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A thin 2.50 kg box rests on a 5.50 kg board that hangs over the end of a table, as shown in (Figure 1).

How far can the center of the box be from the end of the table before the board begins to tilt?

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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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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Answer:

Approximately [tex]0.293\; {\rm kg \cdot m^{2}}[/tex].

Explanation:

Since the wheel started from rest, initial angular velocity will be [tex]\omega_{0} = 0\; {\rm rad \cdot s^{-1}}[/tex]. It is given that the angular velocity [tex]\omega_{1}[/tex] is [tex]14.9\; {\rm rev \cdot s^{-1}}[/tex] after [tex]t = 3.41\; {\rm s}[/tex]. Apply unit conversion and ensure that all angular velocity are measured in radians-per-second:

[tex]\begin{aligned} \omega_{1} &= 14.9\; {\rm rev \cdot s^{-1}} \times \frac{2\, \pi \; {\rm rad}}{1\; {\rm rev}} \\ &\approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}[/tex].

Change in angular velocity:

[tex]\begin{aligned} \Delta \omega = \omega_{1} - \omega_{0} \approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}[/tex].

Since the tangential force is constant and there is no friction on the wheel, the angular acceleration [tex]\alpha[/tex] of this wheel will be constant. Since the change in velocity [tex]\Delta \omega \approx 93.620\; {\rm rad \cdot s^{-1}}[/tex] was achieved within [tex]t = 3.41\; {\rm s}[/tex], the average angular acceleration will be:

[tex]\begin{aligned} \alpha &= \frac{\Delta \omega}{t} \\ &\approx \frac{93.620\; {\rm rad \cdot s^{-1}}}{3.41\; {\rm s}} \\ &\approx 27.45\; {\rm rad \cdot s^{-2}}\end{aligned}[/tex].

At a distance of [tex]r = 0.110\; {\rm m}[/tex] from the axis of rotation, the tangential force [tex]F = 73.0\; {\rm N}[/tex] will exert on the wheel a torque [tex]\tau[/tex] of magnitude:

[tex]\begin{aligned} \tau &= F\, r \\ &= (73.0\; {\rm N})\, (0.110\; {\rm m}) \\ &\approx 8.030\; {\rm N \cdot m}\end{aligned}[/tex].

Let [tex]I[/tex] denote the moment of inertia of this wheel. The equation [tex]\alpha = (\tau / I)[/tex] relates angular acceleration [tex]\alpha[/tex] to moment of inertia [tex]I\![/tex] and net torque [tex]\tau[/tex]. Rearrange this equation to find the moment of inertia:

[tex]\begin{aligned}I &= \frac{\tau}{\alpha} \\ &\approx \frac{8.030\; {\rm N\cdot m}}{27.45\; {\rm rad \cdot s^{-2}}} \\ &\approx 0.293\; {\rm N \cdot m \cdot s^{2}} \\ &= 0.293 \; {\rm kg \cdot m^{2}}\end{aligned}[/tex].

Note that the unit "radians" is typically ignored. Additionally, [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].

Hence, the moment of inertia of this wheel is approximately [tex]0.293\; {\rm kg \cdot m^{2}}[/tex].

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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In a reference frame fixed to Earth, a point on the equator moves zero distance in 1 hour.

A frame of reference in physics is made up of an abstract coordinate system and the collection of physical reference points that fix the coordinate system specifically and regulate measurements inside it.

When we take a reference frame fixed to Earth,  the displacement due to earth's rotation can not be measured. So, any point on earth remains in rest through out the day with respect to the reference frame fixed to Earth. Hence, a point on the equator moves no distance in 1 hour in a reference frame fixed to Earth.

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Answer:

Thank you for helping use

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 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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The following which is not a way that machines provide a mechanical advantage is that it changes power and is denoted as option C.

This is referred to the measure of how much a force is increased by using a tool or machine and is calculated by finding the ratio of the output force to input force.

Changing the power is not a way in which machines provide a mechanical advantage as a result of the parameter not directly involved in how it functions which is therefore the reason why it was chosen as the correct choice.

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1900 J of heat or thermal energy is added to 90 g of water at an initial temperature of 18°C than temperature of the water 21402 K.

The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is performed in the process.

Given in the question 1900 J of heat is added to 90 g of water at an initial temperature of 18°C.

heat = mass*specific heat*change in temperature

1900 = .09 Kg*1*ΔT

ΔT =    21111

final temperature  = 21111 + 291 K

final temperature = 21402 K

1900 J of heat or thermal energy is added to 90 g of water at an initial temperature of 18°C than temperature of the water 21402 K.

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Statement 1 is not correct.

The rate at which velocity changes over time, both in terms of speed and direction, is defined as acceleration. A point or object traveling in a straight path is accelerated if it accelerates or decelerates. Even if the speed is constant, the motion of a circle is accelerated because the direction is constantly changing.

Given that

Acceleration is always a positive number.

Acceleration has SI units of m/s².

The slope on a speed-time graph is the object's acceleration.

To calculate the acceleration of an object, the change in velocity is divided by the length of the time interval over which the change occurs.

When a body accelerates, it is positive; when it slows, it is negative. Positive acceleration is directed along the velocity axis. Negative acceleration, on the other hand, is oriented in the opposite direction as the velocity.

The SI unit of acceleration is m/s2. Because acceleration is defined as velocity in meters per second divided by time in seconds, the SI units for acceleration are m/s2, meters per second squared, or meters per second per second, which essentially means how many meters per second the velocity changes every second.

The acceleration of an item is represented by the slope of a speed-time graph. The slope of a speed-time graph displays the rate of change of the body's speed and is known as acceleration.

To compute an object's acceleration, divide the change in velocity by the length of the time interval over which the change occurs. When you accelerate quickly, you apply a lot of force on an object. Depending on the speed and direction, acceleration can be either positive or negative.

A = velocity change/time

A = (vf- v1)/time

Therefore statement 1 is not correct

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Answer: yes

Explanation:

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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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: 2m

Explanation:

x = v(t)

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

x = 2m

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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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:

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

13.50 g/cm³ value when converted into SI unit is  13500 kg/m³.

SI stands for system international that is standardized to be used for research , various purposes.

1 gram /cm³=1000 kg/m³

so therefore 13.50g/cm³= 13.50 x 10³kg/m³

The reason behind using kg as SI unit of mass is because its defined by taking fixed numerical value of plank's constant.

SI unit of length is metre as it's defined by taking fixed numerical value of speed of light in vacuum. so unit of volume is m³.

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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 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 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 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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There is no acceleration that is imparted to the car, then the forces are balanced.

We know that unbalanced force is that which cause a body to move or to cause a body to stop moving at a constant rate. If that is so, then a body that acted upon by a system of balanced forces would remain at rest or would continue in a state of uniform motion.

Given the fact that we have been told that this car is moving at a constant speed of about +12 m/s, it is now plain that what acts on the car must be forces that are balanced since the  car is not accelerating.

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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.

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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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:

The photon rapidly goes towards the 'singularity' at the center of the black hole, and acceleration will increase tremendously towards center of black hole.

The rate at which an item changes its velocity is known as acceleration, a vector variable. If an object's velocity is changing, it is accelerating. A moving object can occasionally alter its velocity by the same amount every second. a moving object that changes its speed by 10 m/s per second. Since the velocity is changing by a fixed amount every second, this is known as a constant acceleration. It is important to distinguish between an item with a constant acceleration and one with a constant velocity. Be not deceived! An object is accelerating if its velocity is changing, whether by a fixed amount or a variable quantity. Additionally, a moving item with a constant speed is not accelerating.

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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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Answer:

the answer is 0

Explanation: