Find the momentum of a body of mass 5kg moving with velocity of 4.5m/s

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When a fragment of bone containing carbon-14 isotopes of the element carbon is discovered during an archaeological dig, and it is estimated to be approximately 23,000 years old, one can calculate the proportion of the carbon-14 isotopes that remains.

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Fringe Width is the distance between two maxima of the next order on one side of the central maxima.

The distance between two fringes can be calculated by β=λDd where β represents fringe width, λ represents wavelength, D represents the distance between slit and screen, and represents the distance between two slits. The fringe width depends on the wavelength of light, the distance between the slits, and slit separation.

The bright (constructive interference) fringes are at maximum intensity and the dark (destructive interference) fringes are at minimum intensity. Fringe is the alternating light and dark band that forms due to interference. Fringe width is the distance between two successive bright or dark fringes.

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Acceleration.

Acceleration is the time rate o change of velocity

The distance between the 7th and the tenth second is 29.4 m.

We know that we can have to formulate the information that has been given here so as to obtain a proper progression and this would help us to get the common difference of the progression that we are looking at.

Now we know that the progression would look something like; 4.9, 14.7, 24.5 ....

We can see that this is an arithmetic progression that has a common difference of 9.8.

U7 = a + (n - 1)d

a = first term

n = Number of terms

d = common difference

U7 = 4.9 + (7 - 1) 9.8

= 63.7

U10 = 4.9 + (10 - 1) 9.8

U10 = 93.1

Between the 7th and 10th seconds, we have;

93.1 - 63.7

= 29.4 m

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Therefore, the airplane will end up flying approximately 5.116 degrees off course due to the wind.

To determine the angle off course, we can use vector addition. Let's consider the airplane's velocity vector and the wind velocity vector as vectors in a two-dimensional plane.

The airplane's velocity vector points north with a magnitude of 600 km/hr. We'll denote it as A.

The wind's velocity vector blows from the southwest, which is a direction 45 degrees south of west. We'll denote it as W, with a magnitude of 70 km/hr.

To find the resultant velocity vector R, which represents the actual velocity of the airplane, we can add vectors A and W.

By decomposing vector W into its north and west components, we can determine the effect of the wind on the airplane's heading.

The north component of W is given by: \(W_{north}\) = W × cos(45°)

\(W_{north}\) = 70 × cos(45°) = 70 × 0.7071 ≈ 49.497 km/hr (rounded to three decimal places)

The west component of W is given by: \(W_{west}\) = W × sin(45°)

\(W_{west}\) = 70 × sin(45°) = 70 × 0.7071 ≈ 49.497 km/hr (rounded to three decimal places)

Now, let's subtract the westward wind component from the northward airplane component:

Resultant northward component = A - \(W_{north}\)

Resultant northward component = 600 - 49.497 = 550.503 km/hr (rounded to three decimal places)

Since there is no westward component remaining, the airplane will not drift off course in the west-east direction due to the wind.

To find the angle off course, we can use trigonometry. The angle θ can be determined as follows:

θ = arctan(\(W_{west}\) ÷ Resultant northward component)

θ = arctan(49.497 ÷ 550.503)

θ ≈ 5.116 degrees (rounded to three decimal places)

Therefore, the airplane will end up flying approximately 5.116 degrees off course due to the wind.

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As an interstellar cloud of hydrogen gas shrinks in size, its temperature increases due to the principle of conservation of energy.

When the cloud contracts, gravitational potential energy is converted into thermal energy. The gravitational force pulling the gas inward does work on the gas particles, increasing their kinetic energy and causing them to move faster. As the particles move faster, collisions between them become more frequent and more energetic, resulting in an overall increase in temperature. This process is governed by the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant. As the interstellar cloud contracts, its volume decreases, leading to an increase in pressure. According to the ideal gas law, this increase in pressure is accompanied by an increase in temperature.

Therefore, as the interstellar cloud of hydrogen gas shrinks in size, its temperature increases as gravitational potential energy is converted into thermal energy through the increase in kinetic energy of the gas particles.

FULL QUESTION: As an interstellar cloud of hydrogen gas shrinks in size, its temperature increases,

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wings

Lift is generated by every part of the airplane, but most of the lift on a normal airliner is generated by the wings. Lift is a mechanical aerodynamic force produced by the motion of the airplane through the air.

Answer:

the answer will be c

car p has twice the kinetic energy of car Q

Answer:

The value is  \(u = 27.93 \ m/s\)

Explanation:

From the question we are told that

   The  mass of the ball is \(m = 0.250 \ kg\)

    The radius of the circle is  \(r = 0.5 \ m\)

    The force exerted is  \(F = 33 \ N\)

    The  speed of the ball on top of the circle is  \(v = 13.0 \ m/s\)

Gnerally from the law of energy conservation we have that

     \(PE + W = \Delta KE\)

Here PE  is the potential energy of the pitcher which is mathematically represented as

      \(PE = m * g * (2 r )\)

=>   \(PE = 0.250 * 9.8 * (2 * 0.5 )\)  

=>   \(PE = 24.5\ J\)

And  W  is the workdone by the force applied which is mathematically represented as

       \(W = F * \pi * r\)

=>    \(W = 33 * 3.142 * 0.5\)

=>    \(W = 51.84 \ J\)

And   \(\Delta KE\) is the change in kinetic energy which is mathematically represented as

       \(\Delta KE = \frac{1}{2} * m * (v^2 - u^2 )\)

=>    \(\Delta KE = \frac{1}{2} * 0.250 * ( 13^2 - u^2 )\)

=>    \(\Delta KE = 0.125 * ( 13^2 - u^2 )\)

=>    \(\Delta KE = 21.125 - 0.125u^2\)

So

      \(24.5 + 51.84 = 21.125 - 0.125u^2\)

=>    \(u = 27.93 \ m/s\)

The velocity of the ball released at the bottom is required.

The velocity of the ball released at the bottom is 24.56 m/s

F = Force = 33 N

r = Radius of loop = 0.5 m

m = Mass of ball = 0.25 kg

u = Initial velocity of ball = 13 m/s

v = Final velocity

g = Acceleration due to gravity = \(9.81\ \text{m/s}^2\)

The energy balance of the system is given by

\(F\pi r+mg2r=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow v=\sqrt{\dfrac{2}{m}(F\pi r+mg2r)+u^2}\\\Rightarrow v=\sqrt{\dfrac{2}{0.25}(33\times \pi\times 0.5+0.25\times 9.81\times 2\times 0.5)+13^2}\\\Rightarrow v=24.56\ \text{m/s}\)

The velocity of the ball released at the bottom is 24.56 m/s.

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A 0.1mm diameter glass tube is inserted into an ethyl alcohol at 20 degrees centigrade in a cup. The angle of alcohol with the tube is 0 degrees. D = 0.1 mm = 0.001m, alcohol angle is 0 degrees, ethyl alcohol

\(η=1.1×10^-3 N⋅sec/ m², s_g\)

=0.79.

We have to determine how much glycerin rises in the tube. The force that moves the liquid through the tube is the difference between the downward force of gravity on the liquid column and the upward capillary force produced by the surface tension of the liquid against the walls of the tube.

The height to which the fluid rises in a capillary tube may be measured by balancing the capillary force against the force of gravity on the column.

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At 0 degrees Celsius, the speed of sound in air is around 331 m/s. The right answer is (d) 331 m/s.

We can use the formula for the speed of sound in air:

v = 331.3 √(T/273.15 + 1)

where v is the speed of sound in m/s and T is the air temperature in Kelvin.

At 0 C, the air temperature in Kelvin is:

T = 0 + 273.15 = 273.15 K

Substituting this value into the formula, we get:

\(v = 331.3 \sqrt{(273.15/273.15 + 1)}\\v = 331.3 m/s\)

Therefore, the speed of sound in air at 0 C is approximately 331 m/s. So, the correct option is (d) 331 m/s.

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Answer:

a.

445 nm : 4.45 x 10^-19 J

445 nm : 2.78 eV

586 nm : 3.38 x 10^-19 J

586 nm : 2.11 eV

667 nm : 2.99 x 10^-19 J

667 nm : 1.87 eV

b.

E3 to E1 : ΔE = 1.87 eV

E3 to E2 : ΔE = 0.24 eV

E2 to E1 : ΔE = 0.24 eV

Explanation:

a) To calculate the energies of the emitted photons, we can use the formula:

E = hc/λ

where E is the energy of the photon in Joules, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon in meters.

For the first line of wavelength 445 nm:

E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(445 x 10^-9 m) = 4.45 x 10^-19 J

To convert this to electron volts (eV), we can use the conversion factor 1 eV = 1.602 x 10^-19 J:

E = 4.45 x 10^-19 J/(1.602 x 10^-19 J/eV) = 2.78 eV

For the second line of wavelength 586 nm:

E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(586 x 10^-9 m) = 3.38 x 10^-19 J

E = 3.38 x 10^-19 J/(1.602 x 10^-19 J/eV) = 2.11 eV

For the third line of wavelength 667 nm:

E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(667 x 10^-9 m) = 2.99 x 10^-19 J

E = 2.99 x 10^-19 J/(1.602 x 10^-19 J/eV) = 1.87 eV

b) The energy level diagram for an atom of the element which has produced these photons can be drawn as follows:

markdown

     E3

     |

     | ΔE = 1.87 eV

     |

     |

     |

     E2

     |

     | ΔE = 0.24 eV

     |

     |

     |

     E1

The electron transitions which have given rise to the three spectral lines are:

The transition from energy level E3 to E1, which produces the photon of wavelength 445 nm and energy 2.78 eV (ΔE = 1.87 eV).

The transition from energy level E3 to E2, which produces the photon of wavelength 586 nm and energy 2.11 eV (ΔE = 0.24 eV).

The transition from energy level E2 to E1, which produces the photon of wavelength 667 nm and energy 1.87 eV (ΔE = 0.24 eV).

Note that this energy level diagram is just one possible arrangement of energy levels that could produce the observed spectral lines. There may be other possible arrangements of energy levels that could also explain the observed lines.

ChatGPT

The semicircular canals are specialized to assess rotational acceleration of the head, while the otolith organs are specialized to detect linear acceleration and static position of the head relative to the gravitational axis. The semicircular canals contain fluid-filled channels arranged in three perpendicular planes, allowing them to detect angular movements of the head. On the other hand, the otolith organs consist of small calcium carbonate crystals suspended in gelatinous fluid, which respond to linear accelerations and changes in head position relative to gravity.

The semicircular canals are responsible for detecting rotational acceleration of the head. They are three fluid-filled canals positioned in different planes: the horizontal canal, anterior (superior) canal, and posterior (inferior) canal. Each canal has a bulge at one end called the ampulla, which contains hair cells that detect fluid movement. When the head rotates, the fluid within the canals also moves, bending the hair cells and signaling the brain about the rotational acceleration.

The otolith organs, consisting of the utricle and saccule, are specialized in detecting linear acceleration and static position of the head relative to the gravitational axis. These organs contain small calcium carbonate crystals called otoliths that are embedded in a gelatinous layer. When the head accelerates linearly or changes position relative to gravity, the otoliths move, causing the gelatinous layer to shift and stimulating the hair cells. This signals the brain about changes in linear acceleration and head position, including tilting or linear movements such as walking or riding in a vehicle.

In summary, the semicircular canals are designed to detect rotational acceleration, while the otolith organs are specialized in detecting linear acceleration and static position changes of the head relative to the gravitational axis. This division of sensory functions allows for a comprehensive assessment of different types of head movements and orientations in space.

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The phase change that involves the absorption of heat is endothermic phase transitions, such as melting and evaporation.

In endothermic phase transitions, substances absorb heat energy from their surroundings, causing molecular vibrations to increase. During melting, a solid substance gains heat energy, breaking the bonds holding its molecules in a rigid structure, and transitioning into a liquid state.

Similarly, in evaporation, a liquid substance absorbs heat, which enables its molecules to overcome the intermolecular forces, and turn into a gaseous state. In both cases, the absorption of heat energy is crucial for the phase change to occur.

These processes are in contrast to exothermic phase transitions, like freezing and condensation, where heat energy is released.

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Answer:

The force of gravitational attraction between them also decreas

the height of the image of the insect will be the same as the height of the insect, but with a negative sign.

To determine the height of the image of the insect formed by a lens with a focal length of f = -125 mm, we need to apply the sign conventions used in lens imaging.

In the sign convention:

Distances measured from the lens towards the object are positive.

Distances measured from the lens towards the image are negative.

Focal length (f) is positive for a converging lens (convex) and negative for a diverging lens (concave).

Since the given focal length is f = -125 mm (negative value), we know that we are dealing with a diverging lens.

When an object is placed in front of a diverging lens, the resulting image is virtual, upright, and located on the same side as the object. The image formed by a diverging lens is always reduced in size compared to the object.

Since the image is virtual and upright, the height of the image will have the same magnitude as the height of the object, but it will be negative to indicate that it is located on the same side as the object.

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Answer:

is suppoesd to say after 20 yrs

Explanation:

Answer:

they move towards the positive side... that's option 2

Because the refractive indexes of the immersion oil and the glass microscope slide are the same; the oil prevents the refraction of the incoming light rays.

An invisible space is present between the slide and the lens that contains air.

When the light rays fall on the slide and lens they might get refracted due to air present between them.

If the rays get refracted then the diameter of the objective lens will expand further.

So in order to prevent the refraction of light, it is very important to lubricate the surface of the lens because glass and oil have same refractive index and without oil the light rays refract from their original path.

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The clocks on the space vehicle will gain approximately 0.0031 day or 4.5 minutes per day compared to Earth-based clocks due to time dilation. This means that if the space vehicle travels for a year, its clocks would be ahead of Earth-based clocks by approximately 26 hours.

According to the theory of relativity, time is not absolute and can vary depending on the relative motion of two observers. When an object travels at a high velocity relative to another object, time dilation occurs, which means that time appears to move slower for the moving object. In the case of a space vehicle traveling at a high velocity relative to the Earth, its clocks will appear to move slower than Earth-based clocks.

To calculate the exact amount of time dilation, we need to use the formula:

Δt' = Δt * √(1 - v²/c²)

where Δt' is the time interval measured by the moving clock, Δt is the time interval measured by the stationary clock (Earth-based clock), v is the velocity of the moving clock relative to the stationary clock, and c is the speed of light.

Assuming that the space vehicle is traveling at a velocity of 27,500 km/h, which is about the speed of the International Space Station (ISS), we can calculate the time dilation as follows:

Δt' = Δt * √(1 - v²/c²)
Δt' = 1 day * √(1 - (27,500 km/h)²/(299,792 km/s)²)
Δt' = 1 day * √(1 - 0.0002525)
Δt' = 1 day * √(0.9997475)
Δt' = 0.999873 day or 23.9969 hours

Therefore, the clocks on the space vehicle will gain approximately 0.0031 day or 4.5 minutes per day compared to Earth-based clocks due to time dilation. This means that if the space vehicle travels for a year, its clocks would be ahead of Earth-based clocks by approximately 26 hours.

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Answer: Try A. 110 Hz

Explanation:

Answer:

i got 110 Hz..

Explanation:

Explanation: The amount of solar energy available varies by location on earth, the time of the day at that location, the season and weather conditions. Rain, fog, and clouds will lower the amount of solar power available to gather and there are fewer hours of sun available in different seasons (winter for North America). To gather solar power, you also need a large area. This can be lessened by conditions such as snow, so that even on a sunny day, covered panels will collect less energy.

Answer:

.

Explanation:

The quantity of solar energy accessible varies by location on the planet, time of day, season, and weather conditions. Rain, fog, and clouds reduce the quantity of solar power available for collection, and various seasons have less hours of sunlight accessible (winter for North America). A big region is also required to collect solar electricity. Conditions such as snow can reduce this, such that even on a sunny day, covered panels will absorb less energy.

Sunlight at a wavelength of 0.4 μm (micrometers) is absorbed (i.e. reduced) by about 0 to 12 percent by the atmosphere when it gets to the surface of the earth.

Between 0.40 and 0.71 micrometers (m) is the wavelength of visible light. In this region, the sun only emits 44% of its total radiation. The spectrum of solar energy ranges from around 0.1 to 4.0 micrometers. The 0.1 to 0.4 micrometer wavelength range contains around 7% of the sun's radiation (UV). Near infrared radiation from the sun falls in the range of 0.71 to 1.5 micrometers, whereas far infrared radiation falls in the range of 1.5 to 4.0 micrometers.

Extraterrestrial radiation is solar radiation that occurs outside the earth's atmosphere. The extraterrestrial irradiation is typically 1367 W/m2. As the earth circles the sun, this number fluctuates by about 3%.

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Answer: F = 102141N

Explanation: Newton's 2nd Law states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

\(a=\frac{\Delta v}{\Delta t}\)

\(a=\frac{11}{0.14}\)

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

Jenny's car experienced a force of magnitude 102141N.

Answer:

NO

Explanation:

It is just impossible to know if someone is staring at you while they are standing behind you. UNLESS they turned around OR was able to see them though a mirror of some sort. :)

Answer:

batmintan ......

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thanks

The false statement is B) Homolytic bond cleavage yields neutral radicals in which each atom gains one electron.

In homolytic bond cleavage, each atom retains one electron from the shared pair of electrons, resulting in the formation of two neutral radicals, where each atom retains its original number of electrons.

No atoms gain or lose electrons in this process.

In a homolytic bond cleavage, a covalent bond is broken, and the shared pair of electrons is split equally between the two atoms involved in the bond.

This results in the formation of two neutral radicals, with each atom retaining one of the electrons from the shared pair.

A radical is a chemical species characterized by the presence of an electron that is unpaired, meaning it does not have a partner electron with which it forms a complete pair. When a covalent bond is homolytically cleaved, each atom involved in the bond gains one electron, resulting in the formation of two radicals.

These radicals are highly reactive due to the presence of the unpaired electron, which makes them prone to participate in further chemical reactions.

It's important to note that in homolytic bond cleavage, there is no transfer of electrons from one atom to another.

Instead, the bond is broken in a way that allows each atom to retain one of the electrons, leading to the formation of two neutral radicals.

Therefore, statement B, which suggests that each atom gains one electron, is false.

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