explain the relationship among visible light, the electromagnetic spectrum, and sight.

sorry, it's B, not a made a mistake in the approximation

Work done by gravity when it moves from A to C will be 288.6 joule

Applying conservation of energy between point A and B,

7.4xgx5.9 = 1/2x7.4xv2 + 7.4xgx3.2

873.2 = 7.4v2 + 473.6

v = 7.34 m/s

It should be noted that to find velocity at C, which has come down to a height of 2 meter, it means it has travelled by 5.9-2 =3.9 m

so, v2 = 2gh = 20x3.9 =

v = 8.83 m/s

Work done by gravity when it moves from A to C = mxgx(5.9-2) = 288.6 joule

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Answer:

Net force = 419.5N

Explanation:

Given the following data;

Mass = 50kg

Radius = 15m

Time = 2.1 secs

Turns = 1/4 = 0.25

In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.

To find speed;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

\(Speed = \frac{distance}{time}\)

But the distance traveled is given by the circumference of a circle = \( 2\pi r\)

Since he covered 1/4th of a turn;

\( Distance = 0.25 * 2 \pi *r\)

Substituting into the equation;

\( Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}\)

\( Speed, v = \frac {23.565}{2.1}\)

Speed, v = 11.22m/s

To find acceleration;

\( Acceleration, a = \frac {v^{2}}{r}\)

Where, v = 11.22m/s and r = 15m

Substituting into the equation, we have;

\( Acceleration, a = \frac {11.22^{2}}{15}\)

\( Acceleration, a = \frac {125.8884}{15}\)

Acceleration, a = 8.39m/s²

To find the net force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

\( F = ma\)

Where;

Substituting into the equation, we have;

\( F = 50 * 8.39\)

F = 419.5N

Therefore, the net force acting upon the halfback is 419.5 Newton.

Answer:

The answer is 0.042 Hz (rounded)

or 0.0417 Hz

Explanation:

Hope it helps

Answer:

The answer is 0.041Hz

Explanation:

The answer is 0.041Hz

I guess

The y-intercept of a graph is the point at which the graph crosses the y-axis. In this case, the y-intercept of the graph showing the relationship between the motor's force and the boat's acceleration would represent the force required to produce zero acceleration.

In this scenario of the boats being used in an experiment, if the boats have the same mass and the motors are producing force, and the forces are graphed against the acceleration, this will give us a straight line, and the y-intercept of this graph will be the force required to keep the boat at a zero acceleration, this force is known as "friction force" which includes the friction force of the water and the friction force of the bearings in the motor and other mechanical parts in the boats and motor.

It's worth noting that, in practice, getting the y-intercept in this experiment might be very hard as it's very rare to get zero acceleration. Still, one can aim for the lowest possible acceleration, and the force measured can be considered as an approximation for the friction force.

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(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

∑ F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.112 m

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

\( E_{i} = E_{f} \)

\( \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} \)  (1)

Where:

\(m_{o}\): is the mass of the object = 0.250 kg

\(v_{o}\): is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

\( x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m \)

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

\( W = \Delta E = E_{f} - E_{i} \)

\( W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} \)   (2)

Work is also equal to:

\( W = F*d = F*x \)     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

\( F = -\mu N = -\mu m_{o}g \)   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

\(-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}\)

\(\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0\)

\( \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 \)        

Solving the above quadratic equation for x

\( x = 0.112 m \)  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

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I hope it helps you!  

Answer:

Eubacteria Kingdom :)

Explanation:

Answer: eubacteria

Explanation:

Answer:

\(F=BqV\)

• F is the force experienced

• B is magnetic flux density

• V is potential difference

• q is the charge

Now

Force be F

\(\begin{gathered}\\ \bull\tt\longmapsto F=qvBsin\Theta\end{gathered} \)

Answer:

The vertical component of the velocity can be found using the formula:

V₀y = V₀ * sin(θ)

where V₀ is the initial velocity, θ is the angle of projection, and V₀y is the vertical component of the velocity.

Substituting the given values, we have:

V₀y = 48 * sin(53°)

Using a calculator, we can evaluate sin(53°) to be approximately 0.799:

V₀y = 48 * 0.799

V₀y ≈ 38.352

Therefore, the vertical component of the velocity with which the ball was launched is approximately 38 meters/second, which corresponds to option B.

Answer:

B.) 38 meters/second

Explanation:

The speed of the wind is 20 miles per hour.

To solve this problem, we can use the formula:

Speed = Distance/Time

Let's assume that the speed of the wind is x miles per hour.

With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:

460 = Airspeed + x

Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:

420 = Airspeed - x

We can solve this system of equations to find the airspeed of the plane:

460 = Airspeed + x

420 = Airspeed - x

Adding the two equations gives:

880 = 2Airspeed

Dividing both sides by 2 gives:

Airspeed = 440 miles per hour

Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:

460 = Airspeed + x

460 = 440 + x

x = 20

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The magnitude of normal force (N) on motorcycle is 704.537 N

In this type of scenario, we have been told that the mass, denoted by "m," is 350 kilograms, that the speed of the motorcycle, denoted by "v," is 29 meters per second, and that the radius of curvature, denoted by "R," is 105 meters. We can utilize the second law of motion that Newton proposed to compute the magnitude of the normal force, which is as follows:

\(m . g - N = \frac{mv^{2} }{R}\)

By applying the formula above we have:

\(350 x 9.8 - N= \frac{350x29^{2} }{108}\)

3430 - N = 2725.46296

N = 3430 - 2725.46296

N = 704.537 N

Thus, the magnitude of normal force on motorcycle would be 704.537 N

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The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

\($t=\sqrt{\frac{2H}{g}}$\)

\(v_1t= 0.35\)  ...........(i)

\(v_0t= 1.09\)  ...........(ii)

Dividing the equation (ii) by (i)

\($\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$\)

∴ \(v_0=3.11 \ v_1\)

Now loss of energy  = change in the kinetic energy

\($W=\frac{1}{2} m [v_0^2-v_1^2]$\)

\($W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$\)

\($W=7.307\times 10^{-3} \ v_0^2$\)

If f is average friction force, then

(f)(L) = W

(f) (1) = \($7.307\times 10^{-3} \ v_0^2$\)

(f)  = \($7.307\times 10^{-3} \ v_0^2$\)

The Average force of friction is ( F )  = 7.307 * 10⁻³ v₀²

Given data:

Predicted range ( v₁t ) = 0.3503 m

Actual range ( v₀t ) = 1.09 m

mass = 16.3 g

First step : Determine the value of  V₀

\(t = \sqrt{\frac{2H}{g} }\)    ,    v₁t  =  0.3503 ,    ( v₀t ) = 1.09 m

To obtain the value of  V₀  

Divide ( v₀t ) by ( v₁t )  =  1.09 / 0.3503 = 3.11 v₁

∴ V₀ = 3.11 v₁

Next step : Determine the average force of friction ( f )

given that loss of energy results in a change in kinetic energy

W = \(\frac{1}{2} m ( vo^{2} - v1^{2} )\)

    = 1/2 * 16.3 * 10⁻³ * [ v₀² - \((\frac{v_{0} }{3.11} )^{2}\) ]

∴ W = 7.307 * 10⁻³ v₀²

Average force of friction = W / Actual length

                                         = 7.307 * 10⁻³ v₀² / 1  

∴ Average force of friction ( F )  = 7.307 * 10⁻³ v₀²

Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²

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Your question has some missing data below are the missing data related to your question

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

The models provide important tools for understanding the strong force interactions within a proton.

The strong force interactions between quarks and gluons within a proton are described by Quantum Chromodynamics (QCD), which is a fundamental theory of the strong nuclear force in particle physics. QCD is a non-Abelian gauge theory, meaning that the interactions between the quarks and gluons are highly nonlinear and non-perturbative.

One of the most promising theoretical models for describing the strong force interactions within a proton is lattice QCD, which is a numerical approach that uses a discrete grid to represent the space-time continuum. Lattice QCD allows for the calculation of QCD observables from first principles, without resorting to perturbative expansions. This method can handle non-perturbative effects such as confinement and chiral symmetry breaking by allowing for the simulation of the strong interactions on a discrete space-time grid

Another promising model is effective field theory, which provides a way to describe the low-energy behavior of QCD by constructing an effective Lagrangian that contains only the degrees of freedom relevant to a particular energy scale. This allows for the calculation of QCD observables in a systematic expansion in powers of a small parameter, such as the ratio of the quark mass to the QCD energy scale.

Chiral perturbation theory is another effective field theory that focuses on the dynamics of light quarks, which are the building blocks of pions, the lightest hadrons. Chiral perturbation theory provides a systematic expansion for the interactions between pions and nucleons, and can be used to calculate the properties of these particles at low energies.

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Answer:

the answer is 32.0-25.0=7

Explanation:

the absolute pressure at the bottom of the container is 7

Answer:

W = 312.5 J

P = 12.5 W

Explanation:

W = Fd

First find d:

d = vt = (2.5 m/s)(25 s) = 62.5 m

Now find W;

W = (5.0 N)(62.5 m) = 312.5 J

Power = P = W/t = 312.5 J / 25 s = 12.5 Watts

Answer:

Neutronic is not a defined type of energy.

This is actually an advantage of a parallel circuit. In a parallel circuit, the resistance in each branch is the same as if there was just one light bulb in the whole circuit.

In a parallel circuit, the current has multiple pathways to flow through. Each component, such as a light bulb, is connected across its own branch. The total resistance in a parallel circuit is determined by the sum of the reciprocals of the individual resistances in each branch.

Having the same resistance in each branch ensures that the current is distributed equally among the branches. This means that each component receives the same amount of current as if it were the only component in the circuit. As a result, the brightness of the light bulbs in a parallel circuit will be the same.

This characteristic of parallel circuits is advantageous because it allows for independent operation of each component. If one light bulb were to burn out or be removed, the other bulbs would continue to operate unaffected. Additionally, parallel circuits offer lower overall resistance compared to a series circuit, enabling the flow of more current.

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Answer:

v = 73.75 m/s

Explanation:

It is given that,

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.

Let us assume we need to find the velocity of the rocket when it runs out of fuel.

Let v is the final speed. Using the third equation of kinematics as :

\(v^2-u^2=2as\)

u = 0

\(v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s\)

So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s

Answer:

The two main types of bonds formed between atoms are ionic bonds and covalent bonds.

Explanation:

Hope this helped <3

Answer:

D

Explanation:

The motion of an object can be described by its speed, position, and direction so its all of the above

As the charge is negative, the force, which has a magnitude of 33.05 N, is directed to the left, against the electric field.

E is a symbol for the magnitude of the electric field at a specific distance from a point charge. At twice the distance from the point charge, what is the electric field's strength? The field's strength is E/2 at twice the distance. The field's strength is still equal to E at a distance that is twice as great.

E = k*q/r²

r1 = 2.00 cm

r2 = 1.00 cm + 3.00 cm = 4.00 cm

r3 = 1.00 cm

Using these distances, we can calculate the electric field due to each charge:

E1 = kq1/r1² = (9.0 x 10⁹ Nm²/C²) * (1.50 x 10⁻⁶ C) / (0.02 m)² = 168.75 N/C (to the right)

E2 = kq2/r2² = (9.0 x 10⁹ Nm²/C²) * (-2.00 x 10^⁻⁶ C) / (0.04 m)² = -112.50 N/C (to the left)

E3 = kq3/r3² = (9.0 x 10⁹ Nm²/C²) * (6.00 x 10⁻⁶ C) / (0.01 m)² = 5.40 x 10⁶ N/C (to the right)

E = E1 + E2 + E3 = 168.75 N/C - 112.50 N/C + 5.40 x 10⁶ N/C = 5.39 x 10⁶ N/C (to the right)

F = q*E

F = (-6.13 x 10 C) * (5.39 x 10⁶ N/C) = -33.05 N

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The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:

Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)

ω ≈ 115.28 rad/s

The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².

Use the formula for rotational kinetic energy:

Rotational Kinetic Energy (KE_rot) = (1/2) I ω²

Substituting the given values:

KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²

Calculate the value inside the parentheses:

KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)

KE_rot ≈ 1.331 × 10^-3 J

Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.

KE_rot ≈ 4.8 × 10^-4 J

Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

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The minimum thickness of the magnesium fluoride layer that will cause destructive interference of reflected light with a wavelength of 480 nm is 86.9 nm

To find minimum thickness of a non-reflective coating that causes destructive interference we use the formula t = (mλ) / (4n)

The wavelength (λ) of the light is 480 nm

the refractive index (n) of magnesium fluoride is 1.38,

we can calculate the minimum thickness (t) by saying

t = 0 × 480 / 4 × 1.38

t = 0 nm

t = 1 × 480 nm / 4 × 1.38

t = 480 nm / 5.52

t = 86.96 nm

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Answer: The mass of an object can be calculated in a number of different ways: mass=density×volume (m=ρV). Density is a measure of mass per unit of volume, so the mass of an object can be determined by multiplying density by volume. mass=force÷acceleration (m=F/a).

A. The hockey puck moved at a constant speed away from Janine.

The hockey puck is in equilibrium as a result of moving at a steady pace. Dynamic equilibrium is the name given to this form of equilibrium. Hence, if the hockey puck is moving over the ice at a constant pace, it is in equilibrium.

There is no connection between velocity and speed; velocity is the direction that an object moves in. Velocity is the combination of speed and direction. Speed and velocity are very similar to each other.

The sum of the forces exerted on an object must be zero since, in accordance with Newton's first law of motion, an object moving at a constant speed experiences no net external force.

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At pH = 8, the ratio of H+ ions to OH- ions is 1:1, indicating a neutral solution. The concentration of H+ ions and OH- ions is approximately 1 x 10^(-8) mol/L. The calculated and approximated ratios should match.

To calculate the ratio of H+ ions to OH- ions at pH = 8, we need to use the relationship between pH and the concentration of H+ ions. The pH scale is a logarithmic scale that measures the acidity or alkalinity of a solution based on the concentration of H+ ions.

The formula to calculate the concentration of H+ ions (\(C_H\)+) from pH is:

\(C_H\)+ = \(10^(^-^p^H^)\)

Substituting pH = 8 into the formula:

\(C_H\)+ = \(10^(^-^8^))\)

Using the properties of logarithms, we can calculate the concentration of H+ ions:

\(C_H\)+ ≈ 1 x \(10^(^-^8^))\) mol/L

According to the concept of neutrality in water, the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, the concentration of OH- ions (\(C_O_H\)-) is also approximately 1 x \(10^(^-^8^))\)mol/L.

To calculate the ratio of H+ ions to OH- ions, we divide the concentration of H+ ions by the concentration of OH- ions:

Ratio = \(C_H\)+ / \(C_O_H\)-

Ratio = (1 x \(10^(^-^8^))\) / (1 x \(10^(^-^8^))\))

Ratio = 1:1

The ratio of H+ ions to OH- ions at pH = 8 is 1:1, indicating a neutral solution. This means that the concentration of H+ ions is equal to the concentration of OH- ions, resulting in a balanced ratio.

When comparing the calculated ratio of 1:1 to the approximated ratio at pH = 8, they should be the same because the ratio of H+ ions to OH- ions is determined solely by the pH value, which is consistent and mathematically derived. Therefore, the approximated and calculated ratios should match.

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The gravitational potential energy of the rod-sphere system is - (GMm) / L X ln[ (x+l)/x ]. The magnitude of the gravitational force exerted on the sphere by the rod is - (GMm) / x(x+L) and the direction of the gravitational force exerted on the sphere by the rod is towards it.

(a) Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.

Then the equation for the potential energy as related to the small area of the of the rod can be computed as;

= dU = - (GMm)/ L X (dr/r)

where,

G = gravitational constant

Now, integrating this with the limits of x to (x + L), we get

= U = - (GMm) / L X ln[ (x+l)/x ]

(b) By using F = -dU/dt, the magnitude of the gravitational force can be determined as follows:

Here we have,

= F = - d {- (GMm) / L X ln[ (x+l)/x ]} / dt

= F = - (GMm) / x(x+L)

From above, the negative sign indicates an attractive force.

(c) As, the negative sign indicates an attractive force. Thus, the direction of the gravitational force exerted on the sphere by the rod is towards it.

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The force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).  

The hydraulic lift works based on Pascal's principle, which states that the pressure applied to a confined fluid is transmitted equally in all directions throughout the fluid.

Assuming there is no loss of energy due to friction or other factors, the force exerted on the small piston will be equal to the force exerted on the large piston. This can be expressed as:

F1/A1 = F2/A2

where F1 is the force exerted on the large piston, A1 is the area of the large piston, F2 is the force exerted on the small piston (which we want to find), and A2 is the area of the small piston.

We can rearrange this equation to solve for F2:

F2 = (F1/A1) x A2

Given that the area of the large piston is 100 cm², we can calculate the force exerted on the large piston by using the weight of the car and the gravitational acceleration:

F1 = m x g = 1800 kg x 9.81 m/s² = 17,658 N

Substituting the values into the equation, we get:

F2 = (17,658 N / 100 cm2) x 15 cm² = 2,649 N

Therefore, the force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).

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