a plastic rod that has been charged to − 15 nc touches a metal sphere. afterward, the rod's charge is − 1.0 nc .

Answer:

Explanation:

The answer will be the 1.66N

According to the 1 equation of motion:

v=u+at; where v=final velocity, u=initial velocity, a= acceleration, t=time

According to the given value in question

0=5+a*0.60

a= -5/0.60

a=-13.88m/s2

Now according to the Newton’s 2 law

F=ma; where F=force, m=mass

F=0.12*13.88

F=1.66N

So, the average resultant force acting on the stone while it is being fired will be 1.66N.

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The correct answer is Equal to the weight of the block.

We are aware that when an object floats, the water it displaces weighs the same as the weight of the floating body. The upthrust is now equivalent to the body weight in this situation. The beaker is overflowing with water in this instance. The body was then created to float into it. When it reaches equilibrium in water, it expels water that must have been the same weight as it. Over the edge, the liquid spilled, collecting on the tray.

Therefore, the amount of water gathered on the tray must match the weight of the person floating.

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Answer: Given -

To find -

Solution -

[tex]A=\frac{v-u}{t}\\A=\frac{12-0}{36} \\A=\frac{12}{36} \\A=0.33m/s^2[/tex]

Therefore, the acceleration of the ball is 0.33 m/s².

I hope this helps.

The period of oscillation is 0.5 second.

We need to know about period of oscillation to solve this problem. Period is how much time taken to do one oscillation. It can be determined as

T = t / n

where T is period, t is time and n is how many oscillations.

From the question above, we know that

n = 8

t = 4 s

By substituting the parameters, we get

T = t / n

T = 4 / 8

T = 0.5 second

Hence, the period of oscillation is 0.5 second.

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Answer:

The plates move toward each other

Explanation:

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GEO 1405 Chapter 2 Quiz

The current must be along +x axis.

A stream of charged particles, such electrons or ions, traveling through an electrical conductor or a vacuum is known as an electric current. It is calculated as the net rate of passage of electric charge through a surface or into a control volume. Charge carriers, which can be any one of a number of particle kinds depending on the conductor, are the moving particles. Electrons travelling through a wire are frequently the charge carriers in electric circuits. They may be electrons or holes in semiconductors. Ions are the charge carriers in an electrolyte, whereas ions and electrons make up plasma, an ionized gas.

Explanation:

Given,

Velocity of proton,V = [tex]V_{i} i[/tex]

Magnetic force,F = [tex]F_{i} j[/tex]

Magnetic Field= B

(a)

If proton (p+) is moving in ve x-direction.

So,current(I) must be along +x axis.

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At a point 2 cm horizontally east of the wire's center, the magnetic field generated by the vertical wire is pointing south. As a result, option B south is the correct answer.

There are north and south poles on every magnet. The same poles repel one another whereas opposite poles are attracted to each other. The north-seeking poles of the iron's atoms pairing in the same direction when it is touched against a magnet. A magnetic field is produced by the force that the aligned atoms produce. The iron object has changed into a magnet. An electric current has the ability to magnetize specific materials. A magnetic field results from the flow of electricity through a coil of wire. But as quickly as the electric current is cut off, the field around the coil will disappear.

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The level of sound intensity at the late-night is β₂ = 67.0dB

Acoustic measurements are made with a sound level metre, often known as a sound pressure level metre (SPL). A hand-held device with a microphone is the most typical type. The condenser microphone is the best kind of microphone for sound level metres because it combines accuracy with stability and dependability. Sound waves generate changes in air pressure, which the microphone's diaphragm reacts to. The device is sometimes referred to as a sound pressure level metre because of this (SPL).

We are given:

The level of sound intensity in the afternoon is  β₁ = 80dB

No. of cars per min in the afternoon = 100

No. of cars per min in late night = 5

The level of sound intensity in the late night β₂ =?

The level of sound intensity in the late night I² =?

Reference sound intensity, I₀ = 10⁻¹²W/m²

The sound intensity level is given by: β = 10Log(I/I₀)

Numerically evaluating,

β₂ = 10 log (5X10⁻⁶/10⁻¹²)

β₂ = 67.0dB

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The velocity of Sam just before hitting the ground is 294 m/s.

The above situation represents a case of motion in one dimension.

This type of motion is governed by the following three equations of motion,

v = u + at

v² - u² = 2as

S = ut + 1/2 at²

As in the given case, the acceleration and time have been given and the final velocity is to be calculated, therefore the 1st equation can be used,

v = u+ at

As Sam jumped from the plane, his initial velocity is zero.

So,

v = 0 + 9.8(30)

v = 294 m/s.

Thus, Sam's velocity just before hitting the ground is 294 m/s.

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It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

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Answer: a it goes north b goes south

Explanation:

One pith ball's charge, q1, is equal to 30 nC, or 30x10-9C.

the second pith ball's charge, q2 = -30n c= -30x 10-9 c

They are separated by d=8 cm, or 8x10-2 metres.

They repel one another by a force of F = (81/64) x 10-3.

F = 1.266x10^-3 N

The gravitational force is solely an attracting one in both the General Theory of Relativity and Newton's theory of gravity. But quantization of gravity demonstrates that there are opposing gravitational forces as well.

A force between two or more opposite or different charges is called attraction. Two charges that are at odds with one another are attracted to one another. A force between two or more like or comparable charges is called repulsion. Two identical charges separate themselves.

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The problem of early water runoff that engineers might be asked to solve is the issue of Impervious surfaces, or surfaces that are unable to absorb water.

They are:

In the water cycle, runoff is known to be when water just "streaming off" the surface of the ground. The rain need to also drains downhill, just like the water a person use to wash their  car does as a person work are all part of runoff.

Therefore, The problem of early water runoff that engineers might be asked to solve is the issue of Impervious surfaces, or surfaces that are unable to absorb water.

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Magnetic field due to current in wire is 0.1308 Tesla.

The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing those surfaces from moving. Typically, it is represented by the Greek letter mu.

Given:

[tex]Mass\;per\;unit\;length(\frac{m}{l})=1.0\;g/cm=0.1\;kg/m\\[/tex]

[tex]Coefficient\;of\;friction,\;(\mu)=0.2[/tex]

[tex]Current,\;(I)=1.5\;A[/tex]

[tex]According\;to\;the\;question,\;wire\;is\;moving\;on\;a\;surface\;with\;\mu\;and\;force\;that\;is[/tex][tex]moving\;the\;wire\;is\;magnetic\;force\;f_{m} , then,[/tex]

[tex]f_{m}=\mu f_{N} (f_{N}=normal\;reaction\;of\;wire)[/tex]

[tex]Also, f_{m}=B\;L\;I\;sin\propto[/tex]

Therefore,

[tex]B_{b}=\frac{\mu\;m\;g}{L\;I\;sin\propto}[/tex]

[tex]B=magnetic\;field[/tex]

[tex]f_{N}=mass \times accelaration\;due\;to\;gravity[/tex]

Therefore,

[tex]B_{b}=\frac{0.2\times0.1\times10}{1.5\times sin\propto}[/tex]

[tex]sin \propto = angle\;magnetic\;field\;and\;the\;current\;in\;the\;wire=sin\;90^{\circ}=1[/tex]

Therefore,

[tex]B_{b}=0.1308\;T[/tex]

Now, from the figure shown and using Fleming's Right Hand rule in which the fore finger indicates current that is in east direction and the thumb indicates force that is in north direction and then direction, shown by your middle finger shows the direction of induced magnetic field.

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The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

[tex]K.E_m_a_x=E-\phi[/tex]      ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  [tex]ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV[/tex]

As we know that [tex]E=\frac{hc}{\lambda}[/tex]  ....(3)

where Speed of light,[tex]c = 3\times10^8 m/s[/tex] and Planck's constant , [tex]h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs[/tex]

From equation (3) , we get

[tex]\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm[/tex]

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The final velocity of the car, given the data is 32.88 m/s

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

The following data were obtained from the question:

The final velocity of the car can be obtained as follow:

a = (v – u) / t

1.6 = (v – 22) / 6.8

Cross multiply

v – 22 = 1.6 × 6.8

v – 22 = 10.88

Collect like terms

v = 10.88 + 22

v = 32.88 m/s

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Answer:

86.2 m  range

Explanation:

Vertical component of initial shot = 30 sin 55= 24.575 m/s

Now you can find out for how long it is in the air

     when it hits the ground , position = 0

            0 = 24.575 t  - 1/2 ( 9.81) t ^2

                         finds t = 5.01 seconds       (Using quadratic formula)

HORIZONTAL component = 30 cos 55 = 17.21 m/s

    it travels this speed for 5.01 seconds

               5.01 * 17.21 = 86.2 m

The global positioning system (GPS) satellite that moves in a circular orbit with period of 11h 58 min has a speed of: 3872.9848 m/s

To solve this problem the formulas and the procedures we will use are:

Where:

Information about the problem:

By converting the time period of (h) and (min) from (s) we have:

T = (11h *3600 s/ 1 h) + (58 min * 60 s/1 min)

T = 39600 s +3480 s

T = 43080 s

Applying the orbital radius formula we get:

r =[(G* m *T²)/(4 * π²)]

r =[(6.67 × 10⁻¹¹ m³/kgs² * 5.972 × 10²⁴ kg *(43080 s)²)/(4 * (3.1416)²)]

r =[1.8725*10^22 m³]

r= 26.55465166*10^6 m

Applying the tangential velocity formula we get:

v = (2*π * r)/ T

v = (2 * 3.1416 * 26.55465166*10^6 m)/ 43080 s

v = 3872.9848 m/s

It is the maximum of the radial distribution curve of the outermost orbital.

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Answer:

4m/s^2

Explanation:

u find the average of the two velocities then place the outcome in the acceleration formula given time as 6.3 seconds .....dividing the average of the velocities

The sound energy radiated through the windows and doors in 20 min is 332.6 J.

When a force, such as sound or pressure, causes an item or substance to vibrate, the result is sound energy. That energy moves through the substance in waves. Those sound waves are called kinetic mechanical energy. Sound waves are sometimes dubbed mechanical waves because sound waves require a physical medium to propagate. Liquids, gases, or solid materials transfer pressure variations, creating mechanical energy in waves. Like all waves, sound waves include peaks and valleys. The peaks are called compressions, while rarefaction is the term used for the lows. The oscillations between compression and rarefaction move through gaseous, liquid, or solid media to produce energy. The number of compression/rarefaction cycles in a given period determines the frequency of a sound wave.

Sound energy = E = power x time

energy = force x time / area

Given:

sound level, [tex]\beta=101 dB[/tex]

Area, A = [tex]22\;m^{2}[/tex]

Time, [tex]\triangle t=20\;min=1200\;s[/tex]

Intensity, [tex]I_{o}=1 \times 10^{-12}\;W/m^{2}[/tex]

We know that, Sound level is,

[tex]\beta=10 \times log\frac{I}{I_{o} }[/tex]

Solving the above equation for sound intensity,

[tex]I=I_{o}\times 10^{\frac{\beta}{10} }[/tex]

[tex]I=I \times 10^{-12} \times 10^{\frac{101}{10} }[/tex]

[tex]I=0.0126\;W/m^{2}[/tex]

Therefore, The sound energy is,

[tex]E = P \times \triangle t[/tex]

Substitute [tex]P=I \times A[/tex] in the above equation,

[tex]E = I \times A \times \triangle t[/tex]

[tex]E = 0.0126 \times 22 \times 1200[/tex]

[tex]E = 332.6\;J[/tex]

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The cheetah that can accelerate from rest to a speed of 30.0 m/s in 7.00s has an acceleration of: 4.285 m/s²

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

a = (vf - vi) /t

Where:

Information about the problem:

Applying the acceleration formula we have:

a = (vf - vi) /t

a = (30.0 m/s - 0 m/s) /7.00 s

a = (30.0 m/s m/s) /7.00 s

a = 4.285 m/s²

It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2

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Answer:

D. Adapt your communication style.

Explanation:

It is quite difficult to gauge the protocol here given the lack of information about the work environment; but as a general answer, communication skills are always handy. Furthermore, it seems that Marta in particular struggles with this. Assuming Marta is providing a specialized service of a sort, she would need to understand that there is a reason why her services are required at all. Through this, hopefull she'll understand that they are most likely outside of this field of work, and ergo [she] must cater and gear her actions towards the interest of the person(s) in question.

It is to be noted that this is merely my surmise.

The aphelion distance of Hailey's comet from the sun is 36 A. U.

The perihelion distance of Hailey's comet = 0.6 amu

The orbital period of Hailey's comet = 76 years

According to Kepler's law,

The time duration is T.

T = 76 years.

The semi-major axis is a.

The semi-major axis is,

[tex]T ^{2} =a ^{3} [/tex]

[tex]a = \sqrt[3]{(T) ^{2} }[/tex]

[tex]a = \sqrt[3]{(76) ^{2} }[/tex]

= 18 A. U

The aphelion distance of Hailey's comet from the sun is,

= 2 × a

= 2 × 18

= 36 A. U

Therefore, the aphelion distance of Hailey's comet from the sun is 36 A. U.

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It is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Given ,

The motion of a transparent medium influences the speed of light .

The water moves with speed v in a horizontal pipe .

Assume that the light travels in the same direction as the water moves .

The speed of the light with respect to the water is c/n

Where n = 1.33 is the refractive index of water .

Let us assume ,

u' be the speed of light in water , in the frame moving with the water .

u' is related to the refractive index of water ,n as :

u'=c/n

where , c is the speed of light .

let , u be the speed of light in water in the lab frame .

Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)

Here v is the speed of water in the horizontal pipe .

we know the value of u' , so by substituting the value , we will get ,

u= (c/n+ v)/(1+cv/nc^2)

u= c/n(1+ nv/c)/(1+v/nc)

(b) We have , v<<c

v/c<<1 .

so , (1+v/nc )^-1 = (1-v/nc)

Now substituting this , we will get ,

u = c/n(1+nv/c) (1-v/nc)

u≈c/n(1+ nv/c-v/cn)

u≈c/n + v - v/n^2

Hence , it is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

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Disclaimer : incomplete question , here is the complete question .

Question: The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc) . (b) show that for v<<c , the expression from part (a) becomes , to a good approximation , u ≈ c/n + v - v/n^2 .

Answer:

Mass and gravity

Explanation:

[tex]{ \rm{weight = mass \times gravity}}[/tex]

Weight is the force exerted on an object due to gravity. The correct answer is mass and gravity. The correct option is an option  (3).

Weight requires both mass and gravity.

The formula for calculating weight is given by:

Weight = mass × gravitational acceleration

W = m × g

The gravitational acceleration is known by the strength of the gravitational field, which is typically constant at any given point.

The mass of an object and the strength of its gravitational field is its weight. It is the gravitational force that pulls an object toward the center of the Earth.

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Therefore, The correct option is (3).

The complete question is:

What does weight require?

1) gravity and volume

2) only mass

3) mass and gravity

4) mass and volume

Answer:

Velocity ratio of simple machine is the ratio of distance traveled by the effort to the distance traveled by the load in the machine. As velocity ratio or ideal mechanical advantage is a simple ratio of two distances.Hence always remains constant. Option A is correct.

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Answer:

Velocity ratio of simple machine is the ratio of distance traveled by the effort to the distance traveled by the load in the machine. As velocity ratio or ideal mechanical advantage is a simple ratio of two distances.Hence always remains constant. Option A is correct.

Explanation:

EXTRA:Velocity ratio of a lever is 3 means distance travelled by effort is 3 times the distance travelled by the load. Efficiency of the pulley is 60% means 40% of the energy is lost in the machine due to the friction.

EXTRA:Ans:- MA of a lever is 3 it means that the load is 3 times of effort and VR of a lever is 4 it means that effort distance is 4 times of load distance.

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The kinetic energy of the proton is 1982.31 MeV.

We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by

Eo = m₀ . c²

where Eo is rest energy, m₀ is rest mass and c is the speed of light (3 x 10⁸ m/s).

The total energy of object can be described as

E = Eo / √(1 - v²/c²)

where E is total energy, v is the object speed.

Kinetic energy can be defined as energy change from rest energy until total energy. It can be determined as

KE = E - Eo

where KE is kinetic energy.

From the question above, we know that :

m₀ = 1.6 x 10¯²⁷ kg

c = 3 x 10⁸ m/s

v = 0.95c

Find the rest energy

Eo = m₀ . c²

Eo = 1.6 x 10¯²⁷ . (3 x 10⁸)²

Eo = 1.44 x 10¯¹⁰ joule

Eo = 1.44 x 10¯¹⁰ / (1.6 x 10¯¹⁹) eV

Eo = 900 x 10⁶ eV

Eo = 900 MeV

Determine the total energy

E = Eo / √(1 - v²/c²)

E = 900 / √(1 - (0.95c)²/c²)

E = 900 / 0.31

E = 2882.31 MeV

Calculate the kinetic energy

KE = E - Eo

KE = 2882.31 MeV - 900 MeV

KE = 1982.31 MeV

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The density of ice does not affect its melting rate. Adding objects will affect the melting rate.

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The player has to apply a force of 1956.5N to start tackling sled in motion.

Coefficient of static friction μ[tex]_{s}[/tex] = 0.70

Weight of sled = mg

                        = 590N

Weight of Player A = [tex]m_{a}g[/tex]

                               = 100 × 9.8

                              = 980N

Weight of Player B = [tex]m_{b}g[/tex]

                               = 125 × 9.8

                              = 1225N

Combined weight of sled:

W =  (590+980+1225)N

   = 2795N

Now, friction on sled:

f = μ[tex]_{s}[/tex]R

  = 0.70 × 2795

  = 1956.5N

Therefore, the player has to apply force equal to static friction to start tackling sled in motion.

∴ Force = Static friction =  1956.5N

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