Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the outer sphere is 1.62 10-8 C. Find the magnitude of the electric field (in N/C) at 0.012 m.

After time t, the position function of the ball is determined as \(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

The given parameters;

The position function of the ball after time t, is calculated as follows;

\(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.

Thus, after time t, the position function of the ball is determined as \(y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2\)

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(a) The resonance curve of cigar box would be a plot of  vibration amplitude versus frequency; (b) The Fourier spectra of string without considering role of cigar box would show the various harmonic frequencies ; (c) The Fourier spectra of string will show stronger peak at  resonant frequency of cigar box (300 Hz).

Resonance curve is the curve whose abscissas are frequencies lying near to and on both sides of natural frequency of vibrating system and whose ordinates are corresponding amplitudes of near-resonant vibrations.

(a) The resonance curve of cigar box would be a plot of the vibration amplitude versus frequency. It would show that cigar box has resonant frequency of 300 Hz, which means that it will vibrate more easily at this frequency than at others.

(b) The Fourier spectra of the string without considering the role of the cigar box would show the various harmonic frequencies that the string can produce when plucked. It would be a series of peaks at integer multiples of fundamental frequency (C2 = 66 Hz), with decreasing amplitude as frequency increases.

(c) When the string is attached to cigar box, resonance curve of the cigar box will affect the frequencies that are amplified more than others. The Fourier spectra of string will show a stronger peak at the resonant frequency of the cigar box (300 Hz), and possibly some harmonics that are also amplified by the cigar box. The amplitudes of other harmonics may be reduced due to damping, so the overall spectra will be modified by resonant properties of cigar box.

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Answer:

Their not in the door space, and not facing the right way, so even if they were in the door space they would be facing the wrong way

Explanation:

Hope this helps!

Answer:

548.8 J

Explanation:

Answer:

See the explanation below

Explanation:

No matter at what height a body is dropped, the body will always accelerate to the same reason, which corresponds to Earth's gravitational acceleration.

g = 9.81 [m/s²]

That is, in the first second the velocity is 9.81 [m/s] = 9.81 [m/s²] x 1 [s]

Now in the second 2; 19.62 [m/s²] = 9.81 [m/s²] x 2 [s]

And in the third second 29.43 [m/s²] = 9.81 [m/s²] x 3 [s]

And etc.

Answer:

21 m/s

Explanation:

If

m1 = mass of 1875 kg car

u1 initial speed of 187 kg car = 23.0 m/s

m2 = mass of 1025 kg car

u2= initial speed of 1025 kg car = 17.0 m/s

Since they stuck together, they have a common final velocity v

From principle of conservation of linear momentum

m1u1 + m2u2 = (m1 + m2) v

v= m1u1 + m2u2/(m1 + m2)

v= 1875 × 23 + 1025× 17/(1875 + 1025)

v= 43125 + 17425/2900

v= 21 m/s

The final velocity of both cars is 20.87 m/s.

Given that the two cars have a mass of 1875 kg and 1025 kg respectively and the velocity of both the cars is 23.0 m/s and 17.0 m/s.

The common velocity of both cars is given below.

\(v (m_1+ m_2) = m_1u_1 + m_2u_2\)

Where v is the final velocity of both cars, m1 and m2 are the mass of both cars respectively. u1 and u2 are the initial velocities of both cars.

Substituting the values, we get the final velocity.

\(v ( 1875 + 1025) = 1875 \times 23 + 1025 \times 17\)

\(v = 20.87 \;\rm m/s\)

Hence we can conclude that the final velocity of both the cars is 20.87 m/s.

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a.  output voltage is 110 V, the RMS output voltage is approximately 155.56 V, the output (load) current is 15.56 A, the ripple factor is 0.866, and the form factor is 0.866. b. the input active power is 2640 W, the input apparent power is 2640 VA, and the power factor is 1 (or unity).

a. For a single-phase full-bridge uncontrolled (diode) rectifier with a pure resistive load of R = 10 Ohms and neglecting diode volt-drops, we can calculate the following values:

Average Output Voltage:

The average output voltage of a full-bridge rectifier can be calculated as half of the peak input voltage. Since the input voltage is 220 V, the average output voltage will be:

Average Output Voltage = (220 V) / 2 = 110 V

RMS Output Voltage:

The RMS output voltage of a full-bridge rectifier can be calculated as the peak input voltage divided by the square root of 2. In this case, the RMS output voltage will be:

RMS Output Voltage = (220 V) / √2 ≈ 155.56 V

Output (Load) Current:

Since the load is pure resistive, the output (load) current will be the same as the RMS output voltage divided by the load resistance. Therefore:

Output (Load) Current = RMS Output Voltage / R = 155.56 V / 10 Ω = 15.56 A

Ripple Factor:

The ripple factor for a full-bridge rectifier can be calculated as the ratio of the RMS value of the ripple voltage to the average output voltage. In this case, since we are neglecting diode volt-drops, the ripple factor is:

Ripple Factor = √(3/4) ≈ 0.866

Form Factor:

The form factor is the ratio of the RMS value of the output current to its average value. Since the load is purely resistive, the form factor is the same as the ripple factor:

Form Factor = 0.866

b. Now, assuming the load has an inductive nature with L >> R and a load current of 12 Amperes:

Input Active Power:

The input active power can be calculated as the product of the RMS input voltage, RMS input current, and the power factor. In this case, since the load current is flat and equal to 12 Amperes, and we neglect diode losses, the input active power will be:

Input Active Power = (220 V) * (12 A) = 2640 W

Input Apparent Power:

The input apparent power can be calculated as the product of the RMS input voltage and RMS input current. Therefore:

Input Apparent Power = (220 V) * (12 A) = 2640 VA

Power Factor:

The power factor is the ratio of the input active power to the input apparent power. In this case, the power factor will be:

Power Factor = Input Active Power / Input Apparent Power = 2640 W / 2640 VA = 1 (or unity)

Note: Neglecting diode losses implies that we assume the diodes are ideal, without any voltage drops or losses during the rectification process. In practical scenarios, there will be some voltage drops across the diodes, and losses should be taken into account for more accurate calculations.

Therefore, a. For a single-phase full-bridge uncontrolled (diode) rectifier with a pure resistive load of 10 Ohms, neglecting diode volt-drops, the average output voltage is 110 V, the RMS output voltage is approximately 155.56 V, the output (load) current is 15.56 A, the ripple factor is 0.866, and the form factor is 0.866. b. Assuming a load with an inductive nature, L >> R, and a flat load current of 12 A, the input active power is 2640 W, the input apparent power is 2640 VA, and the power factor is 1 (or unity).

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The amount of work done by the gas is proportional to the pressure and the change in volume, as well as the efficiency of the process. If the pressure and volume are known, the work done by the gas can be calculated by multiplying these values by the efficiency of the process.

The amount of work done by a gas when it expands is proportional to the change in volume, pressure, and temperature. According to the first law of thermodynamics, the energy of a closed system is conserved, so the work done by the expanding gas is equal to the energy transferred from the gas to the environment in the form of work. Therefore, the work done by the gas is equal to the change in energy of the system. Assume that the process is 10% efficient. Then, only 10% of the energy available to the system is converted into work. This means that the remaining 90% of the energy is lost to the environment in the form of heat. As a result, the amount of work done by the gas expanding into the atmosphere is given by the formula

W = E x η, where W is the work done by the gas, E is the energy available to the system, and η is the efficiency of the process. The energy available to the system is determined by the difference between the internal energy of the gas before and after the expansion. The internal energy of a gas is determined by its temperature, pressure, and volume.

Assuming that the temperature and pressure are constant, the change in internal energy is proportional to the change in volume. Therefore, the energy available to the system is equal to the product of the pressure and the change in volume: E = P x ΔV, where P is the pressure of the gas and ΔV is the change in volume during the expansion. Substituting this equation into the formula for work, we get W = P x ΔV x η.

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Answer:

yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain

The spring stretched 0.476 meter from equilibrium position.

The work done is given as,

                             \(workdone=\frac{1}{2}kx^{2}\)

Where x is the distance stretched from  equilibrium position.

Given that,   \(k=440N/m,W=25J\)

              \(25=\frac{1}{2}*440*x^{2} \\\\x^{2} =\frac{100}{440}=0.227\\\\x=\sqrt{0.227}=0.476m\)

Therefore, the spring stretched 0.476 meter from equilibrium position.

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The Carbon 14 isotope will take 34200 years to reduce to 1 gram if the half-life of the isotope is 5700 years.

Half-life is the amount of time it takes an unstable isotope to decay to half of its initial value. It is identified as \(t_{1/2}\).

Applying, the half-life formula:

A = A'( \(2^{x/y}\) )

Where A is the original mass of Carbon-14, A' is the final mass of carbon-14 after decaying, x is the total time, and y is the half-life.

We have,

A = 64 g, A' = 1 g, y = 5700 years.

Substitute these values in the equation,

A = A'( \(2^{x/y}\) )

64 = 1 ( \(2^{x/5700}\))

64 =  \(2^{x/5700}\)

\(2^{x/5700}\)  = 2⁶

Therefore,

x/5700 ≈ 6

Multiplying each side of the equation by 5700

x ≈ 6 × 5700

x ≈ 34200 years

Hence, it will take 34200 years for the isotope to reduce to 1 gram.

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Answer:

The very small particles that make up matter are I) Atoms

Matter - Anything that have mass and occupies space is called matter . it is made up of atoms and molecules

Atoms - The smallest part of matter is called atom.

Molecule - Group of atoms combine together to form a molecule.

Answer:

The first law states that if the net force is zero, then the velocity of the object is constant.

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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The rise in temperature of the body as determined from the specific heat capacity, mass, and heat change of the body is 12.24 °C.

The rise in temperature of the body is determined from the specific heat capacity, mass, and heat change of the body.

The formula relating the temperature rise of the body, the specific heat capacity, mass, and heat change of the body is given below:

Heat change = mass * specific heat capacity * temperature rise

Temperature rise = Heat change / mass * specific heat capacity

Temperature rise = 3.6 * 10⁶ J / 84 kg * 3,500 J/kg/°C

Temepartaure rise = 12.24 °C

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We can see from the calculations that the acceleration of the proton is 1.43 * 10^7 ms-2.

Acceleration is the rate of change of velocity with time. We know that the magnetic force is obtained by;

F = qvB

q= 1.6 * 10^-19 C

v = 365 m/s

B =  4.10*10^-5 T

F = 1.6 * 10^-19 C * 365 m/s *  4.10*10^-5 T

F = 2.39 * 10^-20 N

Recall that the mass of the proton is 1.67 × 10−27 kg hence

a = F/m = 2.39 * 10^-20 N/1.67 × 10−27 kg  = 1.43 * 10^7 ms-2

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Option A) 3.9 km is the correct answer. the magnitude of Rhea's displacement vector is approximately 3.92 km.

In order to find out the magnitude of Rhea's displacement vector, we have to add up all of the displacement vectors.

Then we can use the Pythagorean theorem to calculate the magnitude of the resultant vector.

Since Rhea is first driving north for 2.4 km and then west for 3.1 km, we can represent her displacement vectors as follows: Δx = 0 km and Δy = 2.4 km for the first vector, and Δx = -3.1 km and Δy = 0 km for the second vector.

We can then add these vectors together by adding their components: Δx = 0 km + (-3.1 km) = -3.1 km and Δy = 2.4 km + 0 km = 2.4 km.

This gives us a resultant vector of -3.1 km east and 2.4 km north.

Using the Pythagorean theorem, we can find the magnitude of this vector: \(\sqrt{(\(-3.1 km)^{2} + (2.4 km)^{2} ) } = \sqrt{(9.61 + 5.76) km^{2} } = \sqrt{15.37 km^{2} } \approx 3.92 km.\)

Therefore, the magnitude of Rhea's displacement vector is approximately 3.92 km.

Therefore, option A) 3.9 km is the correct answer.

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Explanation:

They speed up, slow down, or change direction.

Answer:In order for object to be accelerating, the net force (sum of all forces) acting on the object has to be non-zero. According to Newton's Second Law of Motion, Force = Mass x Acceleration. ... If the object is sitting motionless on the ground, then the net force (sum of all forces) acting on it is zero.

Explanation:

(a)  The work done by the force of gravity from A to B is 4.41 Joules.

(b)  The work done by the force of gravity from B to C is zero.

(c) The work done by the force of gravity from A to C is 4.41 Joules.

a) To calculate the work done by the force of gravity from A to B, we need to consider the change in potential energy. The potential energy at point A is maximum due to the maximum angle of 35.0∘ to the left of vertical, while at point B, the string is vertical, and the potential energy is zero.

The change in potential energy (ΔPE) is given by:

ΔPE = m * g * h

where m is the mass of the sphere (0.500 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the change in height.

Since the potential energy at point A is maximum, the change in height is equal to the length of the string (0.900 m).

ΔPE = 0.500 kg * 9.8 m/s^2 * 0.900 m = 4.41 J

Therefore, the work done by the force of gravity from A to B is 4.41 Joules.

b) From B to C, the change in height is zero since the string is already vertical. Hence, the work done by the force of gravity from B to C is zero.

c) The total work done by the force of gravity from A to C is the sum of the work done from A to B and from B to C.

Total work = Work from A to B + Work from B to C = 4.41 J + 0 J = 4.41 J

Therefore, the work done by the force of gravity from A to C is 4.41 Joules.

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I think it is the question:

A Pendulum Is Made Up Of A Small Sphere Of Mass 0.500 Kg Attached To A String Of Length 0.900 M. The Sphere Is Swinging Back And Forth Between Point A, Where The String Is At The Maximum Angle Of 35.0∘ To The Left Of Vertical, And Point C, Where The String Is At The Maximum Angle Of 35.0∘ To The Right Of Vertical. The String Is Vertical When The Sphere Is At

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.900 m. The sphere is swinging back and forth between point A, where the string is at the maximum angle of 35.0∘ to the left of vertical, and point C, where the string is at the maximum angle of 35.0∘ to the right of vertical. The string is vertical when the sphere is at point B.

a) Calculate how much work the force of gravity does on the sphere from A to B.

b) Calculate how much work the force of gravity does on the sphere from B to C.

c) Calculate how much work the force of gravity does on the sphere from A to C.

Answer:

The final speed of cart 2 is -1.0 m/s, and the final speed of cart 3 is 0.5 m/s.

To find the final speeds of cart 2 and cart 3 after the collision, we can use the principles of conservation of momentum and kinetic energy.

Given:

Mass of cart 1 (approaching cart): m = 0.20 kg

Initial velocity of cart 1: v0 = 1.5 m/s

Mass of cart 2: 2m

Mass of cart 3: 3m

Let's denote the final velocities of cart 2 and cart 3 as v2 and v3, respectively.

According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.

Initial momentum = Final momentum

(m * v0) + (2m * 0) + (3m * 0) = m * v2 + 2m * v3 + 3m * v3

Simplifying the equation, we have:

m * v0 = m * v2 + 5m * v3    ...(1)

Since the collision is elastic, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.

Initial kinetic energy = Final kinetic energy

(1/2) * m *\(v0^2 = (1/2) * m * v2^2 + (1/2) * 2m * v3^2 + (1/2) * 3m * v3^2\)

Simplifying the equation, we have:

(1/2) * m *\(v0^2 = (1/2) * m * v2^2 + m * v3^2 + (3/2) * m * v3^2\)

m * \(v0^2 = m * v2^2 + 2m * v3^2 + 3m * v3^2\)

\(v0^2 = v2^2 + 2v3^2 + 3v3^2\)    ...(2)

Now, we have two equations (equation 1 and equation 2) with two unknowns (v2 and v3). We can solve these equations simultaneously to find the values of v2 and v3.

From equation 1, we can rewrite it as:

v2 = v0 - 5v3

Substituting this expression into equation 2, we get:

\(v0^2 = (v0 - 5v3)^2 + 2v3^2 + 3v3^2\)

Expanding and simplifying the equation, we have:

\(v0^2 = v0^2 - 10v0v3 + 25v3^2 + 2v3^2 + 3v3^2\)

\(0 = -10v0v3 + 30v3^2\)

Rearranging the equation, we get:

10v0v3 = 30\(v3^2\)

v0 = 3v3

Solving for v3, we find:

v3 = v0/3 = (1.5 m/s) / 3 = 0.5 m/s

Substituting this value of v3 back into the expression for v2, we have:

v2 = v0 - 5v3 = 1.5 m/s - 5 * 0.5 m/s = 1.5 m/s - 2.5 m/s = -1.0 m/s

Therefore, the final speed of cart 2 is -1.0 m/s (indicating it moves in the opposite direction with respect to the initial velocity), and the final speed of cart 3 is 0.5 m/s.

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Answer:

dx = 2000 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

\(dx =v_{o}*t +\frac{1}{2} *a*t^{2}\)

where:

dx =  distance [m]

Vo = initial velocity = 100 [m/s]

a = acceleration = 20 [m/s²]

t = time = 10 [s]

dx = 100*10 + 0.5*20*10²

dx = 2000 [m]

Note: the positive sign in the above equation means that the car is increasing its velocity.

Answer:

A. the motion of an object along a curved path

Answer:

so you have your boxes go by tens on the x-axis okay which is the one going straight up then the second number goes by 5's is your y-axis. 30,6 so 30 is your x and 6 is your y. start at 0,0 which is the little corner piece. find 30 on x and then go over 6 and that's it!

Explanation:

Answer:

Overgrazing, which refers to the overutilization of grazing resources, by livestock promotes the increase in undesirable herbaceous plant species and bush encroachment, which are all indicators of rangeland degradation

Answer:

1/2 M1 V1^2 = 1/2 M2 V2^2       equal Kinetic Energy

M1 V1^2 = M2 V2^2

M1 V1 / (M2 V2) = V2 / V1

Since P (momentum) = M V

P1 / P2 = V2 / V1      Equation (1)

V2 / V1 = (M1 / M2)^1/2  from the first equation

V2 / V1 = (2 / 5)^1/2 = .63

P1 / P2 = .63 from Equation (1)    where 1 refers to 2 kg

Answer:

72 J

Explanation:

Use the Work formula

W= F x d

Given:

F - 48 N

d - 1.5 m

Solution:

W= F x d

W=  48 N x 1.5 m

W= 72 J

Answer:

Explanation:

Answer

How about a thunder storm. I don't know if you live in a city or out in the sticks as I do. Lighting is very obvious and it is not a good idea to be out when experiencing a thunderstorm, especially in an open field. You might be the only thing around that will cause the lightning to be connected to the ground.#  Seconds later, you will hear the thunder which is quite harmless. If you live in the city, observing this is not quite so easy. There are all kinds of buildings around and some of them may block your view.

#The great golfer, Lee Travino, was "hit" by lightening storm twice while playing in tournaments.