A spherical surface surrounds a point charge q. Describe what happens to the total flux through the surface if the following happens:

(a) The charge is tripled.
*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(b) The volume of the sphere is doubled.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(c) The surface is changed to a cube.
*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(d) The charged is moved to another location inside the surface.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(e) The charge is moved outside the surface.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

To find the position along the x-axis where the sum of the forces exerted by the first two spheres on the third one is zero, we can use Coulomb's law. The forces between the charges are given by F = k * (|q1*q2| / r^2). By setting the forces equal to zero and solving for x, we can determine the desired position.

To find the location along the x-axis where a third sphere with charge q3 should be placed so that the sum of the forces exerted by the first two spheres on the third one is zero, we need to consider the forces between the charges. According to Coulomb's law, the force between two point charges is given by:

F = k * (|q1*q2| / r^2)

Where k is the Coulomb's constant, q1 and q2 are the charges of the two spheres, and r is the distance between them. In this case, the force exerted by q1 on q3 is:

F1 = k * (|q1*q3| / (x - x1)^2)

And the force exerted by q2 on q3 is:

F2 = k * (|q2*q3| / (x - x2)^2)

For the sum of the forces to be zero, we must have:

F1 + F2 = 0

Substituting the values of q1, q2, x1, and x2, we can solve for x to find the position where the sum of the forces is zero.

Here, a=2, b=−82, and

c=437. Plug these values into the formula to find x_{3} . You should find two possible positions for x_{3}

Answer:

a) (-367231.63i ,  367231.63i, 0) N/C

b) (0 , 0  , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

[tex]E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i[/tex]

Case x > 2.15

[tex]E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i[/tex]

Case -2.15 < x <+2.15

[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]

b)

Case x < -2.15

[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]

Case x > 2.15

[tex]E =E_{+} - E_{+} \\\\E = 0[/tex]

Case -2.15 < x <+2.15

[tex]E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i[/tex]

Answer:

20 A

Explanation:

Given

Spring pulled from position

x = 5A

we need to calculate total distance of one full cycle  of spring motion

if you see image below, you understand easily

When cycle complete

Its total distance become 20 A

Answer:

The reach of the tongue is 23 cm.

Explanation:

Hi there!

The equation of traveled distance is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = traveled distance at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Let´s calculate the distance traveled by the tongue of the chameleon during the first 20 ms (0.020 s):

The initial position and velocity are zero (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

x = 1/2 · 280 m/s² · (0.020 s)²

x = 0.056 m

Now, let´s find the distance traveled while the tongue moves at constant speed. But first, let´s find the velocity (v) of the tongue after the accelertation interval using the following equation:

v = v0 + a · t     (v0 = 0)

v = 280 m/s² · 0.020 s

v = 5.6 m/s

Then, the distance traveled at constant speed can be calculated:

x = v · t

x = 5.6 m/s · 0.030 s

x = 0.17 m

The reach of the tongue is 0.17 m + 0.056 m = 0.23 m = 23 cm.

In the centripetal movement, what happens with velocity is that it will remain constant, always pointing in its tangential direction of the trajectory. Said speed, although constant, will have a constant direction that will generate an acceleration that will always point towards the center of the circle radius. Both vectors as the turn is performed will always be perpendicular to each other.

The direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.

Circular motion is the motion of an object in a circle at a constant speed. As the object is constantly changing its direction as it moving in a circle at constant speed.

At all instances, the object is moving tangent to the circle. The direction of velocity vector is tangent to the circle, as the direction of velocity vector is towards the direction of object's motion.

An object moving in a circle is accelerating due to its change in direction. The direction of the acceleration is inwards.

Therefore, the direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.

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Answer:

2.43 s

Explanation:

Using newton's equation of motion.

T = (v-u)/g

Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.

Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²

T = (-14-14)/-9.81

T = 2.85 s

Note: We look for the object's speed at 5.0 m.

using

v² = u²+2gs.................................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Given:  u = 14 m/s, g = -9.81 m/s², s = 5.0 m

Substitute into equation 1

v² = 14²+(-9.81×5×2)

v² = 196-98.1

v = √97.9

v = 9.89

We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.

using

v = u+gt

t =(v-u)/g........................... Equation 2

Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s

Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²

t = (14-9.89)/-9.81

t = -4.11/-9.81

t = 0.42 s

Thus,

Time taken to reach 5.0 m above projection point = T-t

=2.85-0.42

2.43 s

It takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.

The time it takes for an object to reach a certain height when thrown upwards can be determined using equations of projectile motion.

In this case, with an initial speed of 14 m/s and a height of 5.0 m, we can use the equation :

[tex]$$y = y_0 + v_{oy}t - \frac{1}{2}gt^2$$[/tex]

Substituting the known values, we get:

[tex]$$5.0 = 0 + 14t - \frac{1}{2}(9.8)t^2$$[/tex]

By rearranging the equation and solving it,

we find that it takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.

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Answer:

R = 207.45 mm , θ_return = 18.47 south west

Explanation:

This vector addition exercise is schematized in the attachment where the displacements are

1    d1 = 120 mm west

2   d2 = 250mm at 45 south east

3   d3 = 280 mm at 30 east of nort.

R  is the final displacement that takes the goat to its initial point (origin)

The analytical way to perform this exercise is to find the components of each displacement and add them

Decompose the displacement using trigonometry

Displacement d1

       d1ₓ = 120 cos 180 = -120 mm

Displacement d2, with the angle measured from the axis this   θ = 270 + 45

     sin 45 = [tex]d2_{y}[/tex] / d2

     cos 45 = d2ₓ / d2

      [tex]d2_{y}[/tex]  = d2 sin45

      [tex]d2_{y}[/tex]  = 250 sin (270 + 45)

      [tex]d2_{y}[/tex]  = -176.77 mm

     d2ₓ = d2 cos (270 + 45)

     d2ₓ = 176.77 mm

displacement d3, for half the angle from the east axis  θ = 90-30 = 60

     sin 60 =  [tex]d3_{y}[/tex]  / d3

     cos 60 = d3ₓ / d3

      [tex]d3_{y}[/tex]  = d3 sin 60

     d3ₓ = d3 cos 60

      [tex]d3_{y}[/tex]  = 280 sin 60 = 242.49 mm

     d3ₓ = 280 cos 60 = 140 mm

Having all the displacement components we can find the total displacement

         Rₓ = d1ₓ + d2ₓ + d3ₓ

         Ry =  [tex]d1_{y}[/tex] + [tex]d2_{y}[/tex] +  [tex]d3_{y}[/tex]  

          Rₓ = -120 + 176.77 +140

         Rₓ = 196.77 mm

         Ry = 0 -176.77 +242.49

         Ry = 65.72 mm

Therefore the displacement you must make to return to the starting point is

         R = RA Rx2 + Ry2)

         R = RA (196.77 2 + 65.72 2)

         R = 207.45 mm

 We used trigonometry

        tan tea = RY / Rx

        tea = tan-1 Ry / Rx

        ea = tan-1 (65.72 / 196.77)

        tea = 18.47

This is the point where the girl is, to return to its origin this path must be serial, but in the opposite direction,

       θ_return = 18.47 south west

The problem involves adding vectors to find the total displacement, then finding the negative of this displacement to calculate the magnitude and direction of the fourth displacement.

This problem involves the mathematical concept of vectors, particularly in determining the resultant vector, which in this context represents the spelunker's path. First, we convert the movements into rectangular coordinates: going 120 mm west is -120i, going 250 mm 45° east of south is -250 cos(45)i - 250 sin(45)j, and going 280 mm 30° east of north is 280 cos(30)i + 280 sin(30)j. Adding these vectors together, we get the spelunker's total displacement vector. The negatives of this sum will represent the fourth displacement needed to get the spelunker back to where she started.

(A) The magnitude of the fourth displacement is the sum of these vectors taken as negatives, because she has to go back, which is equivalent to the length of the path taken. This can be calculated using Pythagoras' theorem for two dimensions.

(B) The direction of the fourth displacement is calculated by finding the angle made by the resulting vector with respect to one of the axes (for instance, the x-axis). For this, we take the inverse tangent of the y-coefficient over the x-coefficient of the vector.

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Answer:

unknown wavelength is 540 nm

Explanation:

given data

wavelengths =  600 nm

10th bright fringe overlap = 9th bright fringe of 600 nm

to find out

unknown wavelength

solution

we get  here first nth bright fringe from the central maximum at distance y

y = n ×[tex]\frac{\lambda D }{d}[/tex]   ...............1

and we have given 10th bright fringe overlap = 9th bright fringe of 600 nm

so here

10 × [tex]\frac{\lambda D }{d}[/tex] = 9 × 600 × [tex]\frac{D}{d}[/tex]    

solve it we get [tex]\lambda[/tex]

[tex]\lambda[/tex] = 9 × [tex]\frac{600}{10}[/tex]

[tex]\lambda[/tex] = 540 nm

so unknown wavelength is 540 nm

The unknown wavelength is equal to 540 nanometers.

Given the following data:

To determine the unknown wavelength:

At the maximum distance, the nth bright fringe is given by the formula:

[tex]Y = \frac{n\lambda D}{d}[/tex]

From the question, we were told that the 10th bright fringe of the unknown wavelength overlaps the 9th bright fringe of the 600 nm light.

Mathematically, this can be expressed as follows:

[tex]Y_{10} = Y_9\\\\\frac{10WD}{d} =\frac{9\times 600D}{d} \\\\10W=5400\\\\W=\frac{5400}{10}[/tex]

W = 540 nanometers.

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Answer:

Explanation:

Check the attachment for solution

The direction of the electric force is exerted by a 40-nC charged particle located at the origin of a cartesian coordinate system on a 15-nC charged particle located at (2.0 m, 2.0 m ) is  45° counterclockwise from the +x axis.

Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.

The given two charges are: 40 nC and 15 nC. As both of them are positive, the electric force is repulsive in nature.

15-nC charged particle located at (2.0 m, 2.0 m ).

So, the direction of repulsive force = tan⁻¹(2.0/2.0) = 45⁰.

Hence, The direction of the electric force is   45° counterclockwise from the +x axis. Option (e) is correct.

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The amount of charge on the plates increases during this process.

Explanation:

The relationship between charge and potential difference through a capacitor is

[tex]C=\frac{Q}{\Delta V}[/tex]

where

C is the capacitance

Q is the charge stored

[tex]\Delta V[/tex] is the potential difference

The capacitance of a parallel-plate capacitor is given by

[tex]C=\frac{\epsilon_0 A}{d}[/tex]

where

[tex]\epsilon_0[/tex] is the vacuum permittivity

A is the area of the plates

d is the distance between the plates

Combining the two equations, we get:

[tex]Q=\frac{\epsilon_0 A \Delta V}{d}[/tex]

In this problem:

- The potential difference between the plates, [tex]\Delta V[/tex], is kept constant

- The area of the plates, A, remains  constant

- The distance between the plates, d, is decreased

Since Q is inversely proportional to d, this means that as the plates are pushed together, the amount of charge on the plates increases during the process.

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Final Answer:

By sustaining a constant voltage across two conducting plates, the parallel-plate capacitor creates an electric field, gathers positive and negative charges, and preserves the battery's electric potential difference.

Explanation:

The amount of charge on the plates remains constant during the process of pushing the plates of a parallel-plate capacitor together. When a constant voltage (potential difference) is applied across the plates by a battery, it establishes an electric field between the plates. When you push the plates together without touching, you are essentially changing the distance between them while keeping the voltage constant.

The key point to understand is that the amount of charge on the plates is determined by the voltage (potential difference) and the capacitance of the capacitor, the charge on the plates is denoted by the equation Q = C * V.

In summary, when the plates of a parallel-plate capacitor are maintained with a constant voltage by a battery and pushed together without touching, the amount of charge on the plates remains constant throughout the process.

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Answer:

B. Yes, at the end of the trajectory.

Explanation:

By the conservation of energy, in order the object to have the same speed, it must have the same potential and kinetic energy, therefore, it must be at the same height as its initial height.

Answer:

b. Yes, at the end of the arch (end of trajectory)

Explanation:

according to conservation of energy and considering no air resistance:

initially the object has some kinetic energy due to speed which gets converted to gravitational potential energy till its maximum height.

since there is no energy loss, the object again gets to same amount of kinetic energy and hence the same speed.

Final answer:

The change in momentum of the basketball after impacting the floor and rebounding at the same speed is 2.268 kg·m/s upward.

Explanation:

You asked what the momentum change was when a basketball with a mass of 567 g hits the floor at a speed of 2 m/s and rebounds at nearly the same speed. To find the change in momentum, we need to consider the initial and final momentums of the basketball. The momentum of an object is given by the product of its mass and velocity. Given the initial velocity (vi) is 2 m/s downward and the final velocity (vf) is 2 m/s upward, and the direction is important in momentum calculations.

Momentum before impact = mass × velocity = 0.567 kg × (-2 m/s) = -1.134 kg·m/s (negative sign indicates downward direction)
Momentum after impact = mass × velocity = 0.567 kg × (2 m/s) = 1.134 kg·m/s (positive sign indicates upward direction)

To find the momentum change, Δp, we subtract the initial momentum from the final momentum:

Δp = pf - pi = 1.134 kg·m/s - (-1.134 kg·m/s) = 1.134 kg·m/s + 1.134 kg·m/s = 2.268 kg·m/s.

Therefore, the momentum change is 2.268 kg·m/s in the upward direction.

Answer:

(a) T1 = 938.3lb

(b) R = 665.5lb

The detail solution to this problem can be found in the attachment below.

This problem was solved by resolving the forces along the vertical and horizontal and equating to Rx and Ry respectively. Rx = 0 because the resultant is directed along the vertical.

Explanation:

See attachment below for the full solution.

Thank you for reading.

Final answer:

To find the tension T1 and the vertical resultant R, use trigonometry, considering that T1 must have an equal and opposite horizontal force component to T2 and a vertical component that combines with T2's vertical component.

Explanation:

Understanding the Tension in Two Portions of a Cable

The student's question pertains to the tensions in a telephone cable clamped at point A to a pole. We are given that the tension in the right-hand portion of the cable (T2) is 1000 lb and are asked to determine two things:

The required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical.

The corresponding magnitude of R.

To solve for the required tension T1 in the left-hand portion so that the resultant force R is vertical, we need to use trigonometry. Given that the resultant R must be vertical and that we already have a horizontal force component from T2, T1 must provide an equal and opposite horizontal force component to cancel it out, as well as contribute to the resultant vertical force. The magnitude of the horizontal component of T1 must equal that of T2, and the vertical component of T1 will combine with T2 to form the vertical resultant R.

To determine the magnitude of the resultant force R, which is vertical, we will sum up the vertical components of T1 and T2. This magnitude can be found using the Pythagorean theorem if necessary, or simply as the sum of the vertical components if T1 is already calculated and has the proper direction to ensure a vertical R.

For this specific scenario, without additional information such as angles or the tension division between T1 and T2, we cannot provide numerical answers but can describe the method to find the solution.

To develop this problem we will proceed to use the principle of energy conservation. For this purpose we will have that the change in the electric potential energy and kinetic energy at the beginning must be equal at the end. Our values are given as shown below:

[tex]m = 9.75g = 0.00975kg[/tex]

[tex]r = 1.3cm = 0.013m[/tex]

[tex]q_1 = 0.1 C/m^3 * \frac{4}{3} \pi r^3[/tex]

[tex]q_1 = 9.2*10^{-7}C[/tex]

[tex]q_2 = -9.2*10^{-7}C[/tex]

Applying energy conservation equations

[tex]U_1+K_1 = U_2+K_2[/tex]

[tex]\frac{k q_1q_2}{d} +0 = \frac{kq_1q_2}{2r}+ \frac{1}{2} (2m)v^2[/tex]

Replacing,

[tex]9*10^{9} (9.2*10^{-7})^2(\frac{1}{0.026}-\frac{1}{0.8}) = v^2[/tex]

Solving for v,

[tex]v = 9.2*10^{-7} (\frac{9*10^9}{0.00975}(\frac{1}{0.026}-\frac{1}{0.8}))^{1/2}[/tex]

[tex]v = 5.4 m/s[/tex]

Therefore the speed of each sphere at the moment they collide is 5.4m/s

To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

[tex]\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}[/tex]

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.

Therefore the correct answer is B.

Answer:

Weight of the object will be 195.6 N

Explanation:

We have given mass of the object m = 20 kg

And acceleration due to gravity at any location is given [tex]g=9.78m/sec^2[/tex]

We have to find the weight of the object in N at the location where value of acceleration due to gravity is [tex]g=9.78m/sec^2[/tex]

Weight of the object is given by W = mg , here W is weight , m is mass and g is acceleration due to gravity

So weight W = 20×9.78 = 195.6 N

So weight of the object will be 195.6 N

The weight of an object is calculated by multiplying its mass with the acceleration due to gravity. Given a mass of 20 kg and gravitational acceleration of 9.78 m/s2, the weight of the object is 195.6 N.

The subject of this question is Physics, specifically dealing with the concept of weight, which is a force. The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity. Given in this problem, the object's mass is 20 kg and the place's gravitational acceleration is 9.78 m/s2.

To find the object's weight, we use the formula: Weight = mass x gravity. Applying the values, we get: Weight = 20 kg x 9.78 m/s2 = 195.6 N.

Therefore, the weight of the object at the given location is 195.6 Newtons.

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Answer:

True, the work is done by the gravitational force on the object being dropped from the second floor of a building.

Explanation:

When an object is dropped  from the second floor of a building then no force is applied by the person on the object but under the influence of gravity the object experience a force equal to its own weight and travels a displacement from the second floor to the ground.

Hence in this case the work is done by the gravitational force but not by the person dropping the object.

As we know that the work done is given as:

[tex]W=F.s[/tex]

where:

F = force on the object getting displaced by a length s

Here:

[tex]F=m.g[/tex]

where, m = mass of the object

g = acceleration due to gravity

Answer:

0.067 seconds

Explanation:

The time needed for a person seated in the middle row to receive a sound from the stage, can be calculated assuming that the sound is moving in a straight line at constant speed, taking into account that the distance traveled will be half of the distance between the stage and the back wall:

t₁ = x₁ / v = 11.5 m / 343 m/s = 0.034 sec

The same sound, after reflecting from the back wall, will travel a distance equal to one and a half the distance between the stage and the back wall:

t₂ = 34.5 m /343 m/s = 0.101 sec

The time elapsed between both sounds will be equal to the difference between t₂ and t₁, as follows:

t₂ - t₁ = 0.101 sec - 0.034 sec = 0.067 sec.

Answer:

B = 1.67 μ T

Explanation:

given,

current, I = 1 x 10⁴ A

r = 120 m

treating lightning bolt as long straight conductor

  [tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

  [tex]B=\dfrac{4\pi \times 10^{-7}\times 1 \times 10^4}{2\pi\times 120}[/tex]

resulting magnitude would be equal to

    B = 16.67 x 10⁻⁶ T

    B = 1.67 μ T

The resulting magnetic field is equal to B = 1.67 μ T

Answer:

1.67 x 10^-5 Tesla

Explanation:

Current, i = 1 x 10^4 A

distance, d = 120 m

The formula for the magnetic field due to long straight current carrying conductor is given by

[tex]B =\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}[/tex]

[tex]B =10^{-7}\times \frac{2\times 10^{4}}{120}[/tex]

B = 1.67 x 10^-5 Tesla

The height that an object reaches in a gravitational field is dependent on its initial speed and the strength of the field. With the same initial speed, an object in a weaker gravitational field will reach a greater height. Therefore, material ejected with the same speed on Io and Earth will reach a greater height on Io due to its weaker gravitational field.

To understand how high material would go on Earth if ejected at the same speed as on Io, we first need to consider that the height that an object reaches in a gravitational field (ignoring air resistance) depends on its initial speed and the strength of the gravitational field. Specifically, the maximum height is given by the equation H = (v^2)/(2g), where v is the initial speed, and g is the acceleration due to gravity.

When an object is thrown upwards, it slows down under the force of the Earth's gravity until its speed decreases to zero. At that point, it begins to fall back down. Therefore, if two objects are thrown upwards with the same speed but under the influence of different gravitational fields (one stronger and one weaker), the object in the weaker gravitational field will reach a greater height.

In the case of Io, Jupiter's moon, the strength of the gravitational field is much less than that on Earth because its mass is much smaller. That's why volcanic material can be ejected to such high altitudes. Now, if we were to eject material with the same speed on Earth, which has a stronger gravitational field compared to Io, it would not reach the same height.

To calculate the specific height that material would reach on Earth assuming the same ejection speed as on Io, we would need to know that ejection speed. Unfortunately, the question does not provide this information.

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material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.

To determine how high material ejected from Io would go on Earth if it were ejected with the same speed, we can use the concept of escape velocity and the energy conservation principle.

The escape velocity [tex](\( v_e \))[/tex] of a celestial body is the minimum speed an object must have to break free from its gravitational pull and escape into space. It is given by the formula:

[tex]\[ v_e = \sqrt{\frac{2GM}{r}} \][/tex]

Where:

- [tex]\( v_e \)[/tex] is the escape velocity,

- ( G ) is the universal gravitational constant[tex](\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]

- ( M \ is the mass of the celestial body,

- ( r ) is the distance from the center of the celestial body.

Let's calculate the escape velocity of Io first:

[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \times 8.93 \times 10^{22} \, \text{kg}}{1821 \times 10^3 \, \text{m}}} \]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 8.93}{1821}} \times 10^{11} \, \text{m/s} \]\[ v_e = \sqrt{2 \times 3.34} \times 10^{11} \, \text{m/s} \]\[ v_e \approx 8.19 \times 10^4 \, \text{m/s} \][/tex]

Now, let's assume that material ejected from Io is given this speed of [tex]\( 8.19 \times 10^4 \, \text{m/s} \)[/tex] on Earth. We can calculate how high it would go using energy conservation principles.

The kinetic energy ( KE ) of the material is given by:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Where:

- \( m \) is the mass of the material, and

- \( v \) is the velocity.

The potential energy (\( PE \)) gained by the material at height \( h \) above the surface of Earth is given by:

[tex]\[ PE = mgh \][/tex]

Where:

- ( g ) is the acceleration due to gravity on Earth[tex](\( 9.81 \, \text{m/s}^2 \))[/tex],

- ( h ) is the height above the surface of Earth.

At the maximum height, the kinetic energy is converted entirely into potential energy. So we have:

[tex]\[ \frac{1}{2} m v^2 = mgh \]\[ \frac{1}{2} v^2 = gh \]\[ h = \frac{v^2}{2g} \][/tex]

Substitute the values:

[tex]\[ h = \frac{(8.19 \times 10^4 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]\[ h = \frac{6696.61 \times 10^8 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]\[ h \approx 3.41 \times 10^8 \, \text{m} \][/tex]

So, material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.

The magnitude of the initial acceleration is 11.52 m/s^2.

Acceleration is a fundamental concept that describes how an object's velocity changes over time. Velocity includes both speed and direction, so acceleration can involve changes in speed, direction, or both. When an object speeds up, slows down, or changes direction, it experiences acceleration.

To find the magnitude of the initial acceleration, we need to use Newton's Law of Gravitation and Coulomb's Law. The weight and electric force acting on the sphere will determine its acceleration. The weight of the sphere is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. The electric force is given by the formula[tex]F = k(q1q2/r^2)[/tex], where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the spheres.

First, let's calculate the weight of the sphere:

[tex]W = (0.19 g)(9.8 m/s^2) = 1.862 N[/tex]

Next, let's calculate the electric force:

[tex]F = (8.99 x 10^9 Nm^2/C^2)(21.0 x 10^-9 C)^2 / (0.10 m)^2 = 0.4083 N[/tex]

Now, we can calculate the acceleration using Newton's second law:

F = ma, so a = F/m

a = (1.862 N + 0.4083 N) / 0.19 g

a = 11.52 m/s^2

Therefore, the magnitude of the initial acceleration is [tex]11.52 m/s^2.[/tex]

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Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5  ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

Answer:

Force acting on the basketball will be 351.85 N

Explanation:

We have given velocity of baseball is changes from 50 m/sec to 45 m /sec in 0.00135 sec

So initial; velocity of the ball u = 50 m /sec

And final velocity of the ball v = 45 m /sec

Mass of the basketball m = 95 gram = 0.095 kg

Time taken t = 0.00135 sec

From third equation of motion v = u+at

So [tex]45=50+a\times 0.00135[/tex]

a = 3703.70 [tex]m/sec^2[/tex]

We have to find the force needed to change the speed of the basketball

According to newtons law force is given by F = ma

So F = 0.095×3703.70 = 351.85 N

Answer: r= 0.04meter

Explanation:

The velocity V and angular speed w is related to the radio by

V=w*r

r=V/w

But the standard unit for velocity is m/s so we convert 57.3cm/sec to m/s

57.3/100= 0.57m/s

Also Angular speed w is rad/secs

So we convert rev/min to rad/sec

0.1rad/sec equals 1 rev/mins

141rev/mins * 0.1rad/secs

=14.1rad/sec

r = V/w

= 0.573/14.1

= 0.04m to 2d.p

The radius of the spool is 12.3 cm.

To find the radius of the spool, we need to use the formula for the linear speed of a point on the edge of a rotating object: v = ωr. Here, v is the linear speed, ω is the angular velocity, and r is the radius. By converting the given information, we can calculate the linear speed:

Linear speed (v) = (57.3 cm/sec) / (100 cm/m) = 0.573 m/sec

The angular velocity can be found by converting the rate of revolution to radians per second:

Angular velocity (ω) = (141 rev/min) * (2 π rad/rev) / (60 sec/min) = 4.676 rad/sec

Now we can rearrange the formula to solve for the radius:

Radius (r) = v / ω = 0.573 m/sec / 4.676 rad/sec = 0.123 m = 12.3 cm

Answer:

Explanation:

ocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itNBself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms ofNGHJGHHG some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.NNNNN

a) The velocity of propogation of the wave  V=w/k

b) The y velocity v_y(x,t) of a point on the string as a function of x v=-wAcos(kx-wt)

A wave can be described as a disturbance that travels through a medium from one location to another location

y(x,t)=Asin(kx−ωt) defines the wave equation.

a)The velocity of propogation of the wave

We are asked to find wave speed (v)

Recall that v = fλ

From the wave equation above,

k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k

ω = 2πf where f is the frequency and ω is the angular frequency.

f = ω/ 2π.

By substituting for λ and ω into the wave speed formulae, we have that

v =( ω/ 2π) × (2π /k)

v = ω/k

b)The y velocity v_y(x,t) of a point on the string as a function of x

y(x,t)=Asin(kx−ωt)

The first derivative of y with respect to x give the velocity (vy)

By using chain rule, we have that

v = dy/dt = A cos( kx −ωt) × (−ω)

v = - ωAcos( kx −ωt)

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Answer:

a)V = 25.1 m/s

b)V = 4.226 m/s

Explanation:

Given that

x(t)=A t + B t​²

A = -4.3 m/s

B = 4.9 m/s​²

x(t)=  - 4.3 t +4.9 t​²

The velocity of the particle is given as

[tex]V=\dfrac{dx}{dt}[/tex]

V=-4.3 + 4.9 x 2 t

V= - 4.3 + 9.8  t m/s

Velocity at point t= 3 s

V= - 4. 3 + 9.8 x 3 m/s

V= - 4.3 + 29 .4 m/s

V = 25.1 m/s

At origin :

x= 0 m

0 =  - 4.3 t +4.9 t​²

0 = - 4.3 + 4.9 t

[tex]t=\dfrac{4.3}{4.9}\ s[/tex]

t=0.87 s

The velocity at t= 0.87 s

V= - 4.3 + 9.8  t m/s

V= - 4. 3 + 9.8 x 0.87 m/s

V= - 4.3 + 8.526 m/s

V = 4.226 m/s

a)V = 25.1 m/s

b)V = 4.226 m/s

The velocity of the particle at t = 3.0s is 25.1 m/s.

The velocity of the particle when it is at the origin is 4.324 m/s.

The given parameters:

The velocity of the particle at t = 3.0s is calculated as follows;

[tex]v = \frac{dx}{dt} = A + 2Bt\\\\v = -4.3 + 2(4.9\times 3)\\\\v = 25.1 \ m/s[/tex]

The velocity of the particle when it is at the origin (x = 0)

[tex]0 = -4.3t + 4.9t^2\\\\0 = t(-4.3 + 4.9t)\\\\t = 0 \ \ \ or \ \ -4.3 + 4.9t = 0\\\\4.9t = 4.3\\\\t = \frac{4.3}{4.9} \\\\t = 0.88 \ s\\\\v = A + 2Bt\\\\ v = -4.3 + (2\times 4.9 \times 0.88)\\\\v = 4.324 \ m/s[/tex]

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Final answer:

The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law.

Explanation:

The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the iron nucleus (+26e) is given as 26 times the charge of an electron (e).

The formula to calculate the electric force is: F = k * |q1 * q2| / r^2, where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.

Using this formula, we can plug in the values: F = (8.99 × 10^9 Nm^2/C^2) * |(26e) * (-e)| / (1.5 × 10^(-12)m)^2. Simplifying this equation will give you the magnitude of the electric force.

The horizontal and vertical components of the velocity vector for the golf ball hit with an angle of elevation of 30° and speed of 20 m/s are determined to be 17.32 m/s and 10 m/s, respectively.

Velocity Vector Components

The components of the velocity vector, which are horizontal and vertical, are calculated by multiplying the speed of the object by cos(θ) for the horizontal component and sin(θ) for the vertical component, where θ is the angle of elevation.

Given that the angle of elevation is 30 degrees and the speed is 20 m/s, we can use these equations to find:

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Answer:

Acceleration of the cart in the beginning is positive

Speed is positive when speeding up and slowing down.

Speed is zero when the cart stops.

Speed is negative when the cart speeds up in opposite direction.

Acceleration is negative when speeding up in opposite direction from rest.

Explanation:

When the cart is speeding up then its speed is increasing with time hence its acceleration is having a positive  value.

Answer:

Light takes 14.12 minutes to travel from the Sun to tatoone.

Explanation:

The equation for the average velocity can be used to estimate the time that light will take to travel from the Sun to tatoone. The average velocity is defined as:

[tex]v = \frac{d}{t}[/tex]   (1)

Where v is the velocity, d is the covered distance and t is the time.  

Therefore, t can be isolated from equation 1:    

[tex]t = \frac{d}{v}[/tex]            

It is necessary to express the speed of light in terms of minutes:

[tex]3x10^{8} \frac{m}{s} . \frac{60 s}{1 min}[/tex] ⇒ [tex]1.8x10^{10} m/min[/tex]

An astronomical unit is defined as the distance between the Earth and the Sun ([tex]1.496 x 10^{11} m[/tex]).        

[tex]d = 1.7 AU \cdot \frac{1.496 x 10^{11} m}{1AU}[/tex] ⇒ [tex]2.5432x10^{11}[/tex]m

Finally, equation 2 can be used:

[tex]t = \frac{d}{v}[/tex]      

[tex]t = \frac{2.5432x10^{11}m}{1.8x10^{10} m/min}[/tex] 

t = 14.12 min                

Hence, light takes 14.12 minutes to travel from the Sun to tatoone.