consider the apparatus used during this experiment. while looking at the vacuum bulb from the front, the electron beam will be going to your right. use your knowledge of the right-hand rule and the fact that f

Answer:

Gamma rays and x rays

Explanation: just took the test

Gamma rays and X-rays have higher frequencies than the waves that make up ultraviolet light. Hence option D is correct.

Electromagnetic spectrum is nothing but the range of wavelengths or frequencies on which electromagnetic radiation extend. it is whole distribution of electromagnetic waves according to frequency or wavelength. Electromagnetic wave is nothing but a photon which carries energy. Photon wave has frequency, wavelength as well as amplitude.

Electromagnetic spectrum is show in figure.

There is inverse relation between wavelength and frequency as υ = c/λ

where υ is the frequency, λ is the wavelength and c is the speed of light.

as wavelength increases, frequency decreases vice verse.

In this figure we can see that radio waves has highest wavelength hence it has lowest frequency. X ray and gamma rays has lowest wavelength hence it has highest frequency.

Hence option D is correct.

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Answer:

you add

Explanation:

you should add the forces since they act in the same direction as that you'll resolve the two forces

The difference in recoil between firing a solid ball and a hollow ball from the same cannon would largely depend on the weight and size of the projectiles.

The basic principle behind recoil is that the force exerted on the projectile in one direction will be equal and opposite to the force exerted on the cannon in the opposite direction.

Assuming that the solid and hollow balls are of the same weight and size, the recoil should be relatively similar. However, if the hollow ball is larger than the solid ball, it will have a larger surface area and therefore experience greater air resistance as it travels through the barrel of the cannon.

This could result in a slightly greater recoil force as the cannon attempts to push the larger, more resistant projectile forward.

On the other hand, if the hollow ball is lighter than the solid ball, it may experience less friction and resistance as it travels through the barrel, resulting in a smaller recoil force. It is also possible that the hollow ball may experience more instability in flight due to its hollowness, which could affect the accuracy of the shot and potentially alter the recoil force as well.

Overall, while the difference in recoil between firing a solid versus a hollow ball from the same cannon may be minimal, factors such as weight, size, and surface area can all play a role in determining the amount of recoil experienced.

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The ratio of PE (potential energy) to DE (kinetic energy) is 1:2, indicating that for every unit of potential energy, there are two units of kinetic energy in the system.

Let's assume the center of the circle O is at the origin (0,0). We can assign coordinates to the points A, B, C, and D as follows:

A: (-a, a)

B: (a, a)

C: (a, -a)

D: (-a, -a)

O: (0, 0)

Since ABCD is an inscribed square, we know that AB and CD are perpendicular bisectors of each other. The midpoint of AB is P, so its coordinates are (0, a).

Now, let's find the coordinates of point E. PD is a diagonal of the square, so its slope is -1.

We can determine the equation of PD as y = -x + 2a. Since E lies on AB, its y-coordinate is 0.

By substituting y = 0 into the equation of PD, we get 0 = -x + 2a, which implies x = 2a.

Therefore, the coordinates of E are (2a, 0).

Finally, we can calculate the ratio PE/DE using the distance formula:

PE = √[(2a - 0)² + (0 - a)²] = √(4a² + a²) = √(5a²) = √5a

DE = √[(-a - 2a)² + (a - 0)²] = √(9a² + a²) = √(10a²) = √10a

The ratio PE/DE = (√5a) / (√10a) = √(5/10) = √(1/2) = 1/√2 = 1:√2 = 1:2.

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Attraction between people is Gravitation while people and earth is Gravity.

Every body in the universe attracts every other body with a force called force of gravitation. The attraction between sun and earth , attraction between table and chair lying in a room ..etc. are examples of gravitation. Gravitation is the weakest of four basic forces in nature. However , its most important force as it has its contribution in initiating the birth of stars and controlling the structure and evolution of entire universe.

Gravity is the special case of gravitation. If one of the attraction body is earth then gravitation is gravity.

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Answer:

star

Explanation:

because that is what our sun is.

Answer:

a is the answer

Explanation:

because when it slows downs

The temperature of the water after 30g of steam has condensed will be approximately 52.14°C.

To solve this problem, we need to consider the energy transfer that occurs when steam condenses into water. The energy released by the condensing steam will be absorbed by the water, resulting in a temperature change.First, let's calculate the energy released when 30g of steam condenses. The specific latent heat of steam is given as 2.26 × 10^6 J/kg, so the energy released by 30g of steam can be calculated as:

Energy released = (30g) × (2.26 × 10^6 J/kg) = 6.78 × 10^7 J

Next, we need to calculate the energy required to raise the temperature of the water from 20°C to the final temperature. The specific heat capacity of water is given as 4200 J/kgK, and the mass of the water is 500g. Therefore, the energy required can be calculated as:

Energy required = (500g) × (4200 J/kgK) × (final temperature - 20°C)

Since the energy released by the steam is equal to the energy required by the water, we can set up the equation:

6.78 × 10^7 J = (500g) × (4200 J/kgK) × (final temperature - 20°C)

Now, we can solve for the final temperature:

(final temperature - 20°C) = (6.78 × 10^7 J) / ((500g) × (4200 J/kgK))

(final temperature - 20°C) = 32.14°C

final temperature = 32.14°C + 20°C

final temperature ≈ 52.14°C

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A) Using Simpson's 1/3 rule (composite formula), the work done with a constant force of 200 N is approximately 1250 J.

B) Using the MATLAB function trapz, the work done is approximately 7750 J.

Let's substitute the given values into the Simpson's 1/3 rule formula and calculate the work done using a constant force of 200 N.

A) Force (F) = 200 N (constant for all t)

Velocity (v) = 4t (0 ≤ t ≤ 5) and v = 20 + (5 - t)² (5 ≤ t ≤ 15)

Step size (h) = 0.25

To find the work done using Simpson's 1/3 rule (composite formula), we need to evaluate the integrand at each interval and apply the formula.

Step 1: Divide the time interval [0, 15] into subintervals with a step size of h = 0.25, resulting in 61 equally spaced points: t0, t1, t2, ..., t60.

Step 2: Calculate the velocity at each point using the given expressions for different intervals [0, 5] and [5, 15].

For 0 ≤ t ≤ 5: v = 4t For 5 ≤ t ≤ 15: v = 20 + (5 - t)²

Step 3: Compute the force at each point as F = 200 N (since the force is constant for all t).

Step 4: Multiply the force and velocity at each point to get the integrand.

For 0 ≤ t ≤ 5: F * v = 200 * (4t) For 5 ≤ t ≤ 15: F * v = 200 * [20 + (5 - t)²]

Step 5: Apply Simpson's 1/3 rule formula to approximate the integral of the integrand over the interval [0, 15].

The Simpson's 1/3 rule formula is given by: Integral ≈ (h/3) * [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ... + 4f(xn-1) + f(xn)]

Here, h = 0.25, and n = 60 (since we have 61 equally spaced points, starting from 0).

Step 6: Multiply the result by the step size h to get the work done.

Work done: 1250 J

B) % Define the time intervals and step size

t = 0:0.25:15;

% Calculate the velocity based on the given expressions

v = zeros(size(t));

v(t <= 5) = 4 * t(t <= 5);

v(t >= 5) = 20 + (5 - t(t >= 5)).^2;

% Define the force value

F = 200;

% Calculate the work done using MATLAB's trapz function

\(work_t_r_a_p_z\) = trapz(t, F * v) * 0.25;

% Display the result

disp(['Work done using MATLAB''s trapz function: ' num2str(\(work_t_r_a_p_z\)) ' J']);

The final answer for the work done using MATLAB's trapz function with the given force and velocity is:

Work done using MATLAB's trapz function: 7750 J

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The energy required to accelerate a particle to speeds of 0.5c, 0.9c, and 0.99c can be calculated as 1.15E₀, 2.29E₀, and 7.08E₀, respectively, where E₀ represents the rest energy of the particle.

Einstein's mass-energy equivalence states that the total energy (E) of an object is equal to its mass (m) multiplied by the speed of light (c) squared, E = mc². To calculate the energy required to accelerate a particle to a specific speed, we need to consider the relativistic effects.

To express the energy in terms of the rest energy (E₀), we divide the total energy (E) by mc², resulting in E/E₀. Thus, for speeds of 0.5c, 0.9c, and 0.99c, we can calculate the energy required as follows:

a) For a speed of 0.5c:

  E/E₀ = (mc²)/(mc²) = 1E₀

  The energy required is equal to the rest energy.

b) For a speed of 0.9c:

  E/E₀ = γmc²/mc² = γ = 1/(1 - v²/c²)^(1/2)

  Here, v = 0.9c, so γ = 2.29

  The energy required is 2.29E₀.

c) For a speed of 0.99c:

  E/E₀ = γmc²/mc² = γ = 1/(1 - v²/c²)^(1/2)

  Here, v = 0.99c, so γ = 7.08

  The energy required is 7.08E₀.

Therefore, the energy required to accelerate a particle to speeds of 0.5c, 0.9c, and 0.99c is 1.15E₀, 2.29E₀, and 7.08E₀, respectively, where E₀ represents the rest energy of the particle.

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The altitude of the mesospheric clouds directly over the observer is approximately 756,000,000 meters, or 756,000 kilometers.

To determine the altitude of the mesospheric clouds directly over the observer, we can use the time it takes for sunlight to reach the observer and illuminate the clouds.

The speed of light is approximately 299,792 kilometers per second (km/s).

However, since the clouds are relatively close to Earth compared to the vast distances of astronomical objects, we can consider the speed of light to be approximately 300,000 kilometers per second for simplicity.

Time taken for light to reach the observer = 42 minutes = 42 × 60 seconds

Now, we can calculate the distance that light travels in this time:

Distance = Speed of light × Time taken for light to reach the observer

Distance = 300,000 km/s × (42 × 60 seconds)

Let's convert the distance to meters for consistency:

Distance ≈ 300,000,000 meters/second × (42 × 60 seconds)

Distance ≈ 756,000,000 meters

Thus, the altitude of the mesospheric clouds directly over the observer is approximately 756,000,000 meters, or 756,000 kilometers.

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Your question seems incomplete, the probable complete question is:

During the summers at high latitudes, ghostly, silver-blue clouds occasionally appear after sunset when common clouds are in Earth's shadow and are no longer visible. The ghostly clouds have been called noctilucent clouds (NLC), which means "luminous night clouds," but now are often called mesospheric clouds, after the mesosphere, the name of the atmosphere at the altitude of the clouds. These clouds were first seen in June 1885, after dust and water from the massive 1883 volcanic explosion of Krakatoa Island (near Java in the Southeast Pacific) reached the high altitudes in the Northern Hemisphere. In the low temperatures of the mesosphere, the water collected and froze on the volcanic dust (and perhaps on comet and meteor dust already present there) to form the particles that made up the first clouds. Since then, mesospheric clouds have generally increased in occurrence and brightness, probably because of the increased production of methane by industries, rice paddies, landfills, and livestock flatulence. The methane works its way into the upper atmosphere, undergoes chemical changes, and results in an increase of water molecules there, and also in bits of ice for the mesospheric clouds. If mesospheric clouds are spotted 42 min after sunset and then quickly dim, what is their altitude if they are directly over the observer?

The magnetic field strength required to scan the mass range between 16 and 300 for singly charged ions is 0.0398 T.

The magnetic field strength required to focus the ion at a particular mass-to-charge ratio is given by the equation:

B = (V × r) ÷ (B² × 2 × (mB ÷ q))

where V is the accelerating voltage, r is the radius of the magnetic sector, B is the magnetic field strength, m is the mass of the ion, and q is its charge.

Since we are dealing with singly charged ions, q = 1. We know the values of V₁ and B₁ for CH⁴⁺ ions. Therefore, we can use the above equation to find the radius r of the magnetic sector:

r = (V₁ × m) / (B₁² × 2 × q)

We can now use this value of r and the above equation to find the magnetic field strength B₂ required to scan the mass range between 16 and 300:

B₂ = The atomic mass of CH₄ is 16 u.

The ions with mass-to-charge ratio of 16 and 300 have masses of 16 u/q and 300 u/q, respectively.

For singly charged ions, we have

m ÷ q = mass ÷ charge = mass.

B₂ = √((V₁ × 16 u) ÷ (2 × r)) ÷ 1.00 + √((V₁ × 300 u) ÷ (2 × r)) ÷ 1.00

√((V₁ × m) ÷ (2 × q × r))

V₁ = 3.00 x 10³ V, B₁ = 0.126 T

Using the above equations, we can calculate the value of r:

r = (V₁ × m) / (B₁² × 2 × q)

= (3.00 x 10³ V × 16 u) / (0.126 T)² × 2 × 1

= 3.08 x 10⁻³ m

Substituting the values of r and V₁ in the equation for B:

B₂ = √((V₁ × 16 u) ÷ (2 × r)) ÷ 1.00 + √((V₁ × 300 u) ÷ (2 × r)) ÷ 1.00

B₂ = √((3.00 x 10³ V × 16 u) ÷ (2 × 3.08 x 10⁻³ m)) ÷ 1.00 + √((3.00 x 10³ V × 300 u) / (2 × 3.08 x 10⁻³ m)) ÷ 1.00

B₂ = 0.0398 T

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The complete question is:

When a magnetic sector mass spectrometer was operated with an initial accelerating voltage (V1) of 3.00 x 103 V, a magnetic field (B1) of 0.126 T was required to focus the CH4 + ion on the detector.

What magnetic field strength would be required to scan the mass range between 16 and 250 for singly charged ions if the accelerating voltage is held constant?

Answer:

(1) 1×10⁻⁴

Explanation:

From the question,

α = (ΔL/L)/(ΔT)............. Equation 1

Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.

Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C

Substitute these values into equation 1

α  = 1×10⁻⁴/10

α = 1×10⁻⁵ °C⁻¹ .

β = (ΔA/A)/ΔT................... Equation 2

Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.

make ΔA/A the subject of the equation

ΔA/A = β×ΔT.......................... Equation 3

But,

β = 2α.......................... Equation 4

Substitute equation 4 into equation 3

ΔA/A = 2α×ΔT................ Equation 5

Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹

Substitute into equation 5

ΔA/A = ( 2)×(1×10⁻⁵)×(5)

ΔA/A  = 10×10⁻⁵

ΔA/A  = 1×10⁻⁴

Hence the right option is (1) 1×10⁻⁴

Answer:

Workdone = 17482.36 Joules

Explanation:

Given the following data;

Force, F = 120 N

Angle, d = 28.0°

Distance, x = 165 m

To find the work done, we would use the following formula;

Workdone = FxCosd

Substituting into the formula, we have;

Workdone = 120*165*Cos(28)

Workdone = 19800 * 0.8830

Workdone = 17482.36 Joules

Answer:

don't know sorry for the irreverent answer..

Answer:

The index of refraction varies with frequency, it doesn't change as light travels from one medium to another, As violet colour has the shortest wavelength and so the refractive index is maximum for it.

She experiences a constant acceleration of 9.8 m/s² (the acceleration due to gravity) at the beginning of her fall.

The acceleration due to gravity is the acceleration that an object experiences when it is in free fall under the influence of gravity.

Here,
As mentioned in the question, At the beginning of her fall, the skydiver has an acceleration at the beginning of her fall. When she first jumps out of the airplane, she is initially at rest and begins to accelerate downwards due to the force of gravity acting on her.

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In the case of the skydiver, the force of gravity is the net force acting on her, and her mass is constant.

Therefore, she experiences a constant acceleration of 9.8 m/s² (the acceleration due to gravity) at the beginning of her fall.

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It requires 9600 joules of work to get the wheelbarrow across the yard in 12 seconds. Rounded to the nearest whole number, the answer is 9600 J.

Power is the rate at which work is done or energy is transferred, and it is measured in watts (W). Work is the amount of energy required to move an object, and it is measured in joules (J). Time is the duration over which work is done, and it is measured in seconds (s). Power is calculated by dividing the amount of work done by the time taken to do it, or by multiplying force by velocity. Mathematically, we can represent the relationship between power, work, and time as: Power = Work / Time

We can use the formula: Work = Power x Time

To find the work required, we can plug in the given values:

Work = 800 W x 12 s

Work = 9600 J

Therefore, it requires 9600 joules of work to get the wheelbarrow across the yard in 12 seconds. Rounded to the nearest whole number, the answer is 9600 J.

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Answer:

Burrowing animals can cause weathering. By digging for food or making space to live in the animal may break apart rock. The growing roots of a tree can also break apart rock.

Answer:

i) Power = Force * Velocity = 1100 * 15 = 16500 W = 16.5 kW(ii)  Find the value of k first: F = 400 + k(15^2)                                              k = 28/9    F = 400 +(28/9)(30^2) = 320

Explanation:

According to the metric system, 1 metric ton (also known as a tonne) = 1,000,000 grams.  In the United States and some other countries, a ton is often used to refer to a unit of weight.

The metric system is a system of measurement used in most of the world that is based on the International System of Units (SI). The SI unit for mass is the kilogram (kg), which is defined as the mass of a specific cylinder of platinum-iridium alloy kept at the International Bureau of Weights and Measures in France.

The metric ton, also known as the tonne, is a unit of bin the metric system that is equal to 1,000 kilograms. This unit is commonly used to measure large masses of objects such as vehicles, cargo, and building materials.

Since 1 kilogram is equal to 1,000 grams, 1 metric ton is equal to 1,000 x 1,000 = 1,000,000 grams. This means that if you have a mass of 1,000,000 grams, you have a mass of 1 metric ton. Similarly, if you have a mass of 2,000,000 grams, you have a mass of 2 metric tons, and so on.

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Explanation:

weight = m g

weight = 1 × 10

weight = 10 N

The appearance of a sight glass filled with vapor and liquid is different, and they can be distinguished based on their transparency or opacity. false

A sight glass that is full of vapor or liquid does not look the same.
In a sight glass, which is a transparent window or tube used to visually inspect the contents of a system, the appearance will vary depending on whether it is filled with vapor or liquid.
When the sight glass is filled with vapor, it will appear as a transparent or translucent gas. The vapor may be less dense and may not fill the entire sight glass, allowing visibility through it.
On the other hand, when the sight glass is filled with liquid, it will appear as a continuous, opaque fluid. The liquid will block visibility through the sight glass, and its level or presence can be clearly observed.
Therefore, the appearance of a sight glass filled with vapor and liquid is different, and they can be distinguished based on their transparency or opacity.

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Using the concepts of fluid,we got that    5.12 m/sec is the fluid speed at a location where the diameter has narrowed to 4 cm for an incompressible fluid flows steadily through a pipe that has a change in diameter.

The volumetric flow rate for an incompressible fluid through a pipe is constant, so we can write:

A₁ .v₁ =A₂ .v₂---------------------(1)

(1)

where

A₁ is the cross-sectional area of the first part of the pipe

A₂ is the cross-sectional area of the second part of the pipe

v₁ is the speed of the fluid in the first part of the pipe

v₂ is the speed of the fluid in the second part of the pipe

Here we have:

v₁=1.28m/sec

r₁=8.0/2=4cm is the radius in the first part of the pipe, so the area is

  A₁=πr₁²=(3.14×4×4)=(3.14×16)=(50.24cm²)

Similarly, r₂ =(4.0/2)=2cm is the radius in the second part of the pipe, so the area is=A₂=πr₂²=(3.14×2×2)=(3.14×4)=(12.56)cm²

Using eq.(1), we find the fluid speed at the second location:

=>v₂=(A₁×v₁)/(A₂)

=>v₂=(50.24×1.28×100)/12.56

=>v₂=(50.24×128)/12.56

=>v₂=(6430.72)/12.56

=>v₂=512cm/sec

=>v₂=5.12m/sec

Hence, an incompressible fluid flows steadily through a pipe that has a change in diameter. the fluid speed at the location where the pipe diameter is 8 cm is 1.28 m/s,the fluid speed at a location where the diameter has narrowed to 4 cm is 5.12m/sec.

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The sound wave with more energy is the sound wave that is louder because it will have more amplitude.

Sound energy occurs when a force, either sound or pressure, makes an object or substance vibrate.

That energy moves through the substance in waves.

The energy transmitted by a sound wave depends on several factors such as;

Mathematically, the formula for energy transmitted by a sound wave is given as;

P = ¹/₂μω²A²v

where;

Thus, a sound wave more amplitude (loudness) will have more energy that a sound wave with lesser amplitude (loudness).

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Based on the motion map, the object first moved to the right, then it moved to the left and finally, it lost movement.

The motion map shows how an object moves, this is possible as it shows with arrows the movement of the object studied. In this case, we know about this object that;

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Q₁ and U₁ when switch S connects V₀ and C₁ is 12.0 V.

New charge and potential across and energy stored on all three capacitors is 32.7 mJ.

a) When switch S connects V₀ and C₁, the charge Q₁ on capacitor C₁ is given by:

Q₁ = C₁ V₀

Substituting the given values:

Q₁ = (4.00 μF) (12.0 V) = 48.0 μC

The potential difference U₁ across capacitor C₁ is given by:

U₁ = Q₁ / C₁

Substituting the value of Q₁ and C₁:

U₁ = (48.0 μC) / (4.00 μF) = 12.0 V

Therefore, the charge Q₁ on capacitor C₁ is 48.0 μC and the potential difference U₁ across capacitor C1 is 12.0 V.

b) When switch S is changed to connect the three capacitors, the equivalent capacitance C of the circuit is given by:

1/C = 1/C₁ + 1/C₂ + 1/C₃

Substituting the given values:

1/C = 1/4.00 μF + 1/6.00 μF + 1/3.00 μF = 0.575 μF

Therefore, the equivalent capacitance of the circuit is:

C = 1 / 0.575 μF = 1.74 μF

The charge Q on the equivalent capacitor is conserved:

Q = Q1 = 48.0 μC

The potential difference U across the equivalent capacitor is given by:

U = Q / C

Substituting the values of Q and C:

U = (48.0 μC) / (1.74 μF) = 27.6 V

The energy stored on each capacitor is given by:

E = (1/2) C V²

where C is the capacitance and V is the potential difference across the capacitor.

For capacitor C₁:

E₁ = (1/2) C1 U²

Substituting the value of C₁ and U:

E₁ = (1/2) (4.00 μF) (27.6 V)² = 8.57 mJ

For capacitor C₂:

E₂ = (1/2) C₂ U²

Substituting the value of C₂ and U:

E₂= (1/2) (6.00 μF) (27.6 V)² = 19.3 mJ

For capacitor C₃:

E₃ = (1/2) C₃ U²

Substituting the value of C₃ and U:

E₃ = (1/2) (3.00 μF) (27.6 V)² = 4.82 mJ

Therefore, the total energy stored on all three capacitors is:

E_total = E₁ + E₂ + E₃ = 8.57 mJ + 19.3 mJ + 4.82 mJ = 32.7 mJ

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Answer:

a.ionic bond is kinds of bonds make up a formula unit of sodium chloride.

Answer:

A

Explanation:

Answer:

John applied a force of approximately \(795\; {\rm N}\) (on average, rounded) on Lucy.

John slows down to a stop after approximately another \(5.37\; {\rm s}\).

(Assuming that \(g = 9.81\; {\rm N\cdot kg^{-1}}\).)

Explanation:

Assuming that the surface is level. The normal force on Johnny will be equal to the weight of Johnny: \(N(\text{John}) = m(\text{John})\, g\). Similarly, the normal force on Lucy will be equal to weight \(N(\text{Lucy}) = m(\text{Lucy})\, g\).

Multiply normal force by the coefficient of kinetic friction to find the friction on each person:

\(f(\text{John}) = \mu_{k}\, N(\text{John}) = \mu_{k}\, m(\text{John})\, g\).

\(f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g\).

Again, because the surface is level, the net force on each person after the first \(1.0\; {\rm s}\) will be equal to the friction. Divide that the net force on each person by the mass of that person to find acceleration:
\(\displaystyle a(\text{John}) = \frac{\mu_{k}\, m(\text{John})\, g}{m(\text{John})} = \mu_{k}\, g\).

\(\displaystyle a(\text{Lucy}) = \frac{\mu_{k}\, m(\text{Lucy})\, g}{m(\text{Lucy})} = \mu_{k}\, g\).

(Note that the magnitude of acceleration is independent of mass and is the same for both John and Lucy.)

\(a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}\).

In other words, after the first \(1\; {\rm s}\), both John and Lucy will slow down at a rate of \(1.962\; {\rm m\cdot s^{-2}}\).

To find the speed of Lucy immediately after the first \(1.0\: {\rm s}\), multiply this acceleration by the time \(t = 8.0\; {\rm s}\) it took for Lucy to slow down to \(0\; {\rm m\cdot s^{-1}}\):

\(\begin{aligned}& (8.0\; {\rm s})\, (1.962\; {\rm m\cdot s^{-2}}) \\ =\; & (8.0)\, (1.962)\; {\rm m\cdot s^{-1}} \\ =\; & 15.696\; {\rm m\cdot s^{-1}}\end{aligned}\).

Thus, in the first \(1.0\; {\rm s}\), Lucy accelerated (from \(0\; {\rm m\cdot s^{-1}}\)) to \(15.696\; {\rm m\cdot s^{-1}}\).

The average acceleration of Lucy in the first \(1.0\; {\rm s}\) would be \((15.696) / (1) = 15.696\; {\rm m\cdot s^{-2}}\). Multiply this average acceleration by the mass of Lucy to find the average net force on Lucy during that \(1.0\; {\rm s}\):

\(\begin{aligned}F_{\text{net}}(\text{Lucy}) &= m(\text{Lucy})\, a \\ &= (45)\, (15.696)\; {\rm N} \\ &= 706.320\; {\rm N}\end{aligned}\).

This net force on Lucy during that \(1.0\; {\rm s}\) is the combined result of both the push from Johnny and friction:

\(F_{\text{net}}(\text{Lucy}) = F(\text{push}) - f(\text{Lucy})\).

Since \(f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g\):

\(\begin{aligned}F(\text{push}) &= F_{\text{net}}(\text{Lucy}) + f(\text{Lucy}) \\ &= F_{\text{net}}(\text{Lucy}) + \mu_{k}\, m(\text{Lucy})\, g \\ &= (706.320) \; {\rm N}+ (0.2)\, (45)\, (9.81)\; {\rm N} \\ &= 706.320\; {\rm N} + 88.290\; {\rm N} \\ &=794.610\; {\rm N}\end{aligned}\).

In other words, Johnny would have applied a force of \(794.610\; {\rm N}\) on Lucy.

By Newton's Laws of Motion, when Johnny exerts this force on Lucy in that \(1.0\; {\rm s}\), Lucy would exert a reaction force on Johnny of the same magnitude: \(794.610\; {\rm N}\).

Similar to Lucy, the net force on Johnny during that \(1.0\; {\rm s}\) will be the combined effect of the push \(F(\text{push})\) and friction \(f(\text{John}) = \mu_{k}\, m(\text{John})\, g\):

\(\begin{aligned}F_{\text{net}}(\text{John}) &= F(\text{push}) - f(\text{John}) \\ &= F(\text{push}) - \mu_{k}\, m(\text{John})\, g\\ &= 794.610\; {\rm N} - (0.2)\, (65)\, (9.81)\; {\rm N} \\ &= 667.080\; {\rm N}\end{aligned}\).

Divide net force by mass to find acceleration:

\(\begin{aligned}\frac{667.080\; {\rm N}}{65\; {\rm kg}} \approx 10.2628\; {\rm m\cdot s^{-2}}\end{aligned}\).

In other words, Johnny accelerated at a rate of approximately \(10.5406\; {\rm m\cdot s^{-2}}\) during that \(1.0\; {\rm s}\). Assuming that Johnny was initially not moving, the velocity of Johnny right after that \(1.0\; {\rm s}\!\) would be:

\((0\; {\rm m\cdot s^{-1}}) + (10.2628\; {\rm m\cdot s^{-2}})\, (1.0\; {\rm s}) = 10.2628\; {\rm m\cdot s^{-1}}\).

After the first \(1.0\; {\rm s}\), the acceleration of both John and Lucy (as a result of friction) would both be equal to \(a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}\). Divide initial velocity of Johnny by this acceleration to find the time it took for Johnny to slow down to a stop:

\(\displaystyle \frac{10.2628\; {\rm {m\cdot s^{-1}}}}{1.962\; {\rm m\cdot s^{-2}}} \approx 5.23\; {\rm s}\).

Answer:

John applied a force of approximately \(795\; {\rm N}\) (on average, rounded) on Lucy.

John slows down to a stop after approximately another \(5.37\; {\rm s}\).

(Assuming that \(g = 9.81\; {\rm N\cdot kg^{-1}}\).)

Explanation:

Assuming that the surface is level. The normal force on Johnny will be equal to the weight of Johnny: \(N(\text{John}) = m(\text{John})\, g\). Similarly, the normal force on Lucy will be equal to weight \(N(\text{Lucy}) = m(\text{Lucy})\, g\).

Multiply normal force by the coefficient of kinetic friction to find the friction on each person:

\(f(\text{John}) = \mu_{k}\, N(\text{John}) = \mu_{k}\, m(\text{John})\, g\).

\(f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g\).

Again, because the surface is level, the net force on each person after the first \(1.0\; {\rm s}\) will be equal to the friction. Divide that the net force on each person by the mass of that person to find acceleration:
\(\displaystyle a(\text{John}) = \frac{\mu_{k}\, m(\text{John})\, g}{m(\text{John})} = \mu_{k}\, g\).

\(\displaystyle a(\text{Lucy}) = \frac{\mu_{k}\, m(\text{Lucy})\, g}{m(\text{Lucy})} = \mu_{k}\, g\).

(Note that the magnitude of acceleration is independent of mass and is the same for both John and Lucy.)

\(a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}\).

In other words, after the first \(1\; {\rm s}\), both John and Lucy will slow down at a rate of \(1.962\; {\rm m\cdot s^{-2}}\).

To find the speed of Lucy immediately after the first \(1.0\: {\rm s}\), multiply this acceleration by the time \(t = 8.0\; {\rm s}\) it took for Lucy to slow down to \(0\; {\rm m\cdot s^{-1}}\):

\(\begin{aligned}& (8.0\; {\rm s})\, (1.962\; {\rm m\cdot s^{-2}}) \\ =\; & (8.0)\, (1.962)\; {\rm m\cdot s^{-1}} \\ =\; & 15.696\; {\rm m\cdot s^{-1}}\end{aligned}\).

Thus, in the first \(1.0\; {\rm s}\), Lucy accelerated (from \(0\; {\rm m\cdot s^{-1}}\)) to \(15.696\; {\rm m\cdot s^{-1}}\).

The average acceleration of Lucy in the first \(1.0\; {\rm s}\) would be \((15.696) / (1) = 15.696\; {\rm m\cdot s^{-2}}\). Multiply this average acceleration by the mass of Lucy to find the average net force on Lucy during that \(1.0\; {\rm s}\):

\(\begin{aligned}F_{\text{net}}(\text{Lucy}) &= m(\text{Lucy})\, a \\ &= (45)\, (15.696)\; {\rm N} \\ &= 706.320\; {\rm N}\end{aligned}\).

This net force on Lucy during that \(1.0\; {\rm s}\) is the combined result of both the push from Johnny and friction:

\(F_{\text{net}}(\text{Lucy}) = F(\text{push}) - f(\text{Lucy})\).

Since \(f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g\):

\(\begin{aligned}F(\text{push}) &= F_{\text{net}}(\text{Lucy}) + f(\text{Lucy}) \\ &= F_{\text{net}}(\text{Lucy}) + \mu_{k}\, m(\text{Lucy})\, g \\ &= (706.320) \; {\rm N}+ (0.2)\, (45)\, (9.81)\; {\rm N} \\ &= 706.320\; {\rm N} + 88.290\; {\rm N} \\ &=794.610\; {\rm N}\end{aligned}\).

In other words, Johnny would have applied a force of \(794.610\; {\rm N}\) on Lucy.

By Newton's Laws of Motion, when Johnny exerts this force on Lucy in that \(1.0\; {\rm s}\), Lucy would exert a reaction force on Johnny of the same magnitude: \(794.610\; {\rm N}\).

Similar to Lucy, the net force on Johnny during that \(1.0\; {\rm s}\) will be the combined effect of the push \(F(\text{push})\) and friction \(f(\text{John}) = \mu_{k}\, m(\text{John})\, g\):

\(\begin{aligned}F_{\text{net}}(\text{John}) &= F(\text{push}) - f(\text{John}) \\ &= F(\text{push}) - \mu_{k}\, m(\text{John})\, g\\ &= 794.610\; {\rm N} - (0.2)\, (65)\, (9.81)\; {\rm N} \\ &= 667.080\; {\rm N}\end{aligned}\).

Divide net force by mass to find acceleration:

\(\begin{aligned}\frac{667.080\; {\rm N}}{65\; {\rm kg}} \approx 10.2628\; {\rm m\cdot s^{-2}}\end{aligned}\).

In other words, Johnny accelerated at a rate of approximately \(10.5406\; {\rm m\cdot s^{-2}}\) during that \(1.0\; {\rm s}\). Assuming that Johnny was initially not moving, the velocity of Johnny right after that \(1.0\; {\rm s}\!\) would be:

\((0\; {\rm m\cdot s^{-1}}) + (10.2628\; {\rm m\cdot s^{-2}})\, (1.0\; {\rm s}) = 10.2628\; {\rm m\cdot s^{-1}}\).

After the first \(1.0\; {\rm s}\), the acceleration of both John and Lucy (as a result of friction) would both be equal to \(a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}\). Divide initial velocity of Johnny by this acceleration to find the time it took for Johnny to slow down to a stop:

\(\displaystyle \frac{10.2628\; {\rm {m\cdot s^{-1}}}}{1.962\; {\rm m\cdot s^{-2}}} \approx 5.23\; {\rm s}\).

Balancing the forces acting on a body is not enough to establish equilibrium because equilibrium also requires the balancing of torques or moments acting on the body.

In physics, equilibrium refers to a state in which an object or system experiences no net force and no net torque. For an object to be in equilibrium, both the forces and the torques acting on it must be balanced.

Balancing the forces means that the vector sum of all the forces acting on the body is equal to zero. This ensures that there is no net force acting on the object, and it will not accelerate in any direction. However, even if the forces are balanced, the object can still rotate or have a tendency to rotate if the torques acting on it are not balanced.

A torque, also known as a moment, is a measure of the tendency of a force to rotate an object about a specific axis. It depends on the magnitude of the force, the distance from the axis of rotation, and the angle between the force and the lever arm. When torques are balanced, the sum of all the torques acting on the object is equal to zero.

To establish equilibrium, both the forces and the torques acting on the body must be balanced. This means that not only should the vector sum of the forces be zero, but also the algebraic sum of the torques should be zero. When both conditions are met, the object will remain at rest or continue to move with a constant rotational motion.

Balancing the forces acting on a body is not enough to establish equilibrium because equilibrium requires the balancing of both forces and torques. Simply balancing the forces ensures that there is no net force acting on the object, but it does not guarantee that the object will be in a state of complete equilibrium. To achieve equilibrium, the torques acting on the object must also be balanced, ensuring that there is no tendency for rotation.

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