A nonconducting wall carries charge with a uniform density of 8.60μC/cm²(b) Does your result change as the distance from the wall varies? Explain.

When an emission line of helium from a galaxy is shifted toward red, it means that the galaxy is moving away from the viewer and is denoted as option D.

This refers to the systems of stars and interstellar matter that make up the universe which are gravitationally bound together and are made up of four main components which are:

When an emission line of helium from a galaxy is shifted toward red, there is a displacement to a longer wavelengths and this alteration is referred to as a doppler shift.

Moving away from the view means that the galaxy will be viewed in the red spectrum of the visible light which is a part of the electromagnetic spectrum.

This is therefore the reason why option D was chosen as the most appropriate choice.

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The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

It is very hard when pulling  water and if a person need to consistently pull air out of the tube before water makes it to the pump, it is very  difficult and some pumps may or cannot do it.

In regards to pushing, when a person is pushing above the Center of Gravity, one is pushing the said object into the ground and thus is increasing the rate of friction.

If a person is pulling you are said to be pulling above the Center of Gravity and lowering friction.

So, The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

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The speed of the wave in the rod is

[tex]The speed of the wave in the rod is $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$[/tex]

5044m/s

A sound wave is the pattern of disruption brought on by the movement of energy moving through a medium as it propagates away from the source of the sound (such as air, water, or any other liquid or solid substance). Pressure waves are produced when an item vibrates, such as a ringing phone, and these waves are known as sound waves. The surrounding medium's particles are disturbed by the pressure wave, and those particles disrupt the particles next to them, and so on. Like ocean waves, the disturbance's pattern causes outward movement in all directions. Usually in all directions and with decreased intensity as it gets further away from the source, the wave transmits the sound energy through the medium.

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Answer:

[tex]{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}[/tex]

Answer:

48.9 m/s

Explanation:

I am ignoring air resistance in this:

horizontal velocity stays constant, so we need to find vertical velocity

v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)

v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s

the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s

Answer:

V = 49.2 m/s

Explanation:

Given:

Vₓ  = 47.4 m/s

t = 1.23 s

___________

V - ?

Vy =  Voy + g·t = 0 + 9.8·1.23  ≈ 12,1 m/s

V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s

Ricardo came up with the three different paper airplane designs. Ricardo plans to fly each paper airplane five times to see which design flies the furthest. Ricardo should calculate the aircraft's flight distance after each flight and record it.

Two of the most popular methods in agricultural surveying involve taking measurements of angles and distances with simple equipment. even when the equipment is simple. A high level of precision can be reached with perseverance and practice.

The primary method for measuring distance is (1) pace. The mileage counter. (3) taping or chaining four phases and (6) methods for electronic distance measurement, optical range finders (EDM).

It will be helpful to first examine the common units of distance (displacement). Common English units are as follows:

A foot is made up of 12 inches (in)

Yd (yard) equals 3 ft (ft)

The length of one rod is 16.5 feet and 5.5 yards (yd)

5280 feet, 1760 yards, and 320 rods make up a mile.

The units cancellation method is very helpful when converting between different units of measurement.

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The electric field on the axis of the uniformly charged ring at the center of the ring at 5.00cm is 2.41×〖10〗^7 N/C

Where a positive charge (q) is given, always note that the charges point away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,

E=(K_e×b)/(r^3×Q)

Where

E is Electric field

Ke is electric field constant

a  is radius of the charge q from the center of the circle

Q is electric charge = 75.0μC

The radius of the charge is a = 10cm= 0.1m

Distance from the electric field = 5cm = 0.05m

Ke is a constant given as  8.99×〖10〗^(9  ) Nm^2/c^2

E=(8.99×〖10〗^(9  ) Nm^2/c^2×0.05)/(〖0.1〗^3×75×〖10〗^(-6) )

E=2.41×〖10〗^7N/C

Therefore, the electric field away from the center of the ring is 2.41×〖10〗^7N/C away from the center of the ring

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Answer:

gogle it

Explanation:

i cant help

The acceleration of Mr. Henderson Porsche is 54545.45 mi/hr².

Velocity is the directional speed of an object, body, or particle in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

We know that acceleration of a body is Change in Velocity per unit time.

[tex]a=\frac{v-u}{t}[/tex]

Here ,

v is the Final Velocity,

u is the Initial Velocity and;

t is the time taken.

In the given question ,

The Porsche accelerated from 0 to 60 miles per hour in four seconds.

u = 0 mi/hr

v = 60 mi/hr

t = 4 sec

[tex]t=\frac{4}{3600}{~}hr\\[/tex]

[tex]t=\frac{1}{900} {~}hr[/tex]

t  = 0.0011 hr

Then,

[tex]a=\frac{v-u}{t}[/tex]

[tex]a=\frac{60-0}{0.0011}[/tex]

[tex]a=54545.45 {~}mi/hr^{2}[/tex]

Hence, the acceleration of the Porsche is 54545.45 mi/hr².

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The dot product of velocity vector and acceleration vector at maximum height is equal to zero.

[tex]$\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

We have a football player who kicked a football in order to do a goal.

We have to determine the relationship between the direction of the velocity and acceleration vectors when the football reaches its maximum height.

Projectile motion is the two - dimensional motion of an object thrown or projected into the air such that it moves under the influence of acceleration of gravity.

According to the question -

Assume that the velocity with which the projectile was fired is u m/s.

Therefore, it can be resolved into two components as -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + u sinθ [tex]a_{y}[/tex].

Now, when the projectile reaches the maximum height, it will no longer cover any vertical height but it will keep moving in the horizontal direction. Therefore - the vertical component of the velocity will become zero. Therefore -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + 0 = u cosθ [tex]a_{x}[/tex]

Now, refer to the figure attached for reference -

At the maximum height, the velocity vector is in the horizontal direction and the vector for the acceleration due to gravity is vertically downwards.

Therefore - the acceleration vector and velocity vector will have an angle of 90° between them. Assume -

Velocity at maximum height = v = u cosθ [tex]a_{x}[/tex]

and acceleration = a = g = 9.8 m/[tex]s^{2}[/tex]

Therefore -

The dot product of velocity vector and acceleration vector is equal to zero.

[tex]\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

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The number of molecules in the given volume is 5.33 x 10⁻¹² moles.

The number of molecules of the gas is calculated as follows;

PV = nRT

n = PV/RT

where;

1 torr = 133 Pa

1 x 10⁻¹⁰ torr =  ?

= 1.33 x 10⁻⁸ Pa

n = (1.33 x 10⁻⁸ x 1) / (8.314 x 300)

n = 5.33 x 10⁻¹² moles

Thus, the number of molecules in the given volume is 5.33 x 10⁻¹² moles.

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A balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

Yes, opposite charges attract each other. When a positive charge and a negative charge interact with each other, their forces act from the direction of positive to the direction of negative charge. As a result opposite charges attract each other while on the other hand, similar charges repel each other because their forces move in the opposite direction so they repel each other.

So we can conclude that a balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

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Option(c) is Kilometer is correct. Kilometer unit for length is greater than a meter.

Worldwide adoption of the International System of Units (SI) as a system of measurement units is widespread. For convenience, this updated version of the metric system is based on the number 10. The metric prefixes or the SI prefixes are a collection of prefixes that have been established. According on the prefixes, the unit is either a multiple or a fraction of base ten. It enables the zeros of very small or very high numbers to be reduced, for example, from 7,500,000 Joules to 1 nanometer and from 0.000000001 meter to 7.5 Megajoules, respectively. Additionally, some SI prefixes have a group of symbols before the unit symbol.

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An astronaut could 6 times higher on the moon than on earth if his takeoff speed is the same in both locations.

From lunar circle, space explorers guided cameras through the window of their space apparatus toward catch photographs of the moon's surface. The nearest look we've had at the moon came from the sendoff of NASA's Apollo program during the 1960s. Somewhere in the range of 1967 and 1972, a progression of missions handled the primary men on the moon. At the point when the following space explorer to arrive at the moon strolls on the lunar surface in 2024, she'll confront radiation levels multiple times higher than on Earth. While Apollo mission space travelers. The prompt surface was extremely fine, as best we could see peering down from 15 feet. Off somewhere out there was an exceptionally clear skyline, perhaps with a rock.

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The expansion of both will be the same. Then the correct option is C.

Thermal expansion, which often excludes phase transitions, is the propensity of matter to alter its form, area, volume, and concentration in response to a variation in heat.

The same metal is used to create two identically sized spheres, but only one is solid and the other is hollow. The same rate of warming is applied to the spheres.

The formula for the thermal expansion is given as,

[tex]\rm \alpha _L = \dfrac{1}{L} \cdot \dfrac{dL}{dT}[/tex]

[tex]\alpha_{L}[/tex] = thermal expansion

L = particular length measurement

dL = change in length

dT = change in temperature

If the material of the sphere is the same, then the coefficient of the expansion of the sphere is the same, then the expansion of both will be the same.

Both will expand at the same rate. Then, C is the right answer.

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The magnitude of the electric field is 0 N/C.

We choose a Gaussian sphere of radius r > R concentric with the solid sphere.

Apply Gauss's law from Equation (1):

[tex]\Phi_{E} = \oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

Since the electric field vector E is parallel to the area vector dA of the curved part the dot product can be the simple product EdA:

                                 [tex]\oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

since the electric field is constant we can put E out of the integral:

                                [tex]E \oint dA = \frac{q_{in}}{\in_{0} }[/tex]

                                    [tex]EA = \frac{q_{in}}{\in_{0} }[/tex]

Substitute for A from Equation (2) and for [tex]q_{in}[/tex] from Equation (3)

recognizing that the enclosed charge by the Gaussian sphere is less than Q:

                              [tex]E(4\pi r^{2}) = \frac{pVgaussian}{\in_{0} }[/tex]

Substitute for V gaussian (volume of the Gaussian surface) from Equation (4):

                                          [tex]E(4\pi r^{2}) = \frac{\rho (\frac{4}{3} \pie r^{2})} {\in_{0} }[/tex]

                                           [tex]E = \frac{\rho r}{3\in_{0} }[/tex]

Where p is the charge per unit volume of the solid sphere which is equal to the total charge Q of the sphere divided by the volume V of the sphere :

                             [tex]E = \frac{(\frac{Q}{V} )r}{3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{(\frac{4}{3} \pi R^{3})3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{4\pi \in_{0}R^{3} }[/tex]

Substitute numerical values:

                                                  [tex]E = \frac{(26\times10^{-6})(0)}{4\pi(8.8542 \times10^{-12})(0.4)^{3}}[/tex]

                                 [tex]E = 0 N/C[/tex]

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Answer:

8.33 m/s^2

Explanation:

The equation for force is Force = mass * acceleration. The force is 500 N, and the mass is 60 kg, so substituting those into the equation you get that 500 = 60 * acceleration. Divide 500 by 60 to get the acceleration, which is 8.33 m/s^2.

Hope this helps! Let me know if you have any confusion!

(i) The largest mass will be of the object B.

(ii) The smallest mass will be of the object C.

(iii) The ratio of the mass of object A and b will be 3:5.

According to the Newton’s 2 law for the same force applied acceleration and mass will be in the inverse ratio .it means the more the acceleration the less will be the mass of the object for the same force applied. As the acceleration of the object B is less so its mass will be more and vice versa. And the ratio of the masses will be the inverse ratio of the acceleration of object.

Hence, the ratio will become inverse ratio of acceleration of both the object that is 3:5.

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The answers for question numbers 3 and 4 is correct. For question number 9, the average speed is 2.5 mi /hr and for question number 10 the speed is 5 yards / s.

We know that,

v = d / t

where,

v = Speed

d = Distance

t = Time

1 ) v = 360 / 6 = 60 km / h

2 ) v = 120 / 3 = 40 mi / h

3 ) v = 18 / 6 = 3 m / s

4 ) v = 1000 / 20 = 50 m / min

5 ) t = 6pm - 5 pm = 1 hr

    v = 2.5 / 1 = 2.5 mi / hr

6 ) v = 1.5 / 0.33 = 4.5 mi / hr ( Since 20 min = 20 / 60 = 0.33 hr )

   d = 4.5 * 1 = 4.5 mi

7 ) d = 20 * 1 = 20 mi ( Since 60 min = 1 hr )

8 ) d = 60 * 2 = 120 mi

9 ) Distance per lap = 0.5 mi

     Total laps = 10

     Total distance = 10 * 0.5 = 5 mi

     v = 5 / 2 = 2.5 mi / hr

10 ) v = 100 / 20 = 5 yards / s

Therefore, the answer for:

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The work required to pump all the water over the top of the tank of radius 5 m and depth 2 m is 5.12 x 10^6 J.

The work required to pump the water over the top of the tank is equal to the gravitational potential energy of the water inside the tank.

The potential energy is given by the following equation,

U= mgh

Here, m is the mass of water present inside the tank.

We know that, Density( Mass/Volume

Hence, m= Density x volume

So, the equation becomes

U= ρ x V x g x h

Where V is the volume of hemispherical tank= 2/3π x r^3

Putting the given values in the above equation,

U= 1000 x 2/3 x 3.14 x 53 x 9.8 x 2 = 5.12 x 10^6 J

Hence, the work required to pump all the water over the top of a tank of radius 5 m and a depth of 2 m is 5.12 x 10^6 J.

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Answer:

•So he stays healthy.

•He has energy and time to do it.

The frictional force is 11 N and the block’s acceleration is 0.14 m/s².

It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.

For part (a):

Taking x component:

F(x) = F cosθ

Taking y component:

F(y) = -F sinθ

Applying Newton’s second law to the y-axis:

F(N) = 15 sin40 + (3.5)(9.8)

F(N) = 44 N

coefficient U = 0.25

f(k) = 11 N

For part (b):

Applying Newton’s second law to the x-axis:

a = ((15) cos40 - 11)/3.5

a = 0.14 m/s²

Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².

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The car's acceleration would be for a mass of 4 kg is 3 m/s²

Given the information below:

12 Newtons of force.

Increasing speed Equals

By using Newton's Second Law of Motion, we can determine the mass of the car:

According to Newton's second law of motion, force is equal to the rate at which momentum changes. Force is defined as mass times acceleration for a constant mass.

force= mass* acceleration

This formula provides Newton's Second Law of Motion mathematically;

12= mass*3

mass=12/ 3

mass=4 kilogram

By changing the parameters in the formula, we obtain;

= 4 kilos of mass.

As a result, the vehicle with a mass of 4 kg would accelerate at 3 m/s².

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Answer: The diaphragm

Explanation:

The diaphragm is a thin dome-shaped muscle which separates the thoracic cavity (lungs and heart) from the abdominal cavity (intestines, stomach, liver, etc.)

Answer:

i think a or b is the answer

Explanation:

i also don't know exactly

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

[tex]B(h) = B(coil) = \frac{\mu_{0} NI}{2R}[/tex]

[tex]B(h) =\frac{4 \pi \times 10^{-7}\times(5)\times(0.6)}{0.3}[/tex]

B(h) = 12.6 μT

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Answer:

YOUR ANSWER IS HERE....

The Big Bang Theory is an example of the universe originated from the explosion and expansion of all matter and energy

The Big Bang Theory is the way that astronomers explain how the universe began as a tiny, dense, fireball that exploded 13.8 billion years ago. It includes Albert Einstein's general theory of relativity along with standard theories of fundamental particles.

The theory was born of the statement that other galaxies are moving away from our own at countless speed in all directions, as if they had all been forced by an ancient explosive force.

The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.

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(ii) The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

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The car’s velocity when it passed the checkpoint was of: 60 m/s

The formula and procedure we will use to solve this exercise is:

vi = [(2 *x)/ t] - vf

Where:

Information about the problem:

Applying the initial velocity formula we have:

vi = [(2 *x)/ t] - vf

vi = [(2 * 225m)/5 s ] - 30 m/s

vi = [(450 m)/5 s ] - 30 m/s

vi = 90 m/s - 30 m/s

vi = 60 m/s

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

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