A marathon runner is running the race with a speed of 15 m/s. She completed the race within 15 minutes. What is the distance covered by the athlete if the speed is doubled?

There are 3.37 × \(10^{22}\) gold atoms per cubic centimeter in the silver-gold alloy.

The density of a binary alloy can be calculated using the following equation:

ρ = w1ρ1 + w2ρ2

where,

ρ = density of the alloy

w1 and w2 = weight fractions of the two components (in this case, gold and silver)

ρ1 and ρ2 = densities of the pure components.

We are given that the alloy contains 10% gold and 90% silver by weight, so we can calculate the weight fractions as:

\(w_{Au}\) = 0.10

\(w_{Ag}\) = 0.90

We are also given the densities of pure gold and silver as:

ρ_Au = 19.32 g/\(cm^{3}\)

ρ_Ag = 10.49 g/\(cm^{3}\)

Now we can substitute these values into the density equation to find the density of the alloy:

ρ = \(w_{Au}\)ρ_Au +\(w_{Ag}\)ρ_Ag

ρ = (0.10)(19.32 g/\(cm^{3}\)) + (0.90)(10.49 g/\(cm^{3}\))

ρ = 11.08 g/\(cm^{3}\)

Next, we need to calculate the number of gold atoms per cubic centimeter in the alloy.

To do this, we can use Avogadro's number and the atomic weights of gold and silver:

\(N_A\) = 6.022 × \(10^{23}\) atoms/mol

Aum = 196.97 g/mol

Agm = 107.87 g/mol

The number of gold atoms:

\(n_{Au}\) = (\(w_{Au}\)ρ/ Aum) × \(N_{A}\)

Substituting the values, we get:

\(n_{Au}\) = (0.10 × 11.08 g/\(cm^{3}\)/ 196.97 g/mol) × 6.022 × \(10^{23}\) atoms/mol

\(n_{Au}\) ≈ 3.37 × \(10^{22}\) atoms/\(cm^{3}\)

Therefore, there are approximately 3.37 × \(10^{22}\) gold atoms per cubic centimeter in the silver-gold alloy.

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The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

This observation that electromagnetic radiation behaves like particles is known as the photoelectric effect.

It provides evidence that electromagnetic radiation exhibits particle-like properties. When EMR with sufficient energy (above a certain threshold) interacts with a metal surface, it can cause the ejection of electrons from the metal.

This behavior indicates that EMR behaves as discrete packets of energy called photons, which transfer their energy to the electrons and cause their release. The photoelectric effect supports the particle nature of EMR and is a fundamental concept in the field of quantum mechanics.

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Answer: (d)

Explanation:

Given

Mass of object \(m=2\ kg\)

Speed of object \(u=5\ m/s\)

Mass of object at rest \(M=3\ kg\)

Suppose after collision, speed is v

conserving momentum

\(\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s\)

Initial kinetic energy

\(k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J\)

Final kinetic energy

\(k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J\)

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

The magnitude of the force applied by the man is equal to the weight of the piano (190 kg) multiplied by the cosine of the ramp angle (10.0 degrees). The magnitude of the force is thus equal to 1647.2 N.

Force is an external influence that will cause an object to move, accelerate, decelerate, remain in motion, or stop. Forces are measured in Newtons (N) and are represented by vectors, which have both a magnitude and a direction. There are different types of forces such as contact forces, non-contact forces, and forces in nature. Examples of contact forces are friction, tension and normal force. Examples of non-contact forces are gravity and magnetism.

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The energy lost to non conservative forces on the hill is

\(4,977.5 J - 34.2 J = 4,943.3 J.\)

The potential energy at the top of the hill is given by mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the height above the base.

Therefore, the potential energy at the top of the hill is \(7.5 kg x 9.8 m/s^2 x 6.5 m = 4,977.5 J\)

The kinetic energy at the bottom of the hill is given by\(1/2 mv^2\), where m is the mass of the object and v is the velocity.

Therefore, the kinetic energy at the bottom of the hill is\(1/2 x 7.5 kg x 9.2 m/s^2 = 34.2 J\)

The energy lost to non conservative forces is the difference between the initial potential energy and the final kinetic energy.

Therefore, the energy lost to non conservative forces on the hill is \(4,977.5 J - 34.2 J = 4,943.3 J.\)

The energy loss is due to the various non conservative forces like friction, air resistance, etc. that the otter has encountered while sliding down the hill.

It's worth noting that the energy lost to non conservative forces is not recoverable and the energy conserved can only be converted between different forms.

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The near point of a human eye is about a distance of 25 cm.

The closest distance that an object may be viewed clearly without straining is known as the near point of the eye.

This distance (the shortest at which a distinct image may be seen) is 25 cm for a typical human eye.

The closest point within the accommodation range of the eye at which an object may be positioned while still forming a focused picture on the retina is also referred to as the near point.

In order to focus on an item at the average near point distance, a person with hyperopia must have a near point that is further away than the typical near point for someone of their age.

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In air, sound travels at a computed speed of 340 m/s.

The speed of sound in air is influenced by a number of factors, including temperature, humidity, and atmospheric pressure. However, under normal atmospheric conditions—a temperature of 20 degrees Celsius or 68 degrees Fahrenheit, a relative humidity of 50%, and a pressure of one atmosphere—the speed of sound in dry air is around 343 metres per second (1,125 feet per second). This number is typically rounded to 340 m/s for convenience.

Speed is the pace at which a distance changes in relation to time. v = dx + dt, etc. Speed may also be described as a function of time and distance. Specifically, speed is equal to distance times time. Its symbol is v, and the SI unit for it is m/s. It has a scalar value. Speed is a measure of how far something can go in a given amount of time. Using the formula Speed = Distance Time, one may determine the dimension of speed. Distance is measured in units of [L1], while time is measured in units of [T1]. Dividing distance by time results in [L1] [T1] = [L1T1]. [L1T]1 is the dimension for speed.

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The mass of a substance is not directly related to latent heat. Instead, latent heat is a parameter that describes the amount of energy required or released during a phase shift of a substance.

Latent heat can be thought of as hidden energy that is supplied or extracted to change the state of a substance without changing its temperature or pressure.

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process—typically a first-order phase transition.

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The organ system is made to connect with the organ and it does the job for the organ by the connection.

What is organ system?

An organ is a group of tissues that work together for the overall function of the organ, and an organ i hain system is a group of organs are  that work together to the  perform a specific function. The human organism of consists of eleven organ system.

An organ system is a group of the organs that work together in the body to perform a  complex function, such as the  pumping blood or the  processing and utilizing the  nutrients. There are 11 of  major organ systems in the human body: The circulatory system. The lymphatic system. In The respiratory system.

Cell can adapt themselves by undergoing 5 different condition

-hyperplasia

-hypertrophy

-atrophy

-metaplsia

-dysplasia

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the missing word is clockwise moment. I hope this helps good luck

The interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

To determine the interval between the first and second echo, we need to consider the time it takes for sound to travel from the student to the first cliff, and then from the first cliff to the second cliff, and finally back to the student.

Let's break down the distances and calculate the time for each part of the journey:

Distance from the student to the first cliff: 330 meters

Time taken: t₁ = distance / speed = 330 m / 330 m/s = 1 second

Distance from the first cliff to the second cliff: 990 meters

Time taken: t₂ = distance / speed = 990 m / 330 m/s = 3 seconds

Distance from the second cliff back to the student: 990 meters

Time taken: t₃ = distance / speed = 990 m / 330 m/s = 3 seconds

Now, we can calculate the total interval between the first and second echo by adding up the individual times:

Interval between first and second echo = t₁ + t₂ + t₃ = 1 s + 3 s + 3 s = 7 seconds

Therefore, the interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

It's important to note that this calculation assumes a straight path for the sound waves and neglects factors such as air temperature and wind that can affect the speed of sound. Additionally, it assumes perfect reflection of sound waves off the cliffs, which may not be the case in real-world scenarios.

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Note the complete questions is:

A student stands 330m from a tall cliff which is 990m from another tall cliff. If the speed of sound between the cliffs is 330m/s.What is the interval between the first and second echo?

Answer: 30 cm.

The situation described is that of two sources of sound waves that are separated by some distance. The two waves interfere with each other constructively at some points and destructively at others. When they interfere constructively, the amplitude (and intensity) of the sound wave is greater than when they interfere destructively.

When the speakers are 40 cm apart, the waves that they produce are in phase at some points on the axis, leading to constructive interference and a maximum in the intensity of the sound. As the distance between the speakers is increased beyond 40 cm, the points of constructive interference move farther apart, and the intensity of the sound decreases. When the speakers are 50 cm apart, the waves that they produce are exactly out of phase at some points on the axis, leading to complete destructive interference and a minimum in the intensity of the sound.

If the separation between the speakers continues to increase, the points of constructive interference will move closer together again, and the intensity of the sound will increase. The separation between the speakers at which the intensity of the sound will again be a maximum can be found using the following equation:

d = λ/2 + nλ

where d is the separation between the speakers, λ is the wavelength of the sound wave, and n is an integer that represents the number of half-wavelengths between the speakers.

At the maximum, the separation is an even multiple of half the wavelength, so we can use the formula above with n = 1. The wavelength can be found from the distance between the speakers at the minimum, which is 50 cm, and the distance at the maximum, which is 40 cm:

λ = 2(d_max - d_min) = 20 cm

Substituting λ and n into the formula gives:

d = λ/2 + nλ = 10 cm + 20 cm = 30 cm

Therefore, the sound intensity will be a maximum again when the separation between the speakers is 30 cm.

The pressure exerted on the block on the ground in N/m² is 200N/m².

The pressure is the amount of force applied per unit area.

Given is a 5000 Newton block rests on the ground over 25 m² of area.

Pressure p = Force/Area

Put the values, we get

p = 5000 /25

p = 200 N/m²

Hence, pressure exerted on the block on the ground is 200 N/m².

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Answer:

I think that a flat fee amount is an amount that is charged or paid that does not change according to the amount of work done, or the number of times something is used

Explanation:

i could be wrong and if i am , i am truly sorry :(

The distance d between the golfer and point b where the ball first lands is 1814.75 ft.

Initial velocity = 140 ft/s

Launch angle θ = 25°

Slope angle Φ = 5°

Let d is the distance on the slope after which projectile lands

The horizontal distance covered by projectile

x = d cos θ

Vertical distance covered by projectile

y = d sin θ

Now Considering horizontal motion,

Initial horizontal velocity, vₓ = v cos 25°

Then position after time t,

d cos5 = 140* cos25 × t

d cos5 = (126.88 t)

d = (126.88 t)/(0.99)

d = 128.16 t   -----(1)

When we consider vertical motion then,

Initial vertical velocity v y = v sin 25°

v y = v sin θ t - 1/2* g t²

-d sin5 = 140 sin25 t - gt²/2

-d sin5 = 59.16 t - 4.9 t²

-d = (59.16 t - 4.9 t²)/0.08

-d = (739.5 t - 61.25 t²)

739.5 t - 61.25 t² + d = 0   -----(2)

Put (1) in (2),

739.5 t - 61.25 t² + 128.16 t = 0

867.66 t - 61.25 t² = 0

61.25 t² = 867.66 t

61.25 t = 867.66

t = 14.16 s, 0 s

Substituting t value in (1),

d = 128.16 t = 128.16* 14.16 = 1814.75 ft

The distance d between the golfer and point b where the ball first lands is  1814.75 ft.

The question is incomplete. The complete question is 'an angle of 25° with the horizontal. knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and point b where the ball first lands.'

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Jonathan needs to maintain a separation of 0.543 mm between the plates to get the desired charge, and a dielectric constant of 92.6 to achieve a separation of 5 mm with a dielectric.

(a) Using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can solve for the capacitance: C = Q/V =\((8.15 x 10^-9 C) / (50 V) = 1.63 x 10^-10 F.\)

Then, using the formula for capacitance of parallel plate capacitors: C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation distance, we can solve for the separation distance: d =\(_{3}OA/C = (8.85 x 10^-12 F/m) x (0.01 m^2) / (1.63 x 10^-10 F) = 0.543 mm.\)

(b) To find the dielectric constant, we can use the formula for capacitance of a parallel plate capacitor with a dielectric: C = εrε0A/d, where εr is the relative permittivity or dielectric constant of the material. Solving for εr, we get: εr = Cd / ε0A = \((1.63 x 10^-10 F)\) x (0.005 m) / \((8.85 x 10^-12 F/m)\) x \((0.01 m^2)\) = 92.6.

Therefore, Jonathan should use a dielectric with a relative permittivity of 92.6 to achieve a separation of 5 mm.

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Answer:another answer is ”a light will not transmit thro it”

Explanation:

Ape.x answers

Answer: Momentum of an object, p = 6 kg-m/s

Explanation: It is given that,

Mass of an object, m = 5 kg

Velocity of that object, v = 1.2 m/s

We need to find the momentum of an object. It is equal to the product of mass and its velocity. Mathematically, it is given by :

p = m v

p = 6 kg-m/s

So, the momentum of an object is 6 kg-m/s. Hence, this is the required solution.

1. The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon.

2. The bright part of the moon appears bright due to the reflection of sunlight.

Introduction: The purpose of the investigation was to study the relationship between the Sun, Earth, and Moon. By creating models of the three celestial bodies, we aimed to understand how their movements and positions influence the phases of the moon and eclipses.The bright part of the moon appears bright due to the reflection of sunlight. As sunlight hits the moon's surface, it bounces back and reflects into space. This reflected light is what we see as the bright part of the moon.

Experimental Methods: To create our model, we used a lamp to represent the Sun, a ball to represent the Earth, and a smaller ball to represent the Moon. We also used a ruler, tape, and a protractor to measure distances and angles.We created our model by placing the lamp at one end of a table, the Earth in the middle, and the Moon at the other end. We attached the Moon to a string and moved it around the Earth to simulate the Moon's orbit around the Earth.

Develop a Model: Our model consists of a lamp, a ball, and a smaller ball on a string. The lamp represents the Sun, the ball represents the Earth, and the smaller ball on the string represents the Moon. As the Moon moves around the Earth, it goes through phases, from a new moon to a full moon and back again.We used diagrams and pictures to label the components of our model and describe causal accounts for the phases of the moon and eclipses.

Use a Model: When the Moon is full, it is in a direct line with the Earth and the Sun. Using our model, we can predict that the Moon would be directly opposite the Sun in the sky during a full moon. A lunar eclipse does not occur every month when there is a full moon because the Moon's orbit around the Earth is tilted at an angle of about 5 degrees to the Earth's orbit around the Sun. Therefore, the Moon is not always in a direct line with the Earth and the Sun during a full moon, which is necessary for a lunar eclipse to occur.

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Answer:

It is 8km/hr.

Explanation:

I just took the test, 11km/hr is wrong.

Edg2020

The speed of the tennis ball is  \(8\ \dfrac {km}{h}\)  when Alana throws the ball behind at the speed of   \(11\ \dfrac{km}{h}\).

The speed will depend upon the point of reference of the Alana. As the frame of the reference changes the speed of the ball also changes.

If  Alana will throw the ball in the front direction, The speed will be the sum of Alana"s speed and the speed of the ball by which it is thrown.

\(19+11=30\ \dfrac{km}{h}\)

If Alana is just leaving  the ball without any effort  the speed

\(19+(-19)=0\ \dfrac{km}{h}\)  

If Alana throws the ball behind her at the speed of 19 km/h. The speed of the ball for a sidewalk observer,

\(19+(-11)=\ 8 \ \dfrac{km}{h}\)

Thus the speed of the tennis ball is  \(8\ \dfrac {km}{h}\)  when Alana throws the ball behind at the speed of   \(11\ \dfrac{km}{h}\).

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Answer:

Explanation:

The work done by an object can be found by using the formula

From the question we have

workdone = 4 × 2

We have the final answer as

Hope this helps you

Take the moment Allen sees the moose to be the origin.

First, convert his speed to m/s.

90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s

(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of

(25 m/s) • (0.85 s) = 21.25 m

(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of

x = 21.25 m + (25 m/s) t + 1/2 a t²

and has a speed v of

v = 25 m/s + a t

It takes him 3.5 s to come to a full stop. Use this to find the acceleration:

0 = 25 m/s + a (3.5 s)

a = - (25 m/s) / (3.5 s)

a ≈ - 7.1 m/s²

After 3.5 s, he will have traveled a total distance of

x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²

x = 152.5 m ≈ 150 m

(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be

a ≈ - 7.1 m/s²

Answer:

I think it's the Cytoplasm

A. The period of the motion is 4.18 s

B. The frequency of the oscillations

Hz is 0.2393 Hz

C. The first time the mass is at the position

x = 0 is 2.09 s

D. The first time the mass is at the site of maximum compression of the spring is 6.27s

a) The period of motion is given by the time it takes for the mass to complete one full oscillation. From the equation given, x = (8.8 cm)cos[2t/(4.18 s)], we can see that the function is in the form x = A cos(ωt + φ). The period of the motion T is given by the reciprocal of the angular frequency ω, which is T = 2π/ω.

In this case, the angular frequency is given by the coefficient of t in the argument of the cosine function, 2t/(4.18 s). So,

T = 2π/(2/(4.18 s)) = 4.18 s

b) The frequency of oscillations, f, is given by the reciprocal of the period, so

f = 1/T = 1/4.18 s^-1 = 0.2393 Hz

c) To find the first time the mass is at the position x = 0, we need to find the value of t when the cosine function is equal to 1.

x = (8.8 cm)cos[2t/(4.18 s)] = 8.8cm * 1 = 8.8cm

so,

cos[2t/(4.18 s)] = 1

t = (4.18 s)/2 = 2.09 s

d) To find the first time the mass is at the site of maximum compression of the spring, we need to find the value of t when the cosine function is equal to -1.

x = (8.8 cm)cos[2t/(4.18 s)] = 8.8cm * (-1) = -8.8cm

so,

cos[2t/(4.18 s)] = -1

t = (4.18 s) + (2.09 s) = 6.27s

It should be noted that the mass will be at the site of maximum compression at t = 6.27s and t = 2.09s + 4.18s = 6.27s + 4.18s = 10.45s and so on.

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Based on the given diagram, we can only reasonably conclude statement c, that the upper plate is positive and the lower plate is negative

What is Electric Field?

Electric field is a concept used in physics to describe the influence that an electric charge exerts on other charges in its vicinity. It is defined as the force experienced by a positive test charge placed in the vicinity of an electric charge, divided by the magnitude of the test charge

The upper plate is positive and the lower plate is negative: This statement is a common convention for charged parallel metal plates in a capacitor. The positive plate is usually considered as the plate with electric field lines originating from it (in this case, the upper plate), and the negative plate is usually considered as the plate with electric field lines terminating on it (in this case, the lower plate). So, this statement is likely to be true based on convention.

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Please find attached photograph for your answer.

The skid mark will be 25.21 m long.

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

Initial velocity (u) = 22.0 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –9.6 m/s²

The displacement can be obtained as follow:

0² = 22² + (2 × –9.6 × s)

0 = 484 – 19.2a

0 – 484 = –19.2s

–484 = –19.2s

\(s = \frac{-484}{-19.2} \\\\\)

Therefore, the skid mark is 25.21 m long

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Answer:

D) The outer planets are often called gas giants because they are composed almost entirely of gas and are much larger than any of the four inner planets.

Explanation:

Answer: its 19 m/sec

Explanation: gotta multiply the speed by 2 so 9.8 X 2 = 19