A flat uniform circular disk (radius = 5.44 m, mass = 150 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 47.0-kg person, standing 1.54 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.80 m/s relative to the ground. Find the resulting angular speed (in rad/s) of the disk.

The maximum velocity is 48π m/s.

The equation: can be used to determine the maximum velocity of a particle in simple harmonic motion.

Vmax = ω * A

Where

The following formula can be used to get the angular frequency:

ω = 2π / T

Where

T is the motion's period.

Given that the period is π/2 seconds (T = π/2) and the amplitude is 12 m (A = 12), we can find the angular frequency:

ω = 2π / (π/2) = 4π rad/s

Now we can calculate the maximum velocity:

Vmax = ω * A = (4π rad/s) * (12 m) = 48π m/s

Therefore, the maximum velocity is 48π m/s.

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Answer:

Using Ohm's Law, we can calculate the voltage of the battery as V = IR = (4 A)(7 Ω) = 28 V.

The total resistance of the circuit is R = R₁ + R₂ + r, where r is the internal resistance of each element.

We know that R₁ = 14 Ω and R = 7 Ω, so we can solve for R₂:

R₂ = R - R₁ = 7 Ω - 14 Ω = -7 Ω

This is a negative resistance, which doesn't make sense physically. However, it indicates that there is an error in the problem or the calculations.

To find the internal EDS of the elements, we can use the equation:

V = ε - Ir

where V is the voltage of the battery, ε is the internal EDS of each element, I is the current flowing in the circuit, and r is the internal resistance of each element.

We know that V = 28 V, I = 4 A, and r = 6 Ω, so we can solve for ε:

ε = V + Ir = 28 V + (4 A)(6 Ω) = 52 V

Therefore, the internal EDS of each element is 52 V.

Answer:

b 4.5

Explanation:

time=distance/speed

Answer:

70

Explanation:

Answer:   Option A

Explanation:

In general, electric field strength E = (F)/q

E = q4πε0d2

⇒ 360 = 9×109×50×10−6d2

solving, we have d2 = 1250

putting d2 = 1250 into the second equation,

E = 9×109×120×10−61250

we have E = 864NC−1

6.07 m

Explanation:

Given:

\(v_0=24.5\:\text{m/s}\)

\(\theta_0 = 35.5°\)

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

\(x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}\)

or

\(t = 1.29\:\text{s}\)

To find the vertical height where the ball hit the wall, we use

\(y = v_{0y}t - \frac{1}{2}gt^2\)

\(\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2\)

\(\:\:\:\:=6.07\:\text{m}\)

Answer: Car collide with man

Explanation:

Given

Speed of car is \(u=30\ m/s\)

Distance of the man from the car is \(s=55\ m\)

Reaction time \(t_r=0.5\ s\)

Rate of deceleration \(a_d=-10\ m/s^2\)

Distance traveled in the reaction time \(d_o=30\times 0.5=15\ m\)

Net effective distance to cover \(d=55-15=40\ m\)

Distance required to stop the car

\(\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m\)

Require distance is more than that of net effective distance. Hence, car collides with the man.

According to the information, the ball will just pass over the net (question A); and the ball hits the ground approximately at 7.04 meters from the player and after a time of flight of approximately 1.31 seconds (question B).

To determine if the ball will clear the net, we compare the vertical displacement of the ball with the net height. The ball's initial height is 1.1 meters, and it reaches its highest point during flight. By calculating the trajectory, we can confirm that the ball's vertical displacement at its highest point will be higher than the net height of 1.81 meters, ensuring it clears the net.

To find when and where the ball hits the ground, we need to calculate the time of flight and horizontal displacement. Using the given initial velocity and angle, we can determine the time it takes for the ball to reach the ground. By calculating the horizontal displacement based on the initial horizontal velocity and total time of flight, we find that the ball hits the ground approximately 7.04 meters from the player. The time of flight is approximately 1.31 seconds. The specific values may vary depending on the given initial conditions.

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Answer:

Prefixed naming convention. All metric units are clearly related to each other using prefixes. ...

The engineer must design an engine that can produce a force of over 3,100 Newtons.

A force is an energy that can cause change of motion. It can be a push or a pull. It also has both magnitude and direction, making it a vector quantity.

From the question;

total mass = mass of person + mass of cart

total mass = 80kg + 75kg

total mass = 155kg

acceleration = 20m/s²

The formula to be used;

F = m x a

F = 155kg x 20m/s²

F = 3,100N

In conclusion, the engineer needs a minimum force of 3,100 N in the engine to be able to power the cart.

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Answer:

0.5

Explanation:

In this picture,

Answer:

C. Taste aversion can develop after only one pairing of a stimulus and response.

Explanation:

Taste aversion is a unique type of learned response where an individual develops a strong aversion or avoidance to a specific taste or food after a single pairing of that taste with a negative reaction, such as nausea or illness. This is in contrast to traditional conditioning, where multiple pairings of a stimulus and response are typically required for learning to occur. Taste aversion demonstrates a unique rapidity and specificity in its development, which deviates from the general principles of conditioning.

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

The displacement of the train after 2.23 seconds is 25.4 m.

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

The displacement is calculated as follows;

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

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The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.

1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).

2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.

3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.

4. However, the given equation does not explicitly mention resistance.

5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).

6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.

7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).

8. Therefore, V = VR + VL.

9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.

10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.

The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.

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The best, most helpful observation you can make is to LOOK steadily and squarely at the PICTURE of the stack.

For some nefarious reason, you decided to prevent us from doing that.

All the equations of motion are as follows, Displacement (s) equation, Final velocity (v) equation, Average velocity (v_avg) equation, Displacement (s) equation with average velocity, and Displacement (s) equation.

In terms of its motion as a function of time, equations of motion define how a physical system behaves. In more detail, the equations of motion define how a physical system behaves as a collection of mathematical functions expressed in terms of dynamic variables.

In conclusion, equations of motion define how a physical system behaves in terms of how its motion changes over time.

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Answer:

θ = Cos⁻¹[A.B/|A||B|]

A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result

Explanation:

We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:

A.B = |A||B| Cosθ

where,

A = 1st Non-Zero Vector

B = 2nd Non-Zero Vector

|A| = Magnitude of Vector A

|B| = Magnitude of Vector B

θ = Angle between vector A and B

Therefore,

Cos θ = A.B/|A||B|

θ = Cos⁻¹[A.B/|A||B|]

Hence, the correct answer will be:

A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result

The relative density of aluminum is 2.7. This means that aluminum is 2.7 times denser than water, which is the reference substance often used for comparing densities

The relative density (also known as specific gravity) of a material is the ratio of its density to the density of a reference substance. In this case, we are given the density of aluminum as 2.7 x 10^3 kg/m^3.

To find the relative density, we need to compare it to the density of the reference substance. The most commonly used reference substance for relative density is water, which has a density of 1000 kg/m^3.

Relative density = Density of the material / Density of the reference substance.Relative density = (2.7 x 10^3 kg/m^3) / (1000 kg/m^3)

Relative density = 2.7

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Answer:

4.246 light years

Explanation:

4.246 light years

The closest star, Proxima Centauri, is 4.24 light-years away. A light-year is 9.44 trillion km, or 5.88 trillion miles. That is an incredibly large distance. Walking to Proxima Centauri would take 950 million years.

HOPE IT HELPS.

PLEASE MARK ME AS BRAINLIEST.

Answer:

4.25 light years

Explanation:

a light year is the distance light travels in one year it is equal to 9.461 x 1012 km. alpha centauri A & B are roughly 4.35 light years away from us. proxima centauri is slightly closer at 4.25 light years.

Answer:

The approximate temperature required to begin fusion in a star is C. 14 million kelvins. This temperature is high enough to overcome the electrostatic repulsion between positively charged atomic nuclei, allowing them to fuse and form heavier elements.

A cell of inter resistance of 0.5 ohm is connected to coil of resistance 4 ohm and 8 ohm joined in parallel.If there is current of 2A in 8 ohm, the electromotive force (emf) of the cell is approximately 14.5 volts.

To find the emf of the cell, we can apply Ohm's Law and Kirchhoff's laws to analyze the circuit.

Given:

Resistance of the coil, R1 = 4 ohm

Resistance of the other resistor, R2 = 8 ohm

Current passing through the 8-ohm resistor, I = 2A

First, let's analyze the parallel combination of the 4-ohm and 8-ohm resistors.

The total resistance of two resistors in parallel can be calculated using the formula:

1/Rp = 1/R1 + 1/R2

Substituting the given values, we have:

1/Rp = 1/4 + 1/8

1/Rp = 2/8 + 1/8

1/Rp = 3/8

Rp = 8/3 ohm

Now, let's consider the total resistance in the circuit, which includes the internal resistance of the cell (0.5 ohm) and the parallel combination of the resistors (8/3 ohm).

R_total = R_internal + Rp

R_total = 0.5 + 8/3

R_total = 1.833 ohm

Now, we can find the emf of the cell using Ohm's Law:

emf = I * R_total

emf = 2 * 1.833

emf ≈ 3.667 volts

Therefore, the emf of the cell is approximately 3.667 volts.

However, it is worth noting that the given current of 2A passing through the 8-ohm resistor does not affect the emf calculation since the emf of the cell is independent of the current in the circuit.

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Answer:

9.461 × 10^12 Km s

..............................

A state of character arises from the repetition of similar activities

Aristotle was a renowned scientist and philosopher. He has made significant contributions to society. He is renowned for his quotable sayings as well. Because of the correlation between the states of character and the distinctions between them, the behaviors we display must be of a specific type. States of character: The traits that "allow us to stand well or poorly in relation to the passions." Because: We are not commended and criticized only for having the ability to feel pleasure, grief, etc., virtues are not capacities.

Here in this quote, similar behaviors give birth to similar states of character

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Answer:

340 kg

Explanation:

F = ma

Rearranging the equation to solve for mass (m):

m = F/a

Substituting the given values:

m = 850 N / 2.5 m/s^2

m = 340 kg

Answer:

T = 3.14 10⁻²  s

Explanation:

The general equation for a traveling transverse wave is

           y = A sin (k x - wt +Ф)

where w is the angular velocity that is related to the period and Ф is the phase of the oscillation

           w = 2π / T          (1)

In this case they indicate that the expression of the wave

         y = 0.003 cos (20 x + 200t)

if we use the relationship of the double angles

          a = kx - wt

          b = Ф

          sin (a + b) = sin a cos b + sin b cos a

for the case Ф = 90 we have

          sin (a + b) = cos b

we substitute

          y = A cos (kx -wt)

with an initial phase of fi = 90º

if we compare the terms of the two expressions

         A = 0.003 m  

         k = 20 m⁻¹

         w = -200  rad/s

the negative sign indicates that the wave goes to the left

if we clear from equation 1

          T = 2π / w

          T = 2π / 200

          T = 3.14 10⁻²  s

The total electric potential energy is \(\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }\).

Electric Potential Energy of a System of Charges :

The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.

Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.

To bring  q₁ no work is done,

\(V_{p} = \frac{kq_{1} }{r_{1} }\)

where, V = electric potential energy.

            q = point charge.

            r = distance between any point around the charge to the point charge.

           k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

\(V_{2} = \frac{kq_{2} }{r_{2} }\)

Work done by q₁ ;

\(W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }\)

Now bring  q₃,

\(V_{3} = \frac{kq_{3} }{r_{3} }\)

Work done on q₃ by q₁ and q₂

\(W= q_{3} [ V_{1} + V_{2} ]\)

    \(=\frac{kq_{1} q_{3} }{r_{13} }\)\(+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }\)

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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The total electric potential energy is \(\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}\)

Electric Potential Energy of a System of Charges

The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.

Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.

No work has been done to bring q1,

\(V_{1} = \frac{kq1}{r1}\)

where, V = electric potential energy.

           q = point charge.

           r = distance between any point around the charge to the point charge.

          k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

\(V_{2} = \frac{kq2}{r2}\)

Work done by q₁ ;

W1 = \(V_{p} q2\) = \(\frac{kq1q2}{r12}\)

Now bring  q₃,

\(V_{3} = \frac{kq3}{r3}\)

Work done on q₃ by q₁ and q₂

W= q3{\(V_{1} + V_{2}\)}

W = \(\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}\)

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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Answer:

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