A train rolls past a stationary observer. To him, the train is moving at a speed of 23 m/s west, and a woman on the train is moving at a speed of 22.4 m/s west. How long does it take the woman to move 13 m relative to the train?

Using Fick's Law of Diffusion, it would take about 394,737 seconds for the scent of vanilla (vanillin) to travel 3 meters in still air, considering its diffusivity in the given conditions.

The time it takes for a scent to travel through air can be estimated using Fick's Law of Diffusion, which relates diffusion time to the diffusivity of the substance, the distance it needs to travel, and the area through which it diffuses.

Diffusion time = (Distance^2) / (2 * Diffusivity)

Given that the distance is 3 meters and the diffusivity (D) of vanillin in air is 0.114 cm^2/s, we need to convert the distance to centimeters before applying the formula:

Diffusion time = (300 cm)^2 / (2 * 0.114 cm^2/s)

Diffusion time ≈ 90,000 cm^2 / 0.228 cm^2/s

Diffusion time ≈ 394,737 seconds

So, it would take approximately 394,737 seconds for the smell of vanilla to reach you from a distance of 3 meters in still air.

The energy of the photon emitted is about 1.60 × 10⁻²⁴ Joule

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

[tex]\large {\boxed {E = h \times f}}[/tex]

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]

[tex]\large {\boxed {E = qV + \Phi}}[/tex]

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem !

Given:

λ = 12.4 cm = 12.4 × 10⁻² m

h = 6.63 × 10⁻³⁴ Js

c = 3 × 10⁸ m/s

Unknown:

E = ?

Solution:

[tex]E = h \times \frac{c}{\lambda}[/tex]

[tex]E = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{12.4 \times 10^{-2}}[/tex]

[tex]E \approx 1.60 \times 10^{-24} ~ Joule[/tex]

Grade: College

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light

The flash travels the round-trip distance approximately 1,000,000 times.

The speed of sound is 336 m/s, and the speed of light (which represents the speed at which the flash travels) is approximately [tex]3\times 10^8 m/s[/tex].

Let's denote the distance between the mirrors as d. The time it takes for the sound to travel the round trip (to the cliff and back) is [tex]2d/336[/tex]seconds. During this time, the flash of light travels at [tex]3\times10^8m/s.[/tex]

To find out how many times the flash of light can travel the round-trip distance before the sound is heard, we calculate:

[tex]\text{Number of round trips}=(3\times10^8\times2d/336)/2d=(3\times10^8)/336\approx1000,000[/tex]

Thus, the flash of the gunshot travels the round-trip distance approximately 1,000,000 times before the echo of the gunshot is heard.

The wavelength of the photon having twice the energy as that of the photon of wavelength [tex]600\,{\text{nm}}[/tex] is [tex]\boxed{300\,{\text{nm}}}[/tex] .

Further Explanation:

The photons are the small packets of energy that move at the speed of light. The photons are considered to remain always in motion. The energy associated with a moving photon is given by:

[tex]E = \dfrac{{hc}}{\lambda }[/tex]

Here,  [tex]E[/tex]  is the energy associated with the photon, [tex]h[/tex] is the Planck’s constant, [tex]c[/tex] is the speed of light and [tex]\lambda[/tex] is the wavelength of the moving photon.

The value of the Planck’s constant is [tex]6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}[/tex] .

The wavelength of the photon is [tex]600\,{\text{nm}}[/tex] .

The energy associated with the photon of wavelength [tex]600\,{\text{nm}}[/tex] is:

[tex]\begin{aligned}{E_1}&=\frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{600 \times {{10}^{ - 9}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6 \times {{10}^{ - 7}}}}\\&= 3.3 \times {10^{ - 19}}\,{\text{J}}\\\end{aligned}[/tex]

The wavelength of photon having energy double of this:

[tex]\begin{aligned}E' &= 2{E_1}\\&= 2 \times\left( {3.3 \times {{10}^{ - 19}}} \right)\,{\text{J}}\\&{\text{ = 6}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}}\,{\text{J}}\\\end{aligned}[/tex]

The new wavelength of the photon will be:

 [tex]\lambda ' = \dfrac{{hc}}{{E'}}[/tex]

Substitute [tex]6.6 \times {10^{ - 19}}\,{\text{J}}[/tex] for [tex]E'[/tex] in above expression.

[tex]\begin{aligned}\lambda ' &= \frac{{\left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{6.6 \times {{10}^{ - 19}}}}\\&=\frac{{1.98 \times {{10}^{ - 25}}}}{{6.6 \times {{10}^{ - 19}}}}\,{\text{m}}\\&= 3.0 \times {10^{ - 7}}\,{\text{m}}\\&= 300\,{\text{nm}}\\\end{aligned}[/tex]

The wavelength of the photon having twice the energy as that of the photon of wavelength [tex]600\,{\text{nm}}[/tex] is [tex]\boxed{300\,{\text{nm}}}[/tex].

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Photon and Energy

Keywords:  Wavelength, photon, energy, E=hc/lamda, 600nm, twice the energy, Planck’s constant, small packets of energy, 300nm, speed of light.

The kinetic energy of the object at t = 0 is approximately 66.27 Joules.

To calculate the kinetic energy of the object, we can use the equation: KE = 0.5 * m * v^2. Given the mass of the object is 5.22 kg and the x-component of velocity is 5.10 m/s, we can substitute these values into the equation:

KE = 0.5 * 5.22 kg * (5.10 m/s)^2

Calculating this gives us a kinetic energy of approximately 66.27 Joules.

Answer:

v =  3321.85 mph

Explanation:

Equivalences :

1 mile = 1609.34 m

1 hour = 3600 seconds

Data

v= 1485 m/s : speed of sound in water

Problem Development

To calculate the speed of sound in mph (mile / hour), we multiply by the conversion factors using the equivalences:

[tex]v= (1485 \frac{m}{s} )*(\frac{1mile}{1609.34 m} )*(\frac{3600s}{hour})[/tex]

We cancel the units in seconds (s) and meters (m) to get the answer in miles per hour (miles / hour or mph)

[tex]v=\frac{1485*3600}{1609.34} \frac{mile}{hour}[/tex]

v= 3321.85 mile/hour

v =  3321.85 mph

The speed of a satellite in a geosynchronous orbit is approximately 2.98 km/s, and the magnitude of the acceleration is approximately 1.92 x 10^-3 m/s^2.

To determine the speed of a satellite in a geosynchronous orbit, we can use the formula:

speed = 2 x π x radius / period

Given that the radius of the Earth is 6.37 * 10^6 m and the altitude of a geosynchronous orbit is 3.58 * 10^7 m, we can use the formula to calculate the speed:

speed = 2 x 3.14 x (6.37 * 10^6 + 3.58 * 10^7) / (24 x 60 x 60)

The magnitude of the acceleration of a satellite in a circular orbit can be calculated using the formula:

acceleration = (velocity)^2 / radius

Using the calculated speed and the radius of the orbit, we can find the magnitude of the acceleration:

acceleration = (2.98 x 10^3)^2 / (6.37 * 10^6 + 3.58 * 10^7)

Therefore, the speed of a satellite in a geosynchronous orbit is approximately 2.98 km/s and the magnitude of the acceleration is approximately 1.92 x 10^-3 m/s^2.

Answer:

35.8 m

Explanation:

Given:

Initial Velocity u = 26.5 m/s

Time period t = 2.7 s

To find:

Maximum height H = ?

Solution:

The toy is projected vertically upward. So the motion is happening in y axis

When a projectile reaches its maximum height, at that point its velocity vill be zero

Using equations of motion we can find the height

[tex]v^{2} =u^{2} -2gH\\\\0^{2} =26.5^{2} -2\times 9.8 \times H\\\\19.6H = 702.25\\\\H = 35.8 m[/tex]

Verification

[tex]H = ut - \frac{1}{2} gt^{2}\\\\H = 26.5 \times 2.7- 0.5 \times 9.8 \times 2.7^{2}\\\\H = 35.8 m[/tex]

A. How much work is being done to hold the beam in place?

Work is the product of Force and Displacement. Since there is no Displacement involved in just holding the beam in place, hence the work is zero.

B. How much work was done to lift the beam?

In this case, force is simply equal to weight or mass times gravity. Hence the work is:

Work = weight * displacement

Work = 500 lbf * 100 ft

Work = 50,000 lbf * ft

C. How much work would it take if the steel beam were raised from 100 ft to 200ft?

The displacement is still 100 ft since 200 – 100 = 100 ft, hence the work done is still similar in B which is:

Work = 50,000 lbf * ft

To find the rate at which oil is flowing through a constricted pipe with different diameters and pressures, apply the principle of continuity.

A horizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m. The pressure in the pipe is 8100 N/m2, and in the constricted section, it is 6075 N/m2. The density of oil flowing in the pipe is 821 kg/m3.

To find the rate at which oil is flowing, we can apply the principle of continuity, which states that the product of the cross-sectional area and the fluid velocity is constant in a pipe with steady flow.

By applying the principle of continuity, you can calculate the rate at which oil is flowing through the pipe.

The ratio of the particle masses is \boxed{\frac{1}{3}} or \boxed3 .

Further explain:

We have to calculate the ratio of the particle masses.

As we know, in the elastic collision between two masses the momentum and the energy both are conserved.

Here, the collision between the masses the head-on it means head to head.

For head on head collision the masses will travel parallel but opposite in the direction.

We have two masses one is heavier and another is lighter.

The mass of massive or heavier particle is [tex]{m_1}[/tex].  

The mass of the lighter particle is [tex]{m_2}[/tex].  

From the conservation of linear momentum total initial momentum is equal to the total final momentum.

Therefore,

[tex]\boxed{\left( {{m_1}v - {m_2}v} \right) = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)}[/tex]

Here, after the collision the massive particle comes into rest.

So, final expression will be,

[tex]\left( {{m_1}-{m_2}}\right)v={m_2}{v_2}[/tex]                                   …… (1)

From the conservation of the energy,

Total kinetic energy before collision is equal to the total kinetic energy after collision.

Therefore,

[tex]\begin{aligned}\frac{1}{2}{m_1}{v^2}+\frac{1}{2}{m_2}{v^2}&=\frac{1}{2}{m_2}{\left( {{v_2}} \right)^2}\\{m_1}{v^2}+{m_2}{v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\left( {{m_1}+{m_2}}\right){v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\end{aligned}[/tex] 

Simplify the above equation,

[tex]\begin{aligned}{m_2}{\left( {{v_2}} \right)^2}&=\frac{{\left( {{m_1}+{m_2}} \right){v^2}}}{{{m_2}}}\\{v_2}&=\left( {\sqrt {\frac{{\left( {{m_1}+{m_2}} \right)}}{{{m_2}}}} }\right)v\\\end{aligned}[/tex]

Substitute the value of [tex]{v_2}[/tex] in equation (1).

[tex]\begin{aligned}\left( {{m_1} - {m_2}} \right)v&={m_2}\left( {\sqrt {\frac{{\left( {{m_1} + {m_2}}\right)}}{{{m_2}}}} } \right)v \\\left( {{m_1} - {m_2}} \right)&=\sqrt {{m_2}\left( {{m_1} + {m_2}}\right)}\\{m_2}\left( {\frac{{{m_1}}}{{{m_2}}} - 1}\right)&={m_2}\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}\\\left( {\frac{{{m_1}}}{{{m_2}}}-1}\right)&=\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}}+ 1}\right)}\\\end{aligned}[/tex]

Substitute [tex]x[/tex] for[tex]\dfrac{{{m_1}}}{{{m_2}}}[/tex] in above equation.

[tex]\left( {x - 1} \right)=\sqrt {\left( {x + 1} \right)}[/tex]

Squaring both the sides in above equation,

[tex]\begin{aligned}{\left( {x - 1} \right)^2}&=\left( {x + 1}\right)\\{x^2} - 2x + 1&=x + 1\\{x^2}-3x&=0\\\end{aligned}[/tex]

Taking [tex]x[/tex] as a common in the above equation.

[tex]x\left( {x - 3} \right)=0[/tex]

On solving above equation

We get,

[tex]x = 3[/tex]

Replace the value of [tex]x[/tex]  

[tex]\boxed{\frac{{{m_1}}}{{{m_2}}} = 3}[/tex]

Or,

[tex]\boxed{\frac{{{m_2}}}{{{m_1}}} = \frac{1}{3}}[/tex]  

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Answer details:

Grade: Senior School

Subject: Physics

Chapter: Impulse and Momentum

Keywords:

Head on collision, two particles, equal speed, ratio of particle masses, momentum, conservation of momentum, energy, conservation of energy, masses, ratio.

The answer is :D. None of the above

Agility and balance is very important in sport activities such as soccer, especially seen when players dribble the ball at high speed. Endurance and muscle control is very important in sport such as gymnastic or boxing. Hand-eye coordination and acuity is important in sports such as golf.

Answer:

the answer is d none of the above

Explanation:

The normal force is calculated by adding the weight of the suitcase and the vertical component of pulling force, while total resistance force is equal to the horizontal component of pulling force. Both forces play a significant role in Renee's effort to move the suitcase at a constant speed.

Here's how to find the normal force and the total resistance force for Renee's suitcase:

Remember, while calculating remember to convert the angle to radians if your calculator is set to radian mode.

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To find the normal force on Renee's suitcase, we resolve the pulling force into its components and subtract the vertical component from the suitcase's weight. As the suitcase is moving at a constant speed, the horizontal component equals the resistance force, which includes both friction and any air resistance.

Renee is pulling her 21-kg suitcase at a constant speed of 0.47 m/s through the airport. To determine the normal force acting on the suitcase, we need to consider the components of the pulling force. The force has a magnitude of 120 N and is exerted at an angle of 38° above the horizontal. We must resolve this force into vertical and horizontal components. The vertical component (Fy) helps support the weight of the suitcase and is calculated as Fy = 120 N × sin(38°). The weight of the suitcase is W = m × g, where m is the mass of the suitcase and g is the acceleration due to gravity (9.8 m/s²).

The normal force is given by N = W - Fy since the vertical component of the pulling force acts upwards, reducing the normal force exerted by the ground. As the suitcase is moving at a constant speed, the net horizontal force must be zero. Therefore, the horizontal component of the pulling force, which is Fx = 120 N × cos(38°), must be equal to the total resistance force (friction + air resistance).

The equations to find the normal force and resistance force are:

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It will take 4.12 s for the flowerpot to fall to the ground.

From the question given above, the following data were obtained:

Height (h) = 85 m

NOTE: Acceleration due to gravity (g) = 10 m/s²

The time taken for the flowerpot to fall to the ground can be obtained as follow:

85 = ½ × 10 × t²

85 = 5 × t²

Divide both side by 5

[tex]t^{2} = \frac{85}{5}\\\\t^{2} = 17[/tex]

Take the square root of both side

[tex]t = \sqrt{17}[/tex]

Therefore, it will take 4.12 s for the flowerpot to fall to the ground.

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The discus, after completing one revolution in 0.90 seconds starting from rest, will be released at a speed of approximately 6.98 rad/s.

This question is related to the concept of rotational motion in physics. As it is stated that the discus thrower takes 0.90s to complete one revolution, and the discus is starting from rest, the rotational speed or the angular velocity (ω) can be calculated using the formula ω = 2π/T, where T is the period of rotation which is the time to complete one revolution. Substituting the given values into the formula gives us ω = 2π/0.90 s which is approximately 6.98 rad/s.

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To determine the speed of a discus at release, given the time for one revolution is 0.90 seconds, we first find the angular velocity to be approximately 6.98 rad/s. Then, using the radius of the circle, we can calculate the linear speed. For a radius of 1 meter, the speed is approximately 6.98 m/s.

When calculating the speed of the discus at release, we first need to determine the angular velocity. Given that the thrower completes one revolution in 0.90 seconds, we can use the formula for angular velocity:

ω = 2π / T

where ω is the angular velocity and T is the period of one revolution. Substituting the given values, we get:

ω = 2π / 0.90 s ≈ ( 2 x 3.14 ) / 0.90s  ≈ 6.98 rad/s

Next, to find the linear speed at release, we use the relationship between linear speed (v), angular velocity (ω), and radius (r):

v = ωr

Assuming we know the radius of the circle in which the discus is being rotated, we can substitute r. If r is, for example, 1 meter, then:

v ≈ 6.98 rad/s * 1 m ≈ 6.98 m/s

Therefore, the speed of the discus at release would be approximately 6.98 meters per second.

Final answer:

The angle between the string and the horizontal is decreasing at a rate of -0.4 radians/second.

Explanation:

In order to solve this problem, we can use trigonometry and the chain rule from calculus. Let's denote the angle between the string and the horizontal as θ. We need to find the rate at which this angle is decreasing (dθ/dt) when 200 ft of string have been let out.

First, we can find the length of the string using the Pythagorean theorem: 100^2 + r^2 = (100+r)^2, where r is the length of the horizontal portion of the string. Solving this equation gives us r = 50 ft.

Next, we can differentiate the equation with respect to time using the chain rule: d/dt (100^2 + r^2) = d/dt ((100+r)^2). Simplifying the equation and solving for dθ/dt gives us dθ/dt = -100/(r+r^2/100).

Plugging in r = 50 ft, we can find dθ/dt = -0.4 radians/second.

The horizontal and vertical components of the tension at the given point in the swing are  281.6 N and 351 N respectively.

Given data:

The length of chain is, L = 2.5 m.

The magnitude of tension on each chain is, T = 450 N.

Distance above the lowest point is, d = 55 cm = 0.55 m.

In problem, first we need to obtain the angle of inclination made by string horizontally.

So, the angle inclined by the string with horizontal is given as,

[tex]cos \theta =\dfrac{L-d}{L}\\\\cos \theta =\dfrac{2.5-0.55}{2.5}\\\\\theta = cos^{-1}(\dfrac{1.95}{2.5})\\\\\theta=38.74^{\circ}[/tex]

Now, the horizontal component of tension force acting on the string is,

[tex]T_{H}=T \times cos \theta\\T_{H}=450 \times cos 38.74\\T_{H}=281.6 \;\rm N[/tex]

And, the vertical component of tension force acting on the string is,

[tex]T_{V}=T \times sin \theta\\T_{V}=450 \times sin 38.74\\T_{V}=351 \;\rm N[/tex]

Thus, the horizontal and vertical components of the tension at this point in the swing are 281.6 N and 351 N respectively.

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Answer:

-496.37 J

Explanation:

P(V2-V1) = 10(.5-.01)

10(.49) =4.9

L x ATM = 4.9 x 101.3= 496.37 J

External pressure means negative therefore its -496.37J

The work done in this scenario can be calculated by multiplying the change in volume, external pressure, and a conversion factor. In this case, the work done is 494.9 J.

The work done in this scenario can be calculated using the formula:
Work = change in volume * external pressure * conversion factor

Given:
Initial volume (V1) = 0.0100 l
Final volume (V2) = 0.500 l
External pressure = 10.00 atm

First, we need to find the change in volume:
Change in volume = V2 - V1 = 0.500 l - 0.0100 l = 0.490 l

Next, we can calculate the work done:
Work = change in volume * external pressure * conversion factor
= 0.490 l * 10.00 atm * 101.3 J/l atm
= 494.9 J

Therefore, the work done in joules is 494.9 J.

50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.

50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.

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The question is about the conservation of momentum. By calculating each player's momentum before the collision and combining them, we find the resultant momentum. The velocity after the collision can then be found.

This is an example of conservation of momentum, a fundamental concept in physics. Whenever objects interact and there is no external force, the total momentum of the system of objects is conserved.

In the scenario given, the linebacker and the halfback can be regarded as a closed system because the only significant forces are their mutual ones. Before the collision, we can calculate the momentum for each player: the linebacker's momentum is mass x velocity = 120 kg x 8.6 m/s = 1032 kg.m/s (north), and the halfback's momentum is 75 kg x 7.4 m/s = 555 kg.m/s (east).

Using the law of vector addition, we can combine them to find the resultant momentum. We then divide the resultant momentum by the total mass (120 kg + 75 kg) to find the velocity of the resulting 'player blob' immediately after the collision.

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