What type of force holds the atoms within a molecule together?
O intramolecular
O electromagnetic
O gravity
O nuclear

The rate at which the radius increases when the radius is 6 cm is approximately 0.056 cm/s.

To determine how fast the radius increases, we can use the given information about the volume of a cylinder and its rate of change. The volume of a cylinder is given by the formula v = πR²H, where R represents the radius and H represents the height.

Given that the radius is three times the height, we can express the height as H = R/3. Substituting this value into the volume equation, we have v = πR²(R/3). Simplifying further, the volume equation becomes v = (π/3)R³.

Now, we are given that the volume increases at a rate of 10 cm/s. By taking the derivative of the volume equation with respect to time, we can determine how the radius changes over time. The derivative, dv/dt, is equal to (π/3)(3R²)(dR/dt), where dR/dt represents the rate of change of the radius.

Simplifying the equation, we have dv/dt = πR²(dR/dt). Substituting the given values, we have 10 cm/s = π(6²)(dR/dt).

Solving for dR/dt, we find that the rate at which the radius increases when the radius is 6 cm is approximately 0.056 cm/s.

Calculus and related concepts to explore the relationships between variables and their rates of change. Understanding these mathematical principles is essential for analyzing dynamic systems and their behaviors.

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Answer:

e = 0.46 m

Explanation:

From the laws of friction, frictional force, F is proportional to normal reaction, R.

F₁ = μR

where μ is coefficient of friction; R = mg and g = 9.8 ms⁻²

Also, from Hooke's law, extension, e, in an elastic spring is proportional to applied force.

F₂ = Ke

where K is force constant of the spring

Since the box is just about to move, the coefficient of friction involved is static friction.

The force on the spring equals the frictional force experienced by the box the box; F₁ = F₂

Ke = μR

e = μR/K

where μ = 0.65; R = 18 kg * 9.8 ms⁻²; K = 250 N/m

e = (0.65 * 18 * 9.8)/250

e = 0.46 m

Answer:

v = 5 [m/s]

Explanation:

Kinetic energy is related to speed, that is, whenever we have kinetic energy we will have a body with speed moving.  This can be calculated using the following equation.

\(E_{k}=\frac{1}{2}*m*v^{2}\\\)

where:

m = mass = 2 [kg]

v = velocity [m/s]

Ek = kinetic energy = 25 [J]

Now replacing:

\(25 = \frac{1}{2}*2*v^{2} \\v=\sqrt{25} \\v = 5 [m/s]\)

The Millenium Falcon is chased by the Imperial Forces. The ship is moving at a speed of 0. 587 c. Han Solo is shooting at the imperial fighters with his newly installed proton cannon purchased at the MSU Surplus Store for $20. 00 plus 6. 00% tax. The cannon emits protons at a speed of 0. 831 c with respect to the ship.

With the use of Relativistic velocity addition formula  we will find the velocity of proton in the resting frame of the movie audience in terms of the speed of the light when the cannon is shot in the forward direction
Formula is given as
v = (u+v')/(1+u*v'/\(c^{2}\))
Where
v = velocity of the protons in the resting frame of the movie audience

u = velocity of the Millennium Falcon with respect to the audience

v' = velocity of the protons with respect to the Millennium Falcon

c = speed of light

By putting all the values we get
v = (0.587c + 0.831c) / (1 + 0.587c*0.831c/\(c^{2}\))

v = (1.418c) / (1 + 0.486)

v = 0.942c

Hence, the velocity of the protons in the resting frame of the movie audience is 0.942 times the speed of light.

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Answer:

Answer in explanation

Explanation:

What we want to do here is to basically put in words in the select pane that could actually beat complete the question.

The higher the distance between the effort and the function of a lever, the higher the force needed to complete the work.

The smaller the distance between the effort force and the fulcrum of a lever, the smaller the mechanical advantage , making the work easier

The Hayashi model is a theoretical model used to describe the temperature distribution and conditions in the protoplanetary disk. However, it does not provide explicit information about the gas-solid interaction regimes or the particle size boundaries.

To determine the particle radius on the boundaries between the different regimes, we need to consider the relevant laws and models related to gas-solid interactions.

Regime (1): Epstein Law

Regime (II): Stokes Law

Regime (III): Ideal Fluid

In the context of gas-solid interactions, these regimes represent different flow regimes based on the size of solid particles and the behavior of the gas surrounding them.

The Epstein Law (Regime 1) applies when the mean free path of gas molecules is greater than the particle radius, and individual gas molecules collide with the particle. In this regime, the gas behaves like individual particles.

Stokes Law (Regime II) applies when the particle size is large enough that gas molecules can no longer individually collide with the particle but instead adhere to its surface, causing a viscous drag. In this regime, the gas behaves like a viscous fluid.

The Ideal Fluid (Regime III) represents the limit where the particle size is large enough that the gas behaves like an ideal fluid, and the viscous drag becomes negligible.

To determine the particle radius on the boundaries between these regimes in the Hayashi model, more specific information about the model is needed. The Hayashi model is a theoretical model used to describe the temperature distribution and conditions in the protoplanetary disk. However, it does not provide explicit information about the gas-solid interaction regimes or the particle size boundaries.

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To find the angle the wheel will have turned through by the time it stops, we can use the following kinematic equation:

ω² = ω₀² + 2αθ

where:

ω = final angular velocity (0 rad/s, as the wheel stops)

ω₀ = initial angular velocity (18 rad/s)

α = angular acceleration (-1.0 rad/s², as the wheel is slowing down)

θ = angle turned

Substituting the known values into the equation, we can solve for θ:

0² = (18 rad/s)² + 2(-1.0 rad/s²)θ

0 = 324 rad²/s² - 2θ

2θ = 324 rad²/s²

θ = 162 rad²/s²

Therefore, the wheel will have turned through an angle of 162 radians by the time it stops.

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a. The force on object A can be determined using Coulomb's law, which states that the force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them. Since object B experiences a force of 0.39 N, the force on object A would be the same, as the forces are equal in magnitude and opposite in direction. Therefore, the force on object A is also 0.39 N.

b. We know that object A has twice the charge of object B. Let's denote the charge on object B as qB. Since object A has twice the charge, the charge on object A would be 2qB.

c. The initial acceleration of object A can be calculated using Newton's second law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. We know the force on object A is 0.39 N and its mass is 550 g (0.55 kg). Therefore, the initial acceleration of object A can be calculated as follows:

Force = Mass × Acceleration

0.39 N = 0.55 kg × Acceleration

Solving for acceleration:

Acceleration = 0.39 N / 0.55 kg

The initial acceleration of object A is approximately 0.709 m/s^2.

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Answer:

stimulus

Explanation:

To solve this problem, we can apply the principle of conservation of angular momentum.Since the final angular velocity is zero, the bar will not rotate after the collision. Therefore, the other ball will not rise after the collision, and it will remain at the same height.

The angular momentum (L) of an object can be calculated as the product of its moment of inertia (I) and angular velocity (ω):

L = I × ω

For the system consisting of the bar and the two balls, the initial angular momentum is zero, and after the collision, the angular momentum is given by:

L = I × ω

The moment of inertia (I) of the system is the sum of the moment of inertia of the bar and the moment of inertia of the two balls.

The moment of inertia of the bar (I\(_{bar}\)) about its center can be calculated as:

I\(_{bar}\) =  × m\(_{bar}\) × L²

where m\(_{bar}\) = 8.0 kg is the mass of the bar and L = 4.0 m is the length of the bar.

The moment of inertia of each ball (I\(_{ball}\)) about the pivot point can be calculated as:

I\(_{ball}\) = m\(_{ball}\) × R²

where m\(_{ball}\)= 5.0 kg is the mass of each ball, and R is the distance from the pivot point to the ball.

Since the balls are attached to the ends of the bar, the distance from the pivot point to each ball is half the length of the bar:

R = \(\frac{L}{2}\)= \(\frac{4}{2}\) = 2.0 m

Now, let's calculate the total moment of inertia :

I\(_{bar}\)= (1/12) × 8.0 kg × (4.0 m)²

= 8/3 kg·m²

I\(_{ball}\)= 5.0 kg × (2.0 m)²

= 20 kg·m²

I\(_{ball}\)= I\(_{bar}\) + 2 × I\(_{ball}\)

= 8/3 kg·m² + 2 * 20 kg·m²

= 8/3 kg·m² + 40 kg·m²

= 8/3 kg·m² + 120/3 kg·m²

= 128/3 kg·m²

After the collision, the system will rotate about the pivot point with an angular velocity (ω). The angular velocity can be calculated from the conservation of angular momentum equation:

L\(_{initial}\) = L\(_{final}\)

0 = I\(_{initial}\) × ω\(_{initial}\) + I\(_{ball}\) × ω\(_{final}\)

Since the initial angular velocity is zero, we can solve for the final angular velocity:

I\(_{ball}\) *ω\(_{final}\)= 0

Now, let's calculate the final angular velocity (ω\(_{final}\)):

ω\(_{final}\)= 0 / (I\(_{total}\) + I\(_{ball}\))

= 0 / (128/3 kg·m² + 20 kg·m²)

= 0 / (128/3 kg·m² + 60/3 kg·m²)

= 0 / (188/3 kg·m²)

= 0

Since the final angular velocity is zero, the bar will not rotate after the collision.

Therefore, the other ball will not rise after the collision, and it will remain at the same height.

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Answer:

Given

mass (m) =5kg

velocity (v) =6m/s

momentum (p) =?

Form

p=mv

5kgx6m/s

p=30kg.m/s

momentum =30kg.m/s

Answer:

(1)540 J

(2)67.5 J/s or 67.5W

(3)look at underline statement in explanation

Explanation:

(1) acceleration = change in speed ÷ time

= (9-5)m/s ÷ 8s

= 0.50 m/s²

resultant force of object during motion

= mass(in kg) × acceleration

= 6 × 0.50 = 3 N

displacement of force applied

= area under speed time graph(when sketched out, it's a trapezium)

= ½ × (5+9) × 8

= 180m

work done is measured in joules

= force × displacement of force applied

= 3N × 180m

= 540 J

detailed explanation:

altho the final unit for the work done should be Nm, but J is equal to Nm.

(2) Power is measured by J/s

therefore power = 540 J ÷ 8s = 67.5 J/s or 67.5 W

(3) the object accelerates constantly along a straight road

Answer:

Approximately \(3.01 \times 10^{-27}\; {\rm m}\) (when measured in a vacuum.)

Explanation:

Apply unit conversion and ensure that the mass of the baseball is in standard units (kilograms):

\(m = 145\; {\rm g} = 0.145\; {\rm kg}\).

The kinetic energy of the baseball will be:

\(\displaystyle E = \frac{1}{2}\, m\, v^{2}\),

Where \(v = 30.2\; {\rm m\cdot s^{-1}}\) is the speed of the baseball.

\(\begin{aligned}E &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, (0.145)\, (30.2)^{2}\; {\rm J} \\ &= 66.12290\; {\rm J}\end{aligned}\).

The energy of a photon of frequency \(f\) is:

\(E = h\, f\),

Where \(h \approx 6.62607 \times 10^{-34}\; {\rm m^{2}\cdot kg \cdot s^{-1}}\) is Planck's constant.

When measured in a vacuum where speed of light is \(c \approx 3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}\), the wavelength \(\lambda\) of this photon will be:

\(\displaystyle \lambda = \frac{c}{f}\).

\(\displaystyle f = \frac{c}{\lambda}\).

Hence, the expression for the energy of this photon can be rewritten as:

\(\displaystyle E = h\, f = \frac{h\, c}{\lambda}\).

Rearrange this equation to find \(\lambda\):

\(\displaystyle \lambda &= \frac{h\, c}{E}\).

Assuming that the energy of this photon to be equal to the kinetic energy of that baseball, \(66.12290\; {\rm J}\):

\(\begin{aligned}\lambda &= \frac{h\, c}{E} \\ &\approx \frac{(6.62607\times 10^{-34})\, (3.00 \times 10^{8})}{(66.12290)}\; {\rm m} \\ &\approx 3.01 \times 10^{-27}\; {\rm m}\end{aligned}\).

The work done by the force of 7 pounds acting at a 45-degree angle to the horizontal in moving the object 9 feet from (0,0) to (9,0) is 23 pounds-feet.

To find the work W done by a force of 7 pounds acting at a 45-degree angle to the horizontal in moving an object 9 feet from (0,0) to (9,0), we need to use the formula W = Fd cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and the displacement. In this case, F = 7 pounds, d = 9 feet, and theta = 45 degrees. To use the formula, we first need to convert theta to radians by multiplying it by pi/180. So, theta in radians is (45*pi)/180 = pi/4. Now, we can plug in the values into the formula: W = (7 pounds)(9 feet)cos(pi/4) = (7)(9)(sqrt(2)/2) = 22.5 pounds-feet. Since the question asks us to round to the nearest whole number, we need to round 22.5 to 23 pounds-feet. Therefore, the work done by the force of 7 pounds acting at a 45-degree angle to the horizontal in moving the object 9 feet from (0,0) to (9,0) is 23 pounds-feet.

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Answer:

6 m/s^2

Explanation:

F = ma

60N = (10 kg)a

a = 6m/s^2

The answer is yes because of conservation of momentum. That is, the momentum before collision is equal to the momentum after collision.

Momentum is the product of mass and velocity. It is a vector quantity. And it is measured in Kgm/s

If two vehicles have a head-on collision, the vehicles essentially stick together and travel a certain distance for 20 seconds before coming to a complete stop. The final velocity for both vehicles immediately after the crash can be obtain by using the formula

v = u - at

Where

If you are able to obtain the mass of both vehicles, the initial velocity for Vehicle A, and the final velocity for both vehicles immediately after the crash, then you can determine the momentum of both vehicles before the collision by calculating the momentum of both after the collision because momentum is always conserved.

Or by first calculating the initial velocity for vehicle B by using the formula below

\(M_{1}U_{1} - M_{2}U_{2} = (M_{1} + M_{2} )V\)

where

After we obtain initial velocity for vehicle B, we can calculate the momentum of both vehicles before the collision

Therefore, the momentum of both vehicles before the collision can be determined.

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The kinetic energy of everyday-sized objects and the potential energy that results from gravity are both forms of mechanical energy.

Kinetic energy and potential energy are the two types of mechanical energy.

The quantity of potential energy a thing possesses and the amount of kinetic energy it is capable of producing determine mechanical conversion.

Regardless of potential, creating power requires energy from motion, without which many energy-generating sources would not function.

Mechanical energy is the combination of stored (potential energy) and moving (kinetic energy), and it depends on the position and velocity of an object. In other words, mechanical energy is produced when an object's kinetic and potential energy are combined.

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Explanation:

the intersection behavior

The magnitude of the electric field at the point on the x-axis with x-coordinate a/2 is (η * r * a) / (4 * π * ϵ0 * a³), where η represents the linear charge density, r is the distance from the point to the rod, a is the length of the rod, and ϵ0 is the permittivity of free space.

To calculate the magnitude of the electric field at the specified point on the x-axis, we can use the formula for the electric field generated by a uniformly charged rod. The electric field at a point on the x-axis due to a uniformly charged rod can be expressed as E = (η * r) / (4 * π * ϵ0 * r³), where η is the linear charge density, r is the distance from the point to the rod, and ϵ0 is the permittivity of free space.

In this case, the point is located at x = a/2, which is half the distance of the rod. Since the rod is uniformly charged, the linear charge density η is constant throughout the rod. The distance from the point to the rod is r = a/2.

By substituting these values into the formula, we can determine the magnitude of the electric field:

E = (η * r) / (4 * π * ϵ0 * r³)

 = (η * a/2) / (4 * π * ϵ0 * (a/2)³)

 = (η * a) / (4 * π * ϵ0 * a³)

 = (η * r * a) / (4 * π * ϵ0 * a³)

Therefore, the magnitude of the electric field at the point on the x-axis with x-coordinate a/2 is (η * r * a) / (4 * π * ϵ0 * a³), where η represents the linear charge density, r is the distance from the point to the rod, a is the length of the rod, and ϵ0 is the permittivity of free space.

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Answer:

–40o

Explanation:

If x(n) is passed through an ideal d/a (digital-to-analog) converter, the reconstructed signal ya(t) will be an analog signal that approximates the original discrete-time signal x(n). The d/a converter converts the discrete-time samples into a continuous-time signal.

To reconstruct the signal ya(t), the d/a converter needs to perform the following steps:

1. Sample and Hold: The d/a converter first samples the discrete-time signal x(n) at regular intervals. This means that it takes snapshots of the signal at specific points in time.

2. Quantization: The sampled values are then quantized, which means they are approximated to a limited set of values. The number of bits used for quantization determines the resolution of the reconstructed signal.

3. Digital-to-Analog Conversion: The quantized values are converted into corresponding analog voltage levels. This process involves reconstructing a continuous-time signal that closely resembles the original waveform.

The reconstructed signal ya(t) is a continuous-time approximation of the original signal x(n). It is important to note that the accuracy of the reconstructed signal depends on the sampling rate and the resolution of the d/a converter. A higher sampling rate and resolution generally result in a more accurate reconstruction.

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Answer:

Speed is 0 m/s.

Distance is 200 meters.

Explanation:

In this scenario, we can treat speed as the absolute value of velocity and distance as absolute value of displacement since we are dealing with one-dimensional motion.

There’ll only exist one component of vector, by applying magnitude formula, it’ll return value as shown below:

Suppose we have a one-dimensional vector, \(\displaystyle{\vec v = 5\hat i}\), this represents a vector with magnitude of \(\displaystyle{|\vec v | = \sqrt{5^2} = 5}\).

Theorem I - Magnitude of One-Dimensional Vector

Suppose we have a vector \(\displaystyle{\vec v = a\hat i}\) then the magnitude is \(\displaystyle{|\vec v| = \sqrt {a^2} = |a|}\).

Speed and distance both are scalar quantity which means they only have magnitude but lack the direction. In one-dimensional motion, speed is defined to be the absolute value of velocity and direction is defined to be the absolute value of displacement, the reason is shown above regarding one-dimensional vector.

An example would be, consider a car moves at velocity of -4 m/s, this means that a car just moves to negative direction or opposite direction of positive at 4 m/s. The negative and positive sign refer to the direction that an object moves, to know how fast they are moving, we consider its magnitude which is speed.

This can also be said same to displacement and distance, but be aware that this can only be applied to straight horizontal or vertical line movement or else the motion will become two-dimensional.

From the problem, there’s no specific distance or displacement’s graph so we will assume that the car is moving in straight horizontal line. The car is moving at speed of 40 m/s and decelerates at 4 m/s².

This means that our acceleration is -4 m/s² since it’s decelerating and is moving at 40 m/s from start. Therefore, we can apply a general physic formula which is:

\(\displaystyle{v = u+at}\)

Where v is final velocity, u is initial velocity, a is acceleration and t is time. As said, we can treat speed = positive velocity in this scenario so substitute u = 40 m/s and a = -4 m/s² in the formula.

\(\displaystyle{v=40-4t}\)

Now substitute t = 10 in the formula:

\(\displaystyle{v = 40-4(10)}\\\\\displaystyle{v=40-40}\\\\\displaystyle{v=0 \ \sf{m/s}}\)

Therefore, the speed of car at 10 seconds is 0 m/s, it’s not moving at this moment.

Next, we find distance which we can use the following formula:

\(\displaystyle{s = ut + \dfrac{1}{2}at^2}\)

Substitute u = 40, t = 10 and a = -4 in:

\(\displaystyle{s=40(10)+\dfrac{1}{2}(-4)(10)^2}\\\\\displaystyle{s = 400-2(100)}\\\\\displaystyle{s=400-200}\\\\\displaystyle{s=200 \ \sf{m}}\)

Therefore, in 10 seconds, it’ll be able to travel 200 meters.

An electrically charged object can be used to attract any object with an opposite charge.

This is due to the fundamental principle that opposites attract and repel in physics.

Electric charge is a fundamental property of matter that gives rise to electromagnetic interactions. An electric charge, whether positive or negative, produces an electric field that surrounds it. This field exerts a force on any other charge in its vicinity that is either attracted to or repelled from it. Electric charge is a fundamental property of matter that produces a variety of electric phenomena. When the charge is concentrated in a localized region of space, the object is electrically charged. When there is a net accumulation of charge in an object, it becomes electrically charged. An electrically charged object produces an electric field in its vicinity, which exerts a force on other charged objects. An electrically charged object can be used to attract objects with an opposite charge or repel objects with the same charge.

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The actual volume of the cone is 160/3 cm3, and it increases from the center of the base to the tip. The form of a cone with a dotted line.

The amount of quantity that is gained by the solid or object when it is located in three-dimensional space is referred to as the volume of the solid.

The volume of the cone may be calculated using the formula below, which is as follows:

\(V=\frac{\pi r^2 h}{3}\)

The radius of the cone's base is denoted by the symbol (r), while the height of the cone is denoted by the symbol (h).

The picture of the cone, the volume of which has to be determined, is attached further down.

The height of the cone is 10 centimeters, while the radius measures 4 centimeters. Thus, enter these numbers into the formula shown above as,

\(& V=\frac{\pi(4)^2(10)}{3} \\& V=\frac{160}{3} \pi \mathrm{cm}^3\)

As this is the case, the precise volume of the cone in which the outline of a cone with a dotted line rises from the center of the base to the tip is 160/3 cm3.

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