A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)

Required:
a. Qualitatively, indicate the polarization of the plastic.

1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.

b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.

1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.

c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.

1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.

Answer:the answer is 3

Explanation:

After pushing the pumpkin hard, you find yourself reversing direction at a speed of 4.0 m/s. 3.0 seconds after being pushed, the pumpkin will slide 12 m. Assume you weigh 60.0 kg.

We can use the conservation of momentum to solve this problem. After the push, the momentum of the system is given by:

p = (449 kg + 60 kg) * v

where v is the speed of the pumpkin and you after the push. Since you end up sliding backward at 4.0 m/s, we have:

v = -4.0 m/s

Substituting this into the expression for momentum, we find:

p = (449 kg + 60 kg) * (-4.0 m/s) = -2036 kg·m/s

The negative sign indicates that the momentum of the system is in the opposite direction of your motion.

During the sliding motion, the net force on the system is given by:

Fnet = (449 kg + 60 kg) * g * sin(θ)

where g is the acceleration due to gravity (9.81 m/s^2) and θ is the angle of the ramp. Since the ramp is smooth and horizontal, θ = 0 and Fnet = 0. Therefore, there is no net force to change the momentum of the system.

Using the equation for motion with constant acceleration, we can find the distance the pumpkin slides in 3.0 seconds:

x = x0 + v0t + (1/2)at²

Since the initial speed of the pumpkin is -4.0 m/s and there is no net force acting on it, its speed remains constant during the slide. Therefore, v0 = -4.0 m/s and a = 0. Substituting these values, we find:

x = x0 + v0t = (-4.0 m/s) * (3.0 s) = -12 m

The negative sign indicates that the pumpkin slides in the opposite direction to your motion. Therefore, the pumpkin slides 12 meters backward (i.e., towards you) in 3.0 seconds after the push.

Learn more about speed here:

https://brainly.com/question/29100366

#SPJ1

When L = L₁, the circuit is a series RL circuit. The voltage across the inductor is given by Vₗ = XL₁i, where XL₁ = ωL₁ is the inductive reactance and i is the current flowing through the circuit. The voltage across the resistor is given by VR = Ri. The voltage across the capacitor is zero since it is connected in series with the inductor and the resistor.

The total voltage across the circuit is given by v = V√2 cos(ωt). By Kirchhoff's voltage law, we have v = Vₗ + VR,

V√2 cos(ωt) = XL₁i + Ri

The current i can be written as i = (1/Z) V√2 cos(ωt - φ), where Z = √(R² + XL₁²) is the impedance of the circuit and φ is the phase angle between the current and the voltage. Substituting i into the equation above, we get:

V√2 cos(ωt) = XL₁/Z V√2 cos(ωt - φ) + R/Z V√2 cos(ωt - φ)

Equating the coefficients of cos(ωt) and cos(ωt - φ), we get:

1 = XL₁/Z cos φ + R/Z sin φ

XL₁ sin φ = Z - R cos φ

tan φ = XL₁ / (Z - R cos φ)

The voltage across the inductor is given by:

Vₗ = XL₁ i = XL₁/Z V√2 cos(ωt - φ)

Vₗ/V = XL₁/Z cos φ

Substituting tan φ into this equation, we get:

Vₗ/V₁ = XL₁/Z₁ √[1 - (XL₁/Z₁)²] ... (1)

When L = L₂, the circuit is a series RC circuit. The voltage across the capacitor is given by VC = XC₂i, where XC₂ = 1/(ωC₂) is the capacitive reactance.  The total voltage across the circuit is given by v = V√2 cos(ωt). By Kirchhoff's voltage law, we have v = VC + VR, which gives:

V√2 cos(ωt) = XC₂i + Ri

Following the same steps as in the previous case, we can show that:

Vₗ/V₂ = XC₂/Z₂ √[1 - (XC₂/Z₂)²] ... (2)

Learn more about Kirchhoff's here:

https://brainly.com/question/30400751

#SPJ1

Answer:

2.753*10^-11N

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

Hence the gravitational force between the asteroid is 2.753*10^-11N

The properties of the young universe are:

Which is the youngest universe?

Light can only move so quickly. Telescope viewing of the universe's farthest objects reveals light that is billions of years old and is only now catching up to us. When you look far enough back, you can see the light coming from the very first galaxies. Hubble Space Telescope users revealed in March 2016 that they had discovered the furthest galaxy ever observed. It was created 400 million years after the Big Bang and is known as GN-z11. Young blue stars in the cosmos are a million times brighter than the sun. As it travels across the expanding cosmos, their light expands, changing from shorter blue wavelengths to longer red wavelengths.

GN-z11 appears clumsier than the typical spiral shapes seen for most galaxies. The scattered arrangement of its stars in Hubble photos resembles ink on paper. Scientists can learn more about the formation and evolution of such large structures by studying distant galaxies like GN-z11. To understand more, look through the photographs and video.

To learn more about GN-z11 visit:

https://brainly.com/question/29641305

#SPJ4

Answer:

(a) To solve for the initial speed of the rock at launch, we can use the kinematic equation:

v = v0 + at

Where:

v = final velocity (15m/s)

v0 = initial velocity (what we're solving for)

a = acceleration due to gravity (-9.8m/s^2)

t = time (2.0s)

Plugging in the values, we get:

15m/s = v0 - 9.8m/s^2 (2.0s)

v0 = 34.6m/s

Therefore, the initial speed of the rock at launch was approximately 34.6m/s.

(b) To solve for the speed of the rock 5.0s after launch, we can use the same kinematic equation:

v = v0 + at

But this time, we need to add the additional time and distance that the rock traveled after the initial 2.0s. To do this, we'll use the equation:

d = v0t + 1/2at^2

Where:

d = distance traveled

v0 = initial velocity (34.6m/s)

a = acceleration due to gravity (-9.8m/s^2)

t = time (5.0s - 2.0s = 3.0s)

Plugging in the values, we get:

d = (34.6m/s)(3.0s) + 1/2(-9.8m/s^2)(3.0s)^2

d = 103.8m - 44.1m

d = 59.7m

So, the rock traveled 59.7m in the additional 3.0s after the initial 2.0s. Now we can find its speed using the kinematic equation:

v = v0 + at

Where:

v0 = final velocity from before (15m/s)

a = acceleration due to gravity (-9.8m/s^2)

t = time (3.0s)

Plugging in the values, we get:

v = 15m/s - 9.8m/s^2 (3.0s)

v = -12.6m/s

Note that the velocity is negative because the rock is now moving downward. Therefore, the speed of the rock 5.0s after launch is approximately 12.6m/s.

Answer:

486,750 kg*m/s

Explanation:

Momentum is mass*velocity

M = m*v

M = 8850kg*55m/s

M = 486,750 kg*m/s

The momentum of an 8850 kg medium truck that is traveling with a velocity of 55 m/s west on the highway is 486750 kg m / s.

Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force acting on the particle equals the time rate of change of momentum.

According to Newton's second law, if a particle is subjected to a constant force for a specific amount of time, the result of the force and time (referred to as the impulse) is equal to the change in momentum.

Given:

The mass of the truck is, m = 8850 kg,

The velocity of the truck is, v = 55 m/s,

Calculate the momentum of the truck as shown below,

Momentum = m × v

Momentum = 8850 × 55

Momentum = 486750 kg m / s

Thus, the Momentum of the truck is 486750 kg m / s.

To know more about momentum:

https://brainly.com/question/904448

#SPJ2

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

The weight of the same object on a planet where the gravitational attraction has been reduced to 1/10 of the earth's pull is equal to the mass of the object .

The weight of an object on the earth's surface is equal to the product of  mass of the object and gravity of the earth.

So , w= m x g

we assume that g= 9.8 ms⁻² on the earth ,so the weight of the object on the earth = mass x 9.8 ms⁻²

Since we are given the condition that the weight of the same object on a planet where the gravitational attraction has been reduced to 1/10 of the earth's pull, the weight of the object on the earth

= mass x 9.8 ms⁻²/ 10,

Since the gravity in earth can be considered approx 10 ms⁻² , then

= mass x 10ms⁻²/ 10,

= mass of the object

which is approx that the weight of the object is equal to the mass of the object.

To know more about weight refer to the link https://brainly.com/question/10785551?referrer=searchResults.

#SPJ9

Which of the following are characteristics of noble gases?

\({ \bf{ \underbrace{Answer :}}}\)

\(\sf\red{B. \:They're\: inert.}\) ✅

\(\sf\purple{D.\: They \:don't \:react\: with\: other\: elements.}\)✅

\(\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘\)

Answer:

1- 66mL

2-16mL

3-7.8mL

Explanation:

they are numbered from left to right, and have the number of the values. Dont forget the measurements, because sometimes if they are forgotten, they are counted wrong.

The temperature gradient in the flow of direction is 294525 W.

A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.

Q= T/( L/ KA)

Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)

   = 294525 W

Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.

Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.

Therefore, The temperature gradient in the flow of direction is 294525 W.

To learn more about Temperature gradient, refer to the link:

https://brainly.com/question/13020257

#SPJ9

If both runner A and runner B start the race together, they will 1.125 m apart at the finish of the race

v = d / t

v = Velocity

d = Distance

t = Time

d = 4.5 m

For runner A,

t = 4.5 min

v = 4.5 / 4.5

v = 1 m / min

For runner B,

t = 6 min

v = 4.5 / 6

v = 0.75 m / min

Runner A finished the race first at 4.5 min. At that time the runner B will have covered,

0.75 = d / 4.5

d = 0.75 * 4.5

d = 3.375 m

Difference in distances = 4.5 - 3.375

Difference in distances = 1.125 m

Therefore, they will 1.125 m apart at the finish of the race

To know more about velocity

https://brainly.com/question/19979064

#SPJ9

19.3 gcm3 19.3 g c m 3 is equivalent to 19300 kgm3 due to the decimal place shifting due to the conversion being in metric measurements.

The amount of objects, including people, animals, plants, and other living things, that are present in a region is known as the density. The number of objects is divided by the area to determine density. The number of inhabitants in a nation divided by its size, measured in square kilometers or miles, is the population density of that nation.

In reference to the amount of matter contained in a thing, density refers to how much volume the object or substance occupies.

To know more Density visit:

https://brainly.com/question/29775886

#SPJ1

We are given two points on the distance-time graph: (1, 40) and (3,130)

This means that:

At time 1 hour, the distance traveled was 40 km

At time 3 hours, the distance traveled was 130 km

We want to find the average speed during this 2 hour interval (from 1 hour to 3 hours).

Average speed is defined as:

Average Speed = Change in Distance / Change in Time

The change in distance is the distance traveled from 1 hour to 3 hours, which is 130 km - 40 km = 90 km

The change in time is 3 hours - 1 hour = 2 hours

Average Speed = 90 km / 2 hours

= 45 km/hr

Therefore, the average speed during this interval of time is 45 km/hr.

The work done on the box from y=0 to 6.0m is 120 J.

To calculate the work done on the box from y=0 to 6.0m, we need to find the area under the force vs. position graph over that interval.

First, we can find the work done from 0 m to 2 m. Since the force is constant at 40 N over this interval, the work done is simply:

W = F * d = 40 N * 2 m = 80 J

From 2 m to 6 m, the force is constant at -20 N, so the work done is:

W = F * d = (-20) N * 4 m = -80 J

Note that the negative sign indicates that the work is done by the box on the force (since the force is in the opposite direction of the displacement).

Therefore, the total work done on the box from y=0 to 6.0m is:

W_total = 80 J - 80 J = 0 J

To know more about force, here

brainly.com/question/13191643

#SPJ1

Answer:

The final temperature of the bullets is 327.3 ºC.

Explanation:

Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:

\(K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0\) (1)

Where:

\(K_{A,o}\), \(K_{A}\) - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.

\(K_{B,o}\), \(K_{B}\) - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.

\(U_{A,o}\), \(U_{A}\) - Initial and final internal energies of the 15-g bullet, measured in joules.

\(U_{B,o}\), \(U_{B}\) - Initial and final internal energies of the 7.75-g bullet, measured in joules.

By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:

\(\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0\) (2)

Where:

\(m_{A}\), \(m_{B}\) - Masses of the 15-g and 7.75-g bullets, measured in kilograms.

\(v_{A,o}\), \(v_{A}\) - Initial and final speeds of the 15-g bullet, measured in meters per second.

\(v_{B,o}\), \(v_{B}\) - Initial and final speeds of the 7.75-g bullet, measured in meters per second.

\(c\) - Specific heat of lead, measured in joules per kilogram-Celsius degree.

\(T_{o}\), \(T\) - Initial and final temperatures of the bullets, measured in Celsius degree.

Now we clear the final temperature of the bullets:

\((m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})]\)

\(T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}\)

\(T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}\) (3)

If we know that \(T_{o} = 30\,^{\circ}C\), \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\), \(v_{A,o} = 295\,\frac{m}{s}\), \(v_{B,o} = 375\,\frac{m}{s}\), \(v_{A} = 0\,\frac{m}{s}\), \(v_{B} = 0\,\frac{m}{s}\) and \(c = 128\,\frac{J}{kg\cdot ^{\circ}C}\), then the final temperature of the collided bullets is:

\(T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}\)

\(T = 852.534\,^{\circ}C\)

Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:

\(\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0\) (4)

\(U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)\)

If we know that \(T_{o} = 30\,^{\circ}C\), \(T = 327.3\,^{\circ}C\), \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\), \(v_{A,o} = 295\,\frac{m}{s}\), \(v_{B,o} = 375\,\frac{m}{s}\), \(v_{A} = 0\,\frac{m}{s}\), \(v_{B} = 0\,\frac{m}{s}\) and \(c = 128\,\frac{J}{kg\cdot ^{\circ}C}\), then latent heat received by the bullets during impact is:

\(U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)\)\(U = 331.872\,J\)

The maximum possible latent heat (\(U_{max}\)), measured in joules, that both bullets can receive during collision is:

\(U_{max} = (m_{A}+m_{B})\cdot L_{f}\) (5)

Where \(L_{f}\) is the latent heat of fusion of lead, measured in joules per kilogram.

If we know that  \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\) and \(L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}\), then the maximum possible latent heat is:

\(U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)\)

\(U_{max} = 557.375\,J\)

Given that \(U < U_{max}\), the final temperature of the bullets is 327.3 ºC.

Answer:

7.65x10^3 m/s

Explanation:

The computation of the satellite's orbital speed is shown below:

Given that

Earth mass, M_e = 5.97 × 10^24 kg

Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg

Orbital radius, r = 6.80 × 10^6m

Based on the above information

the satellite's orbital speed is

V_o = √GM_e ÷ √r

= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6

=  7.65x10^3 m/s

The velocity time graph would have the velocity used instead of the position in this position time graph.

This is referred to a plot between velocity and time and it helps to show the motion of the object that moves in a straight line.

We should note that the velocity-time graph reveals the speed of an object (and whether it is slowing down or speeding up), while the position-time graph describes the motion of an object over a period of time which is therefore why it was chosen as the correct choice.

Read more about Velocity time graph here https://brainly.com/question/24191409

#SPJ1

Answer:

the sphere

Explanation:

From the given information,

A free flow body diagrammatic expression  for the small uniform disk and a small uniform sphere which are released simultaneously at the top of a high inclined plane  can be seen in the image attached below.

From the diagram;

The Normal force mgsinθ - Friction force F  = mass m × acceleration a

Meanwhile; the frictional force

\(F = \dfrac{I \alpha }{R}\)

where

\(\alpha = \dfrac{a}{R}\)  in a rolling motion

Then;

\(F = \dfrac{I a }{R^2}\)

∴

The Normal force mgsinθ - F  =  m ×  a     can be re-written as:

\(\mathtt{mg sin \ \theta- \dfrac{Ia}{R^2} = ma}\)

making a the subject of the formula, we have:

\(a = (\dfrac{mg \ sin \theta}{m + \dfrac{I}{R^2}})\)

Similarly;

I = mk²  in which k is the radius of gyration

∴

replacing I = mk² into the above equation , we have:

\(a = (\dfrac{mg \ sin \theta}{m + \dfrac{mk^2}{R^2}})\)

where;

the uniform disk \(\dfrac{k^2}{R^2 }= \dfrac{1}{2}\)  

the uniform  sphere \(\dfrac{k^2}{R^2 }= \dfrac{2}{5}\)

∴

\(a = \dfrac{2}{3} \ g sin \theta \ for \ the \ uniform \ disk\)

\(a = \dfrac{5}{7} \ g sin \theta \ for \ the \ uniform \ sphere\)

We can now see that the uniform sphere is greater than the disk as such the sphere will reach the bottom first.

In this case, the weight in air is 30.4 N and the weight in water is 17.0 N the density can be found using the formula:

density of object = (weight in air - weight in water) / volume of object

The volume of the statue is required to calculate the density using the formula:

density of object = (weight in air - weight in water) / volume of object

To determine the density of the ceramic statue, you can use the concept of buoyancy. The buoyant force on an object submerged in a fluid is equal to the weight of the fluid that is displaced by the object. This can be calculated using the equation:

density of object = (weight in air - weight in water) / volume of object

In this case, the weight in air is 30.4 N and the weight in water is 17.0 N. To calculate the volume of the object, you will need to measure it or have the information available. Once you have the volume, you can plug it into the above equation and calculate the density of the ceramic statue.

Learn more about buoyancy here:

https://brainly.com/question/19168433

#SPJ4

Answer:

Approximately \(0.10\), assuming that \(g = 9.81\; {\rm N\cdot kg^{-1}}\) and a level surface.

Explanation:

Under the assumption, the normal force between the ice and the hockey puck is equal to the weight of the puck:

\(\begin{aligned}m\, g &= (0.16\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}) \\ &\approx 1.57\; {\rm N}\end{aligned}\).

The friction on the puck is considered "static" when as long as the puck is not moving relative to that surface. In this question, the maximum value of this static friction is \(0.157\; {\rm N}\). When the external horizontal force exceeds \(0.157\; {\rm N}\!\), the puck would start moving relative to the ice.

Divide maximum static friction by the normal force to find the coefficient of static friction:

\(\begin{aligned}\mu_{s} &= \frac{(\text{maximum static friction})}{(\text{normal force})} \\ &\approx \frac{0.157\; {\rm N}}{1.57\; {\rm N}} \\ &\approx 0.10\end{aligned}\).

Answer:

Explanation:

An absorption line spectrum occurs when light passes through a cloud of cool gas in space, such as an interstellar molecular cloud, a planetary atmosphere, or a circumstellar disk. The light from a star behind the cloud is absorbed by the atoms and molecules in the gas, which have a range of temperatures and densities.

When the light passes through the gas, the atoms and molecules in the gas absorb light at specific frequencies that correspond to the energy levels in the atoms. This results in missing or reduced light at those frequencies, which appears as dark lines against the otherwise continuous spectrum of the star.

The specific wavelengths of light that are absorbed depend on the chemical composition of the gas and its physical conditions, such as temperature and pressure. By analyzing the pattern of absorption lines in the spectrum, astronomers can learn about the composition, density, and temperature of the gas, as well as the velocity of the gas along the line of sight.

Absorption line spectra can also provide information about the physical conditions in the star itself, such as its temperature, surface gravity, and chemical composition, by analyzing the strength and shape of the lines. This information can help astronomers to better understand the processes occurring in stars, as well as in the interstellar medium, and the conditions necessary for the formation of stars and planetary systems.

Answer: the rarest element is Francium. J is not on the periodic table. also Dmitri Mendeleev proposed the periodic table.

Explanation: Kinda looked the last one up.

Answer:

a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c)  λ = 0.486 m , d)     λ = 0.35 m

Explanation:

a) The speed of a wave on a string is

          v = √T /μ

also all the waves fulfill the relationship

          v = λ f

they indicate that the fundamental frequency is f = 980 Hz.

The wavelength that is fixed at its ends and has a maximum in the center

          L = λ / 2

          λ = 2L

we substitute

           v = 2 L f

let's calculate

           v = 2  0.243  980

           v = 476.28 m / s

b) The tension of the rope

             T = v² μ

the density of the string is

            μ = m / L

            T = v² m / L

            T = 476.28²   0.717 / 0.243

            T = 6.69 10⁵ N

c)          λ = 2L

            λ = 2  0.243

            λ = 0.486 m

d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air

          v = λ f

          λ= v / f

          λ = 343/980

          λ = 0.35 m

Answer: This (random) thermal motion of the particles due to the temperature is also called Brownian motion. ... The higher the temperature, the faster the diffusion will be, because the stronger the molecule movement and thus the “mixing”.

Explanation:

1. (NF)- splits the nucleus

2 lighter atoms

(BOTH)- releases lots of energy

2.(NF)- produces a heavy atom

combines 2 atoms

i’ve been up for 20 hours my dritydrops