what is costant error​

A. The output voltage, Vout, as a function of time, t, in a series RL circuit can be determined using the equation: Vout(t) = Vmax * (1 - e^(-t/τ)), where τ = L/R.

B. When the time interval T is very short (T → 0) and the maximum voltage Vmax is quite high (VmaxT ≈ Φimp), we can approximate the output voltage Vout using the equation: Vout(t) ≈ Φimp * e^(-t/τ), where τ = L/R.

A. To determine the output voltage Vout as a function of time t in a series RL circuit, we use the following equation:

Vout(t) = Vmax * (1 - e^(-t/τ))

Here, Vmax is the maximum input voltage, τ = L/R is the time constant of the circuit (where L is the inductance and R is the resistance).

B. When the time interval T is very short (T → 0) and the maximum voltage Vmax is quite high (VmaxT ≈ Φimp), we can make the following approximation:

Vout(t) ≈ Vmax * e^(-t/τ)

In this case, we substitute VmaxT with Φimp, which is the total magnetic flux in the circuit.

Rearranging the equation, we get:

Vout(t) ≈ Φimp * e^(-t/τ)

This approximation is valid when the time interval T is very small compared to the time constant τ of the circuit and when the maximum voltage is sufficiently high.

The time constant τ is determined by the values of inductance (L) and resistance (R) in the circuit. It represents the characteristic time scale over which the current and voltage in the circuit change in response to a voltage or current input.

Therefore, in the given scenario, when T is very small and Vmax is high, we can approximate the output voltage Vout(t) in the series RL circuit by the equation: Vout(t) ≈ Φimp * e^(-t/τ), where τ = L/R.

Note: The symbol Φimp in the equation represents the total magnetic flux in the circuit.

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Answer:

100km/h

Explanation:

96km•2.5=240km

240÷3/5=240•5/3=400km

400-240=160km

4.1-2.5=1.6h

160÷1.6=100km/h

The displacement of the object after 1 second, is 3 m.

The displacement of the object after 3 seconds, is 9 m.

The displacement of the object after 5 seconds, is 15 m.

The displacement of an object is the change in the position of the object.

The displacement of the object after 1 second, x = 3 m/s x 1 second = 3 m

The displacement of the object after 1 second, x = 3 m/s x 3 s = 9 m

The displacement of the object after 1 second, x = 3 m/s x 5 s = 15 m

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1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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The sled's speed can be calculated by considering the acceleration, frictional force, and time. After substituting the given values and performing the calculations, the final speed is determined to be 12.25 m/s.

To calculate the speed of the sled after 2.50 seconds, we can use the equations of motion and consider the forces acting on the sled.

Let's denote the initial speed of the sled as v0, the final speed as vf, the acceleration as a, the time as t, and the coefficient of friction as μ.

Initially, the rocket sled is accelerating, so we can use the equation:

vf = v0 + at

Since the sled is skidding along its rails after the rocket engine stops, the only horizontal force acting on the sled is the force of friction. The frictional force can be calculated using the equation:

frictional force = coefficient of friction * normal force

Since the sled is moving horizontally, the normal force is equal to the weight of the sled, which can be calculated as:

weight = mass * gravity

Now, we can determine the acceleration of the sled using Newton's second law:

frictional force = mass * acceleration

Combining the equations and substituting the values, we have:

vf = v0 + (frictional force / mass) * t

To find the frictional force, we need to calculate the weight of the sled and then multiply it by the coefficient of friction:

frictional force = (mass * gravity) * coefficient of friction

Substituting this back into the previous equation, we get:

vf = v0 + ((mass * gravity * coefficient of friction) / mass) * t

Simplifying further, we have:

vf = v0 + (gravity * coefficient of friction) * t

Now we can substitute the given values into the equation. Assuming the acceleration due to gravity is approximately 9.8 m/s², the coefficient of friction is 0.5, the initial speed is 0 m/s (since the sled starts from rest), and the time is 2.50 s, we can calculate the final speed:

vf = 0 + (9.8 * 0.5) * 2.50

vf = 12.25 m/s

Therefore, the sled is moving at a speed of 12.25 m/s after 2.50 seconds.

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Answer:

C.

Explanation:

That is the main part of a cell, apart of the nervous system and controls all the other functions of the cell.

I hope this helped

Brainliest if it did? No pressure!

Have a great day! :D

A circuit that is closed prevents the flow of electrons. So, the correct option is A.

An electrical circuit is defined as an interconnection of electrical components or a model of such interconnection, consisting of electrical elements. An electrical circuit is described as a network consisting of a closed loop, which provides a return path for current.

A closed circuit is required to carry current. When there is a break anywhere in the path, the circuit will open and no current will flow, and the metal atoms in the wire will quickly become cool and electrically neutral.

Thus, a circuit that is closed prevents the flow of electrons. So, the correct option is A.

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The well that will have most of the water will be well A.

The underground water supply is defined as a type of water that exists underground in saturated zones beneath the land surface.

From the two wells represented in the diagrams above, Well A has water supply from underground which is lacking in well B.

Therefore, well A will have most of the water more than B.

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The maximum speed of the cart if the amplitude of the motion is 3.00 cm is  0.036 m/s

The velocity of the cart when the position is 2.00 cm is 0.1414 m/s.

The kinetic and potential energies of the system when the position of the cart is 2.00 cm​ is 5×10⁻³ J and 4×10⁻³ J resp.

a) To find maximum speed potential energy of the spring gets converted into kinetic energy of the cart in the oscillator motion, Hence,

1/2mv² =1/2kA²

1/2×0.5×v² = 1/2 ×20× 0.03²

v² = 40×9×10⁻⁴

v = 0.036 m/s

b) For the velocity at a given point is calculated by the formula,

v = ±ω√(A² - x²)

v = ±ω√(0.03² - 0.02²)

v = ±ω × 0.0223

v = ±√k/m × 0.0223

v = ±√20/0.5× 0.0223

v = 0.1414 m/s

c)

kinetic energy of the system,

K = 1/2 mv²

K = 1/2 ×0.5×0.1414²

K = 5×10⁻³ J

Potential energy of the system

P = 1/2 kx²

P = 1/2 × 20× 0.02²

P = 4×10⁻³ J

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The length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

Given data: r = 0.2 mm = 0.2 x 10^-3m Resistivity = 3 x 10^-6 ohm R = 15.552 ohm

Formula Used: Resistivity (ρ) = (RA)/L

Where, R is resistance, A is the area of cross-section, L is the length of the wire.

Resistance (R) = ρ (L/A)

Multiplying A on both sides, we get

Resistance (R) x A = ρ L ... equation (1)

Area of the cross-section of a wire of radius (r) is given by, A = πr^2

where, π is a constant whose value is 3.14

Substituting the given values, we get

A = πr^2= π (0.2 x 10^-3m)^2= 1.2566 x 10^-7 m^2

Substituting the values of R, A and ρ in equation (1), we get

Length of wire (L) = (Resistance x Area) / Resistivity= (15.552 ohm x 1.2566 x 10^-7 m^2) / (3 x 10^-6 ohm)= 6.5268 m

Therefore, the length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

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The entire park area is considered to be a semi-arid desert, but distinct habitats are located at different elevations along the 8,000-foot elevation gradient. Near the Colorado River, riparian vegetation and sandy beaches prevail.

The forces on one of the wires will be 0.000264 N.It is the field sensed around a moving electric charge that gives rise to the magnetic force.

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.

The given data in the problem is;

I is the current = 6.50 A

B is the magnetic field=f 2.71 X 10⁻⁵ T

The force on the wire is;

\(\rm F= BIL \\\\\ F=2.71 \times 10^{-5} \ T \times 6.50 \ A \times 1.50 \ m \\\\\ F=0.000264225 \ N\)

Hence, the force on one of the wires will be 0.000264 N.

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Explanation:

carbon dioxide, chlorophyll, water and light

Explanation:i think its nuetrons

The mass of the object is 0.4 kg.Mass is different from weight, which is the force exerted on an object due to gravity.

What is Mass?

Mass is a fundamental physical property of matter that measures the amount of substance in an object. It is a scalar quantity, which means it has a magnitude but no direction. The mass of an object is typically measured in kilograms (kg) or grams (g).

To calculate the mass of an object that is experiencing a net force of 200 N and an acceleration of 500 m/s^2, we can use Newton's second law of motion:

F = ma

Rearranging this formula, we can solve for the mass of the object:

m = F / a

Substituting the given values, we get:

m = 200 N / 500 m/s^2

m = 0.4 kg

Therefore, the mass of the object is 0.4 kg.

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The momentum in a coliision is conserved, since this is a perfect inelastic collision the balls will stick together after the colission then in this case we have:

where V is the final velocity of the system. Since one ball is moving to the right and the other is moving to the right one of their velocities (the one moving to the right) is positive and the other one is negative, then we have:

Therefore their final speed is zero.

Answer:

the nucleus of an atom is held together by a strong nucleus force that binds together protons and neutrons.all though the strong nucleus force is stronger than the four fundamental forces.it acts over very short typically nucleus substances

The velocity of car 2, from the frame of reference of car 1, is 20 m/s west. (Answer: B)

When considering the frame of reference of car 1, we need to subtract the velocity of car 1 from the velocity of car 2 to determine the relative velocity. Since car 1 is moving at 45 m/s west and car 2 is moving at 25 m/s east, we can subtract these velocities to find the relative velocity of car 2 with respect to car 1.

45 m/s (west) - 25 m/s (east) = 20 m/s (west)

Therefore, car 2 has a velocity of 20 m/s west when observed from the frame of reference of car 1. It is important to note that velocities are vector quantities, which means they have both magnitude and direction. In this case, the direction is west because car 2 is moving in the opposite direction of car 1. (Answer: B)

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Answer:

The total kinetic energy must be greater than the momentum of either ball

Explanation:

In order to solve this problem, we must clarify the following conditions, the balls are in motion that is, there is kinetic energy, the kinetic energy is a scalar magnitude, therefore this energy is greater than zero.

But however momentum is equal to zero, momentum is a vector quantity, that is, these velocities have a direction, momentum is defined as the product of mass by Velocity.

P = m*v (momentum lineal)

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Since the masses of the balls are equal, we have:

m1 = m2

therefore:

m1*v1 + (m2*v2) = 0 (total momentum)

For the sum to be equal to zero, besides the equivalation of the masses, the velocities must also be equal, but the velocities must be of opposite sign, to meet equality to zero.

m1*v1 - m2*v2 = 0

And the kinetic energy of a body is defined by means of the following equation

Ek = 0.5*m*v²

Since the magnitude of the velocities is greater than zero, so that there is movement, we can say that the kinetic energy is much greater than zero, as well as greater than the total linear momentum

Answer:

42

Explanation:

p=mv

momentum=mass*velocity

x=2.8*15

x=42

Answer:

36.29 m/s

Explanation:

We know from theory that \(v= v_0 +a\Delta t\). Let's replace the value we know in our equation and solve for the only value left.

\(0 = v_0 - 9.55 \cdot 3.8 \\ v_0 = 9.55 \cdot 3.8 = 36.29 m/s\)

Answer:

  A. separation is increasing at about 9.90 m/s

  B. about 9.90 m/s

  C. see attached

Explanation:

A. Both stones are accelerated at the same rate, so the difference in their velocities is the velocity stone A had when stone B was dropped. The distance between the stones increases linearly at that velocity, about 9.90 m/s.

__

B. The speed when accelerated from rest over some distance d can be found as ...

  v = √(2dg) = √(2(5 m)(9.8 m/s²)) = 7√2 m/s

The speed of stone B at 5 meters is about 7√2 ≈ 9.90 m/s.

__

C. See the attachment for a velocity vs. time graph. Stone B has traveled 10 m when its velocity is -14 m/s

Answer:

KE = 25

Explanation:

KE = 1/2 m v^2

KE = 1/2 (50) (1)

KE = 25

Answer:

Explanation:

The electric flux through a surface is given by the equation:

Φ = EAcos(θ)

where Φ is the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

We are given Φ = 4.44 N⋅m2/C, A = 6.66×10−4 m2, and θ = 60.0∘. Substituting these values into the equation above and solving for E, we get:

E = Φ / (Acos(θ))

= 4.44 N⋅m2/C / (6.66×10−4 m2cos(60.0∘))

= 1.62×10^4 N/C

Therefore, the magnitude of the electric field is 1.62×10^4 N/C.

The magnitude of the electric field is 13,320 N/C.

The electric flux through a surface is defined as the product of the electric field and the area of the surface projected perpendicular to the electric field. Mathematically, we can write:

Φ = EAcos(θ)

where Φ is the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

Here in the Question,

We are given the electric flux Φ = 4.44 N·m^2/C, the area A = 6.66×10^-4 m^2, and the angle θ = 60.0°. We can solve for the magnitude of the electric field E by rearranging the equation as follows:

E = Φ / (A*cos(θ))

Substituting the given values, we get:

E = 4.44 N·m^2/C / (6.66×10^-4 m^2*cos(60.0°))

Simplifying the denominator, we get:

E = 4.44 N·m^2/C / (6.66×10^-4 m^2*0.5)

E = 13,320 N/C

Therefore, 13,320 N/C is the magnitude of the electric field.

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Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

\(m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}\)

Energy

\(\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})\)

\(m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}\)

Where:

\(m_{1}\), \(m_{2}\) - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

\(v_{1,o}\), \(v_{2,o}\) - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

\(v_{1}\), \(v_{2}\) - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If \(m_{1} = 0.400\,kg\), \(m_{2} = 0.900\,kg\), \(v_{1,o} = +5.86\,\frac{m}{s}\), \(v_{2,o} = 0\,\frac{m}{s}\), the system of equation is simplified as follows:

\(2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}\)

\(13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}\)

Let is clear \(v_{1,f}\) in first equation:

\(0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}\)

\(v_{1,f} = 5.86-2.25\cdot v_{2,f}\)

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

\(13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}\)

\(13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}\)

\(13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}\)

\(2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0\)

\(2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0\)

There are two solutions:

\(v_{2,f} = 0\,\frac{m}{s}\) or \(v_{2,f} = 3.606\,\frac{m}{s}\)

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (\(v_{2,f} = 3.606\,\frac{m}{s}\))

\(v_{1,f} = 5.86-2.25\cdot (3.606)\)

\(v_{1,f} = -2.254\,\frac{m}{s}\)

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Answer:

Monecious is a plant or invertebrate animal having both the male and female reproductive organs in the same individual; hermaphrodite.

Answer:

false

Explanation:

Because the sun has ultraviolet rays

Answer:

False.

Explanation: The sun is mainly composed of hydrogen, it's 92% composed of hydrogen.

An angle '\(\Theta\)' on the circle Corresponds to an angle 2\(\Theta\) on the part.

On the Mohr, the circle has a positive counterclockwise rotation, similar to the physical space convention for shear stress. is the clockwise shear stress and both are plotted. Mohr's circle is a graphical representation of the general state of stress at a point.

This is a graphical method for evaluating principal and maximum shear stresses. Normal and tangential stress in any plane. Soil mechanics use Mohr's circles to visualize the relationship between normal and shear stresses, and to estimate the maximum stress based on three or more soil samples taken from a site.

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Answer:

Explanation:

A) The output force required to lift a 15 kg object would be equal to the weight of the object, which is given by:

Output force = Weight of object = m * g

where m is the mass of the object and g is the acceleration due to gravity. Assuming that g is equal to 9.81 m/s^2, we have:

Output force = 15 kg * 9.81 m/s^2 = 147.15 N

Therefore, the output force required to lift a 15 kg object would be 147.15 N.

B) In this case, the input force is the force that you are pushing down with the lever, which is given as 20 N.

C) The mechanical advantage of the ramp is given by the ratio of the output force to the input force. In this case, the output force is the weight of the object (40 N) and the input force is the force that you are pushing with (10 N). Therefore, the mechanical advantage of the ramp would be:

Mechanical advantage = Output force / Input force = 40 N / 10 N = 4

So, the ramp multiplies your force by a factor of 4.

Note that in all of these calculations, we have assumed that the system is ideal and that there are no losses due to friction or other factors. In practice, these losses will reduce the mechanical advantage of the system and make it more difficult to lift or move objects.