how much work does the tension in the ropes do as the child swings from the initial position to the bottom? express your answer in joules.

The distance travelled by Arthur is 7 km and his displacement is 5 km.

The distance travelled by Arthur during the entire motion is determined by summing the entire path covered during the motion.

Distance = 3 km + 4 km

Distance = 7 km

The displacement of Arthur during the entire motion is obtained by calculating the length of the shortest path between the initial and final position.

d = √(a² + b)

where;

d = √(3² + 4²)

d = 5 km

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The direction of the electrical field at the remaining corner of the square is d. along the diagonal connecting this corner and another charge, away from the other charge.

Explanation:

Step 1: An electric field is generated by negative charges.

Step 2: Since three equal negative charges are located in the three corners of the square, they all create an electric field.

Step 3: Each electric field in the remaining corner will point away from the respective negative charge because negative charges repel each other.

Step 4: The total electric field in the remaining corner will be the vector sum of the individual fields.

Step 5: When combining these individual fields, they will point along the diagonal away from the negative charges.

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Blood travels throughout the body as follows:

Circulation is the continuous movement of blood throughout the body by the circulatory system. It consists of the heart, blood vessels (arteries, veins, and capillaries), and blood. The heart acts as a pump, sending oxygen-rich blood through the arteries to the body's cells, and then returning oxygen-depleted blood through the veins back to the heart.

The exchange of oxygen, nutrients, and waste products between the blood and cells occurs through the capillaries. This continuous flow of blood helps to maintain the body's functions and overall health.

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The object D is made up of material Lead. The correct option is D.

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.

Q = m C ΔT

Qc > Qd

The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

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Part (a) To find the total capacitance of the circuit, we need to use the formula for capacitors in parallel and series. In this case, the accelerationare in both parallel and series, so we need to break it down into steps.

First, we can find the equivalent capacitance of C1 and C2 in parallel: C12 = (C1 x C2) / (C1 + C2) = (7.8 ?F x 13.2 ?F) / (7.8 ?F + 13.2 ?F) = 4.96 ?F.

Then, we can find the total capacitance of C12 and C3 in series: C123 = 1 / ((1 / C12) + (1 / C3)) = 1 / ((1 / 4.96 ?F) + (1 / 4.9 ?F)) = 2.45 ?F.

Therefore, the total capacitance of the circuit is 2.45 ?F.

Part (b) To find the total energy stored in the capacitors, we can use the formula U = 0.5 x C x V^2, where U is the energy stored, C is the capacitance, and V is the voltage.

For C1, U1 = 0.5 x 7.8 ?F x (12 V)^2 = 673.92 ?J.
For C2, U2 = 0.5 x 13.2 ?F x (12 V)^2 = 1,411.2 ?J.
For C3, U3 = 0.5 x 4.9 ?F x (12 V)^2 = 352.8 ?J.

The total energy stored in the capacitors is U = U1 + U2 + U3 = 2,437.92 ?J.

Therefore, the numerical value of the total capacitance of the circuit is 2.45 ?F and the numerical value of the total energy stored in the capacitors is 2,437.92 ?J.

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There is an equation sheet for AP Physics. The equation sheet is a reference sheet provided by College Board (the organization that administers the AP Physics exam) that contains a list of commonly used equations and formulas that students are expected to know and use in the exam. The use of this equation sheet can greatly help students in solving the problems they encounter during the exam.

In AP Physics, the equation sheet is an essential tool for students to have with them during the exam. The sheet contains a list of equations and formulas that students will need to apply in order to solve the problems they will face in the exam. This can include equations related to motion, energy, forces, and many other topics that are covered in AP Physics.

In conclusion, the equation sheet for AP Physics is a valuable resource for students taking the exam. It provides a comprehensive list of the equations and formulas that are commonly used in the exam, and helps students to better understand the concepts covered in the course. By preparing and using the equation sheet, students can increase their chances of success in the AP Physics exam.

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Answer:

Independent variable

Explanation:

A

Answer:

normal force that 20 kg box exerts on 10 kg box. (a) The weight of the box depends on the value of g: FG = m2g = (20.0 kg)(9.80 m/s2) = 196 N.

Answer:

As altitude increases and atmospheric pressure decreases, the boiling point of water decreases.

We are given the following data: A train has a length of 108 m and starts from rest with a constant acceleration at time t=0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t=13.6 s, the car just reaches the front of the train.

Ultimately, however, the train pulls ahead of the car, and at time t=39.1 s, the car is again at the rear of the train. To solve this problem, we can assume that the train starts moving with a constant acceleration and at a time t=0 s, the train is at rest.

The distance between the car and the train is 108 m. At time t = 13.6 s, the car reaches the front of the train. So, the distance covered by the car in 13.6 seconds is the same as the length of the train: 108 m = (u × 13.6) + (0.5 × a × 13.6²)

The above equation gives us the value of u + 96

a. Similarly, when the car reaches the back of the train at time t = 39.1 s, the distance travelled by the car is the same as the length of the train:108 m = (u × 39.1) + (0.5 × a × 39.1²)

The above equation gives us the value of u + 760.55a. Now we have two equations and two unknowns. Therefore, we can solve the equations simultaneously to obtain the values of u and a.

108 m = (u × 13.6) + (0.5 × a × 13.6²)108 m = (u × 39.1) + (0.5 × a × 39.1²)

Solving the above equations simultaneously, we get: u = 28.01 m/sa = 0.114 m/s²

Now that we have found the value of acceleration, we can calculate the car's velocity at any given time. Since the car is moving with a constant velocity, its velocity will remain the same. Therefore, the car's velocity is: v = 28.01 m/s

The train's acceleration is: a = 0.114 m/s²

The problem is related to the motion of a train and a car moving on a straight path. We are given the length of the train, and the car is moving with a constant velocity. At time t = 0, the train starts moving with a constant acceleration. We need to find the magnitudes of the car's velocity and the train's acceleration. To solve this problem, we can use the equations of motion, which relate the distance travelled by an object to its initial velocity, acceleration, and time. We know that the train starts from rest, so its initial velocity is 0.

The car, on the other hand, is moving with a constant velocity, which means its acceleration is 0. Using the equations of motion, we can relate the distance covered by the car to the distance covered by the train. At time t = 13.6 s, the car reaches the front of the train. Therefore, the distance covered by the car in 13.6 seconds is the same as the length of the train. Similarly, when the car reaches the back of the train at time t = 39.1 s, the distance travelled by the car is again the same as the length of the train. We can use these two equations to solve for the car's velocity and the train's acceleration.

We found that the car's velocity is 28.01 m/s, and the train's acceleration is 0.114 m/s². This means that the train is accelerating at a very slow rate, and it takes a long time for the train to catch up to the car. Ultimately, however, the train pulls ahead of the car, indicating that its acceleration is greater than the car's velocity.

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Answer:

the unit of force in the centimeter-gram-second system equal to the force that would give a free mass of one gram an acceleration of one centimeter per second per second.Explanation:hope its a good answer

In order to have the same final temperature, the mass of the ice must be increased by a factor of 2.

First, we have to obtain the final temperature of the ice;

From;

H = mc(θ2 - θ1)

Where;

m = mass of the ice

c = specific heat capacity of the ice

θ2 = final temperature of the ice

θ1 = Initial temperature of the ice

H = heat added to the ice

Substituting values;

3500 = 3.4 × 2093 × (θ2 - (-7.6))

3500 = 7116.2θ2  +  54083.12

3500 - 54083.12 = 7116.2θ2

θ2 = - 7.1°C

Since the heat added to the ice block is doubled;

H = 2 × 3500 = 7000 J

H = mc(θ2 - θ1)

7000 = m × 2093 × ((- 7.1) - (-7.6))

m = 7000/2093 × ((- 7.1) - (-7.6))

m = 6.8 Kg

The factor by which the mass must be increased in obtained as; 6.8 Kg/3.4 kg

=  2

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Answer:

= 1440 kg·m/s

Explanation:

Step by Step Solution to find Momentum of Mass = 288 kg & Velocity = 5 m/s:

Given that,

mass (m) = 288 kg

velocity (v) = 5 m/s

Substitute the value into the formula

p = 288 x 5

p = 1440 kg·m/s

∴ Momentum (p) =1440 kg·m/s

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Assuming an inductor is connected to a 180-v ac line and the inductor has an induced voltage of 120 v, there are 60 volts available to push the current through the wire resistance of the coil.

To determine the voltage that pushes the current through the wire resistance of the coil, you'll need to consider the voltage across the inductor and the applied voltage from the AC line. Given that the induced voltage across the inductor is 120 V and the AC line voltage is 180 V, you can calculate the voltage across the wire resistance by using the formula:

Voltage across wire resistance = AC line voltage - Induced voltage across the inductor

Voltage across wire resistance = 180 V - 120 V = 60 V

So, there are 60 volts available to push the current through the wire resistance of the coil.

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Given:

The speed of the train is,

The actual frequency of the whistle is,

The speed of sound in air is,

To find:

The frequency heard by the observer

Explanation:

The apparent frequency when the train approaches the observer is given by,

Hence, the frequency heard by you is 209.2 Hz.

Answer:

We know ForceF=mass×acceleration----------(1)

And given ForceF=

r

mv

2

----------(2)

Equating 1 and 2 dimensionaly we get ,

ma=

r

mv

2

MLT

−2

=ML

2

T

−2

L

−1

MLT

−2

=MLT

−2

If the electric potential at point a is \(V_a\) = 55 V and the electric potential at point c is \(V_c\) = 5 V, the electric potential at point b (\(V_b\)) is 38.3 V.

Electric potential is a scalar quantity that gauges how much energy is present per unit of charge at a specific location inside an electric field. It is sometimes referred to as voltage.

When a charged particle is positioned at a specific location in an electric field, it would have a potential energy known as electric potential. The distribution of charges in the area around a point in space and their separation from that point determine the electric potential at that location.

Let us consider the given question:

In an uniform electric field (See Picture(2))

potential difference:

V=Ed

Distance between two points = d

now,

\(V_A - V_C = Ed\)

⇒ 55-5 = E(6)

intensity of electric field

E = \(\frac{50}{6}\) V/m

Let \(V_B\) be potential at B,

∴ \(V_A- V_B = E (x)\)

⇒ \(55- V_B=(\frac{50}{6}*2)\)

⇒ \(V_B = 55- \frac{50}{3}\)

⇒ \(V_B = 38.3 V\)

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The complete question is:

If the electric potential at point a(See Picture) is \(V_a\) = 55 V and the electric potential at point c is \(V_c\) = 5 V. What is the electric potential at point b (\(V_b\))?

Given the following information:Mass of the system, m = 20 kg.Damping coefficient, b = 6 Ns/m.Force, F = 90 N.Frequency of applied force, f = ?Applied force angular frequency, w = 6 rad/s.Forced vibration equation:F(t) = F0 sin(wt)where F0 = 90 N and w = 6 rad/s.Under the action of the force F, the mass m will oscillate.The equation of motion for the mass-spring-damper system is given by:$$\mathrm{m\frac{d^{2}x}{dt^{2}}} + \mathrm{b\frac{dx}{dt}} + \mathrm{kx = F_{0}sin(\omega t)}$$where k is the spring constant.x(0) = 0 and x'(0) = 0.As we have the damping coefficient (b), we can calculate the damping ratio (ζ) and natural frequency (ωn) of the system.Damping ratio:$$\mathrm{\zeta = \frac{b}{2\sqrt{km}}}$$where k is the spring constant and m is the mass of the system.Natural frequency:$$\mathrm{\omega_{n} = \sqrt{\frac{k}{m}}}$$where k is the spring constant and m is the mass of the system.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$At resonance, the amplitude of the system will be maximum when forced by a sinusoidal force of frequency equal to the resonant frequency.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$$$\mathrm{\omega_{d} = \sqrt{(6.57)^{2}-(-2.88)^{2}} = 6.98 rad/s}$$Hence, the frequency of applied force at which resonance will occur is 6.98 rad/s.

The frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

To determine the frequency of the applied force at which resonance will occur, resonance happens when the frequency of the applied force matches the natural frequency of the system. The natural frequency can be determined using the formula:

ωn = √(K / M),

where ωn is the natural frequency, K is the spring constant, and M is the mass of the system.

Substituting the given values of K = 400 N/m and M = 20 kg into the equation, we can calculate the natural frequency ωn.

ωn = √(400 N/m / 20 kg) = √(20 rad/s²) = 2√5 rad/s.

Therefore, the frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

The correct question is given as,

M= 20kg

Fo = 90 N

ω = 6 rad/s

K = 400 N/m

C = 125 Ns/m

Determine the frequency of applied force at which resonance will occur?

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Answer:

\( \sf t = \dfrac{V_f - V_i}{a} \)

Explanation:

\( \rm V_f = V_i + at \\ \rm V_f - V_i = at \\ \rm at = V_f - V_i \\ \rm t = \dfrac{V_f - V_i}{a} \)

Answer:

I don't know sorry brother/sister

Explanation:

please mark as brilliant

Answer:

0.006 km / m/s

Explanation:

divide 343 m/s and 2 km

The radius of the cylindrical barrel is 0.78 m.

The circular area of the cylindrical barrel is calculated as follows;

P = F/A

A = F/P

where;

A = mg/P

A = (50 x 9.8) / (260)

A = 1.885 m²

A = πr²

r² = A/π

r = √(A/π)

r = √(1.885/π)

r = 0.78 m

Thus, the radius of the cylindrical barrel is 0.78 m.

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Answer:

represents the energy the object possesses by virtue of its motion. ... This type of energy is generally known as kinetic energy. Thus, Equation (16) states that any work done on point object by an external force goes to increase the object's kinetic energy.

Explanation:

   The relation between some bodies are involved by Object's position and motion, i.e., through potential and kinetic energy in which it actually explained by Newton's second law of motion.

Definition:

      Newton's second law states that force is equal to the rate of change of momentum, i.e., the mass and velocity.

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The force required to keep the block and brick moving at constant velocity is 46 N.

Given data:

Force needed to keep the block moving with constant velocity = 14 N

Weight of the wooden block = 40 N

Weight of the brick = 20 N

We have to calculate:

a) Coefficient of sliding friction between the block and the table.

b) Force needed to keep the block and brick moving at constant velocity.

Calculation:

a) Coefficient of sliding friction between the block and the table:

Let μ be the coefficient of sliding friction between the block and the table and n be the normal force between the block and the table.

μ = Force of friction / Normal force

We know that normal force is equal to the weight of the block.

n = 40

N = Weight of the block

Force of friction = 14 N (as the block is moving at a constant velocity)

μ = 14 / 40

μ = 0.35

Therefore, the coefficient of sliding friction between the block and the table is 0.35.

b) Force needed to keep the block and brick moving at constant velocity:

For the block and brick to move at a constant velocity, the force required to move the block and brick together should be equal to the force of friction acting on the block and table.

Forces acting on the block and brick:

1) Weight of the block and brick acting downwards

2) Force of friction acting upwards

Net force acting on the block and brick = (Weight of the block + Weight of the brick) - Force of friction

Net force acting on the block and brick = (40 + 20) N - 14 N

Net force acting on the block and brick = 46 N

Force required to keep the block and brick moving at constant velocity = 46 N

Therefore, the force required to keep the block and brick moving at constant velocity is 46 N.

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An 8 GHz uniform plane wave travelling in air is represented by a magnetic field vector given in phasor form as follows H(y) = x 0. 015e^-j beta y + z 0. 03e^j(pi - beta y) mA - m^-1 . The total time-averaged power transferred to each eardrum in 1.0 second is 3.972x10^-7 J.

The magnetic field vector for the uniform plane wave can be represented as:

H(y) = x0.015e^(-jβy) + z0.03e^(j(π-βy)) mA/m

where β is the propagation constant, and has units of rad/m.

The wave frequency can be determined from the wavelength λ, which can be calculated using the propagation constant:

λ = 2π/β

The frequency can then be determined using the relation:

f = c/λ

where c is the speed of light in air, which is approximately 3x10^8 m/s.

To find β, we can equate the phase of the x-component of H(y) to the phase of the z-component of H(y):

-jβy = j(π - βy)

Solving for β, we get:

β = π/(2y)

Substituting y = 1/(2β), we get:

β = πy

Substituting this value of β in the expression for H(y), we get:

H(y) = x0.015e^(-jπy) + z0.03e^(jπy) mA/m

To find the electric field vector, we can use the relation:

E(y) = Z0H(y)

where Z0 is the impedance of free space, which has a value of approximately 377 Ω.

Substituting the values of H(y) and Z0, we get:

E(y) = x5.655e^(-jπy) + z11.31e^(jπy) mV/m

The time-averaged power density carried by the wave can be calculated using the relation:

P = 1/2Re(E(y) x H*(y))

where H*(y) is the complex conjugate of H(y).

Substituting the values of E(y) and H(y), we get:

P = 1/2(0.015)(5.655)(cos(πy) + jsin(πy)) + 1/2(0.03)(11.31)(cos(πy) - jsin(πy))

Simplifying, we get:

P = 0.236cos(πy) W/m^2

To find the total time-averaged power transferred to a surface of area A, we can integrate P over the surface:

P_total = ∫∫ P dA

Assuming the surface is perpendicular to the direction of propagation, and has a diameter of 8.4 mm, we get:

A = π(0.0042)^2 = 5.538x10^-5 m^2

Substituting the value of A and integrating P over the surface, we get:

P_total = 0.236∫∫ cos(πy) dA

P_total = 0.236cos(πy)∫∫ dA

P_total = 0.236cos(πy)(5.538x10^-5)

Substituting the value of y = 30 m, we get:

P_total = 0.236cos(πx30)(5.538x10^-5) = 3.972x10^-7 W

Therefore, the total time-averaged power transferred to each eardrum in 1.0 second is 3.972x10^-7 J.

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Two Sets of solid double yellow lines with spaced two or more feet apart indicates that Do not drive on or over the barrier.

Solid yellow lines - it is marked at the center of a road and used for two-way traffic.

Broken yellow lines - It indicate that you may pass if the broken line is next to your driving lane.

Two solid yellow lines - It indicates no passing or Do not drive on or over the barrier.

Two sets of solid double yellow lines are considered a barrier , make a left turn, or a U-turn to cross it.

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Answer:

For use in multimedia projects, motion graphics are segments of animation or digital video that provide the impression of movement or rotation. Though they can sometimes be presented using manually driven equipment, motion graphics are often displayed via electronic media technology.