamma (γ) ray.

²H + ¹H --> ³He + γ

From this point, the reaction chain can follow one of several paths, but always leads to a helium-4 nucleus emitting a total of two protons.

³He + He --> ⁴He + ¹H + ¹H

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When there is no other force is acting other than the given forces the magnitude of acceleration of mass is 6 m/[tex]s^{2}[/tex].

Magnitude is defined simply as “distance or quantity.” It depicts the absolute or relative direction or size in which an object moves in the sense of motion.

Acceleration is the rate of change of the velocity of an object with respect to time.

Acc. to the given data :

Object’s mass = 2kg

Forces acting = 3N, 4N, 5N

Now to find is the magnitude of acceleration of mass.

We use the following equation of Newton’s law

m = F net / a

rearranging the above equation

a = F net / m

now net force F= f1+f2+f3

net F= 3N+4N+5N=12N

so a=Fnet/m

    a=12N/2kg

    a=6 m/s2

Hence, the magnitude of acceleration of mass is 6 m/[tex]s^{2}[/tex].

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Answer:

[tex]0.60\; {\rm m \cdot s^{-1}}[/tex], assuming that the friction between the skates and the ice is negligible.

Explanation:

Under the assumptions, the total momentum of the two skaters will be conserved. In other words, the sum of the momentum of the two skaters will be the same before and after the collision.

When an object of mass [tex]m[/tex] moves at velocity [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].

The [tex]m_{b} = 50\; {\rm kg}[/tex] skater was initially moving with a velocity of [tex]v_{b} = 1.5\; {\rm m\cdot s^{-1}}[/tex]. The momentum of this skater will be:

[tex]m_{b}\, v_{b} = 50\; {\rm kg }\times 1.5\; {\rm m\cdot s^{-1}} = 75\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the [tex]m_{a} = 75\; {\rm kg}[/tex] skater was initially not moving (velocity is [tex]v_{b} = 0\: {\rm m\cdot s^{-1}}[/tex],) the momentum of that skater will be:

[tex]m_{a}\, v_{a} = 75\; {\rm kg }\times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Thus, the total momentum of the two skaters was [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex] before the collision.

Since the two skaters held on to each other, the two will travel at the same velocity after the collision. Let [tex]v[/tex] denote this velocity. The total momentum of the two skaters after the collision will be [tex]m_{a}\, v + m_{b}\, v = (m_{a} + m_{b})\, v[/tex].

Under the assumptions, momentum will be conserved in the collision. Hence, the total momentum after collision [tex](m_{a} + m_{b})\, v[/tex] should be equal to the total momentum before the collision, [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex]. In other words:

[tex](m_{a} + m_{b})\, v = 75\; {\rm kg \cdot m\cdot s^{-1}}[/tex].

Rearrange this equation and solve for the velocity [tex]v[/tex] of the two skaters after the collision:

[tex]\begin{aligned}v &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{m_{a} + m_{b}} \\ &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{75\; {\rm kg} + 50\; {\rm kg}} \\ &= 0.60\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In other words, the two skaters will travel at approximately [tex]0.60\; {\rm m\cdot s^{-1}}[/tex] after the collision.

Answer:

A

Explanation:

A constant velocity means the position graph has a constant slope. It's a straight line sloping up.

The net work done on the block over the given distance is 39.4d (joules)

The net work done on the block over the given distance is calculated by applying the following equation as shown below;

W(net) = F(net) x d

where;

F(net) = F - μmgcosθ

where;

F(net) = 50 N - (0.25 x 5 x 9.8 x cos30)N

F(net) = 50 N - 10.6 N

F(net) = 39.4 N

The net work done on the block over the given distance is calculated as;

W = 39.4 N x d

where;

W = 39.4d (joules)

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We first apply the data to the problem.

Data:

[tex] \bold{V = 90km/h}[/tex]

[tex] \bold{T = 20s}[/tex]

[tex] \bold{D = ?}[/tex]

Now, we convert km/h to m/s.

Conversion:

[tex] \bold{90km/h * (1000m/1km) * (1h/3600s)}[/tex]

[tex] \boxed{ \boxed{ \bold{V = 25m/s}}}[/tex]

Then, we apply the formula that is.

Formula:

[tex] \bold{D = V * T}[/tex]

To determine the distance traveled we develop the problem.

Developing:

[tex] \bold{D = (25m/s) * (20s)}[/tex]

[tex] \boxed{\boxed{ \bold{D = 500m}}}[/tex]

Its speed is 25 meters per second and the distance traveled is 500 meters.

Answer:

Explanation:

A = 6i - 4j + 8k

B = 2i + 4j - 14k

A + B = (6+2)i + (-4+4)j + (8-14)k

A + B = 8i +0j -6k

A + B = 8i - 6k

(1/2)(A + B) = 4i - 3k

d = √ (4² + 3²) = 5

e = (4/5)i -(3/4)j

Answer: yes

Explanation: