The primary advantage of a bipolar scheme is that when all the voltages are added together after a long transmission, there should be a total voltage of ____.

Answer:

Will experience a force due to electric field.

Explanation:

According to Coulomb's law this force is mathematically given as:

[tex]F=E.q[/tex]

and, electric field due to a charge is given as:

[tex]E=\frac{1}{4\pi.\epsilon_0}.\frac{q}{r^2}[/tex]

where:

permittivity of free space[tex]\epsilon_0=8.85\times 10^{-12}\ m^{-3}.kg^{-1}.s^4.A^2[/tex]

q = magnitude of charge

r = radial distance from the charge

Answer:

The question has a diagram attached to it which I have done in the explanation.

The answer = The currents are equal in magnitude, the algebraic signs of the current values are opposite.

Explanation:

What is applied here is the Kirchoff's junction role or kirchoff's current law which states that the algebraic sum of the current entering any junction must be equal to the algebraic  sum of the current leaving the junction. this is what is applied in the diagram.

The attachment shows the explanation

Final answer:

The relationship between id and ic currents depends on their definitions, such as in capacitors within AC circuits where ic leads the voltage by 90 degrees, or in specific phenomena involving image currents that are equal in magnitude but opposite in sign.

Explanation:

The relationship between the currents id and ic depends on the specific context they are used in, but often these terms relate to currents in electronic components, such as diodes (id) and capacitors (ic). For instance, in a capacitive AC (alternating current) circuit, the current through the capacitor (ic) leads the voltage across the capacitor (vc) by 90 degrees in phase. This relationship is represented by ic(t) = C dv/dt, where C is the capacitance and dv/dt is the rate of change of voltage.

This means at any instant, the magnitude of the current is tied to the rate at which the voltage changes. In circuits involving superconductivity or other specific phenomena, such as image currents, you might encounter situations where an image current I' is equal in magnitude but opposite in sign to the driving current I, as indicated by I' = - I.

Answer:

C. 72

Explanation:

Transformer: A transformer is an electromagnetic device that uses the property of mutual inductance to change the voltage of alternating supply.

In a ideal transformer,

Vs/Vp = Ns/Np ............................................. Equation 1

Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.

Making Ns the subject of the equation,

Ns =(Vs/Vp)Np .......................................... Equation 2

Given: Vs = 24 V, Vp = 115 V, Np = 345.

Substitute into equation 2

Ns = (24/115)345

Ns = 72 turns.

Thus the number of turns in the secondary = 72 turns.

The right option is C. 72

Answer: Electrons

Explanation: For a neutral atom the number of positively charged protons inside the nucleus must be equal to the number of electrons in the orbital shells. Both charges will cancel out having a charge of 0 which makes an atom electrically neutral.

Answer:

Velocity is zero.

Explanation:

Given:

We know that acceleration is defined as the rate of change in velocity and hence we integrate the expression of acceleration.

Now the velocity can be given as:

[tex]v=\int\limits {a} .\, dt=-k.\frac{t^3}{3}[/tex]

Put t = 0 we get:

[tex]v=0\ m.s^{-1}[/tex]

Explanation:

This difference is because of the difference in arrangement of carbon atoms both graphite and Diamond.

Carbon atoms in graphite are arranged in layered form in an infinite array of layers. These layers are held together by a weaker force of attraction called vander waal's force of attraction such that layer's can slip over one another. Whereas in diamond carbon atoms are arranged tetrahedrally. Each carbon atom is attached to four carbon atoms with a bond angle of 109.5°. It is strong rigid three dimensional structure that results in infinite array atoms. This accounts for hardness of the diamond.

Final answer:

Diamond and graphite exhibit vastly different physical properties due to their respective carbon atom structures: diamond's strong three-dimensional covalent bonds make it extremely hard, while graphite's layered structure with weak interlayer forces makes it soft and slippery.

Explanation:

Diamond and graphite are two forms of the same element, carbon, but they have vastly different physical properties due to the way their atoms are bonded together. In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms in a strong three-dimensional network, which is what makes diamonds so hard and durable. This covalent bonding extends throughout the crystal, making it an excellent insulator but very hard to break.

Conversely, graphite is composed of layers of carbon atoms bonded in a hexagonal pattern, with weaker interlayer attractions known as London dispersion forces. This allows the layers to slide past each other easily, which is why graphite can be used as a lubricant and as the 'lead' in pencils – rubbing off onto paper with ease. The strong covalent bonds within the layers give graphite its high melting point, but the weak interactions between layers contribute to its softness.

Furthermore, graphite's structure enables it to conduct electricity parallel to the planes due to delocalized π (pi) bonds, while diamond does not conduct electricity. Graphite's black color results from the absorption of light by its delocalized electrons, whereas pure diamond is colorless.

Answer:

[tex]1.06085\times 10^{-10}\ m[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

n = 1 (ground state)

Angular momentum is given by

[tex]L=mvr[/tex]

From Bohr's atomic model we have

[tex]L=\dfrac{nh}{2\pi}[/tex]

[tex]mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}[/tex]

The centripetal force will balance the electrostatic force

[tex]\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m[/tex]

The diameter is [tex]2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m[/tex]

To determine the outer diameter of the spacers that yields the most economical and safe design, consider the average normal stress in the bolts and spacers. The outer diameter of the spacers that satisfies the conditions is approximately 34.854 mm.

To determine the outer diameter of the spacers that yields the most economical and safe design, we need to consider the average normal stress in the bolts and spacers. The average normal stress in the bolts must not exceed 216 MPa, and in the spacers, it must not exceed 143 MPa. Given that the bolts have a diameter of 16 mm and the spacers fit snugly inside, we can use the equation for calculating stress in a cylindrical object: stress = force/area.

Let's assume the outer diameter of the spacers is 'd'. The area of the spacers can be calculated as follows: area = pi/4 * (d^2 - (d-16)^2), where 'd-16' is the inner diameter of the spacers. To achieve the most economical and safe design, we want to maximize the area of the spacers while keeping the normal stress within the limits.

By substituting the given stress limits and solving for 'd', we can find the outer diameter that satisfies the conditions. After calculation, the outer diameter of the spacers that yields the most economical and safe design is found to be approximately 34.854 mm.

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The economic and safe design requires an outer spacer diameter of 34.854 mm.

To determine the outer diameter of the spacers that yields the most economical and safe design, we need to ensure that the average normal stress does not exceed the maximum allowable stress in both the bolts and the spacers.

The diameter of the bolt is 16 mm, so the cross-sectional area (Abolt) is:

Given the maximum normal stress is 216 MPa:

The maximum force (Fbolt) applied on the bolt:

The outer diameter (D) of the spacer is what we need to determine.

The cross-sectional area of the spacer (Aspacer) should ensure that the normal stress does not exceed 143 MPa:

Using the same maximum force from the bolt (as it transfers to the spacer):

The cross-sectional area

Since the inner diameter (d) is 16 mm (bolt diameter),

After solving for D using the above relationship:

Therefore, the most economical and safe outer diameter for the spacers is 34.854 mm.

Answer:

[tex]m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )[/tex]

Explanation:

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

[tex]V_t=\frac{4}{3} \pi.r_2^3[/tex]

And the volume of the hollow space in the sphere:

[tex]V_h=\frac{4}{3} \pi.r_1^3[/tex]

Therefore the net volume of material required to make the sphere:

[tex]V=V_t-V_h[/tex]

[tex]V=\frac{4}{3} \pi(r_2^3-r_1^3)[/tex]

Then the mass of the material used is:

[tex]m=\rho.V[/tex]

[tex]m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )[/tex]

Complete Question:

The complete question is on the all the uploaded image

Answer:

Explanation:

This solution were gotten by examining the diagrams and figuring out the rules that are broken

For A

we see that the two charges are positive so rule 1 is broken,on the same diagram we see that there is no radial symmetry close to the charge so rule 3 is broken then looking again at the diagram A we can see that the electric field is not tangent to any electric line at any point hence rule 5 is broken

For B:

We see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken

secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

For C:

We see that the number of line is not proportional to the magnitude of the charge hence rule 2 is broken.

For D:

we see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken

secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

For E:

We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.

For F

Looking at the diagram we see that none of the rules are broken

For G:

looking  at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken

For H:

We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.

taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken

Looking again at the diagram we see that the at any point on the electric field line that the electric field itself is not tangent to the electric field line hence rule 5 is broken

Answer

given,

Electric field,E = 490 N/C

time, t = 54 ns

for electron

Mass of electron me = 9.1 x 10⁻³¹ kg

Charge of electron e = -1.6 x 10⁻¹⁹  C

electrostatic force

F = E q

F = 490 x 1.6 x 10⁻¹⁹

F = 784  x 10⁻¹⁹ N

now, using newton second law

[tex]a = \dfrac{784\times 10^{-19}}{9.1\times 10^{-31}}[/tex]

  a = 8.62 x 10¹² m/s²

using equation of motion

v = u + a t

v = 0 + 8.62 x 10¹² x 54 x 10⁻⁹

v = 4.65 x 10⁵ m/s

velocity of electron is equal to v = 4.65 x 10⁵ m/s

For Proton  

Mass mp = 1.67 x 10⁻²⁷ kg  

Charge p = 1.6 x 10⁻¹⁹ C

Electric field E = 490 V/C

from above solution

F = 784  x 10⁻¹⁹ N

now, acceleration

[tex]a = \dfrac{784\times 10^{-19}}{1.67\times 10^{-27}}[/tex]

  a = 4.69 x 10¹⁰ m/s²

using equation of motion

v = u + a t

v = 0 + 4.69 x 10¹⁰ x 54 x 10⁻⁹

v = 4.65 x 10³ m/s

velocity of electron is equal to v = 4.65 x 10³ m/s

Answer:

Explanation:

Electric field, E = 490 N/C

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

charge of electron or proton = 1.6 x 10^-19 C

time, t = 54 ns = 54 x 10^-9 s

initial velocity, u = 0 m/s

Force on each particle, F = q E = 1.6 x 10^-19 x 490 = 7.84 x 10^-17 N

acceleration of electron = Force / mass of electron

ae = (7.84 x 10^-17) / ( 9.1 x 10^-31) = 8.6 x 10^13 m/s²

Let the velocity of electron is ve.

use first equation of motion

ve = u + ae x t

ve = 0 + 8.6 x 10^13 x 54 x 10^-9

ve = 4.65 x 10^6 m/s

acceleration of proton = Force / mass of proton

ap = (7.84 x 10^-17) / ( 1.67 x 10^-27) = 4.69 x 10^10 m/s²

Let the velocity of electron is vp.

use first equation of motion

vp = u + ap x t

vp = 0 + 4.69 x 10^10 x 54 x 10^-9

ve = 2535.1 m/s

Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Put the value in the equation

[tex]1.80=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{1.80\times2}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{1.80\times2}{9.8}}[/tex]

[tex]t=0.606\ sec[/tex]

We need to calculate the velocity

Using formula of velocity

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

Put the value into the formula

[tex]v=\dfrac{2.52}{0.606}[/tex]

[tex]v=4.15\ m/s[/tex]

We need to calculate height

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]h=\dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{4.15^2}{2\times9.8}[/tex]

[tex]h=0.878\ m[/tex]

Hence, The height of the water slide is 0.878 m.

Answer:

a) [tex]\mu_s=0.65[/tex]

b) [tex]\mu_k=0.54[/tex]

Explanation:

Static friction is when the body is at rest or is about to move while kinetic friction is when the body is already in motion. According to Newton's second law:

[tex]\sum F_y:N=mg\\\sum F_x:F_f=F_x[/tex]

a)  In this case, the static friction must be equal to the horizontal force to set the clock in motion:

[tex]F_f=\mu_sN=\mu_smg\\\mu_smg=F_x\\\mu_s=\frac{F_x}{mg}\\\mu_s=\frac{670}{106kg(9.8\frac{m}{s^2})}\\\mu_s=0.65[/tex]

b) In this case, the kinetic friction is equal to the horizontal force that keep the clock moving with constant velocity:

[tex]\mu_kmg=F_x'\\\mu_k=\frac{F_x'}{mg}\\\mu_k=\frac{557}{106kg(9.8\frac{m}{s^2})}\\\mu_k=0.54[/tex]

Answer:

mass of the meter stick=0.063 kg

or

mass of the meter stick=63.3 g

Explanation:

Given data

m₁=41.0g=0.041kg

r₁=(39.2 - 23)cm

r₂=(49.7 - 39.2)cm

g=9.8 m/s²

To find

m₂(mass of the meter stick)

Solution

The clockwise and counter-clockwise torques must be equal if the meter stick   is in rotational equilibrium

[tex]Torque_{cw}=Torque_{cw}\\F_{1}r_{1}=F_{2}r_{2}\\ m_{1}gr_{1}=m_{2}gr_{2}\\(0.041kg)(9.8m/s^{2} )(0.392m-0.23m)=m_{2}(9.8m/s^{2})(0.497m-0.392m)\\0.0651N.m=1.029m_{2}\\m_{2}=0.063 kg\\or\\m_{2}=63.3g[/tex]

Answer:

force between masses will be same as  [tex]3.24\times 10^{-7}N[/tex]

Explanation:

Let the masses are [tex]m_A\ and\ m_B[/tex] and distance between them is r

According to gravitational law force between two mass is given by

[tex]F=\frac{Gm_Am_B}{r^2}[/tex]

So [tex]3.24\times 10^{-7}=\frac{Gm_Am_B}{r^2}[/tex]

Now mass of both are doubled and distance between them is tripled

And now we have to find that from what factor the force between masses are changed

So [tex]F_{new}=\frac{G3m_A3m_B}{(3r)^2}=\frac{9Gm_Am_B}{9r^2}=\frac{Gm_Am_B}{r^2}=F[/tex]

So force between masses will be same as  [tex]3.24\times 10^{-7}N[/tex] as masses and distance between them both has the same effect

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

                                           =2 feet

                                           =2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

                     Distance Jumped=2+2

                     Distance Jumped=2*2

                                                   =4 feet

Similarly after 7 jumps

                    Distance Jumped=2+2+......+2

                    Distance Jumped=2*7

                                                 =14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

              Distance Jumped=(Distance Jumped after 7 jumps)+3

                                           =14+3

                                           =17 feet

The frog takes 8 jumps to reach top of well                

The frog needs 8 leaps each of 2 feet and a final leap of 3 feet to escape a 17-foot well. Therefore, it takes the frog a total of 9 leaps to escape the well.

In this mathematical problem, we need to determine how many leaps it takes for a frog to escape a 17-foot well, given that each leap propels the frog 3 feet up, but it slides back 1 foot before the next leap. Given thefa, each leap results net gain of 2 feet (3 feet up minus 1 foot slide back). However, for the last leap, the frog won't slide back, so the final leap has a net gain of 3 feet.

Therefore, the frog need to leap 15 feet using jumps with a net gain of 2 feet and then make a final leap of 3 feet out of the well. Each of the first leaps covers 2 feet, so the number of such leaps needed is 15/2 = 7.5. Since it's not possible to make half a jump, we round up to 8 jumps. Then, the frog makes its final jump of 3 feet. So in total, the frog needs 9 leaps to escape the well.

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

[tex]F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\   F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2}  }{(6.50*10^{-10} )^{2}  } \\F=5.46*10^{-10}N[/tex]

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

[tex]a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}[/tex]

For(b) Acceleration of proton

[tex]a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}[/tex]

Acceleration of the body is rate of change of the increasing velocity with respect to the time.  is The acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

A proton and an electron are released when they are 6.50×10⁻¹⁰ m apart.

The the distance of proton and electron is 6.50×10⁻¹⁰ m.

Now it is known that,

        [tex]m_e=9.109\times10^{-31}[/tex]

        [tex]m_p=1.673\times10^{-27}[/tex]

        [tex]Q_e=-1.602\times10^{-19}[/tex]

        [tex]Q_e=1.602\times10^{-19}[/tex]

Acceleration of the body is rate of change of the increasing velocity with respect to the time. Acceleration of a body is the ratio of the force to the mass of the body.

The coulomb force between the two charges can be given as,

[tex]F=k\dfrac{q_1q_2}{r^2} [/tex]

Here k is the coulombs constant, r is the distance and q is the charge of proton and electron.Put the values,

[tex]F=9\times10^9\times\dfrac{(1.602\times10^{-19})^2}{(6.5\times10^{-10})^2} [/tex]

[tex]F=5.46\times10^{-10}[/tex]

Now it is known that the acceleration of a body is the ratio of the force to the mass of the body.

[tex]a_e=\dfrac{F}{m_e} [/tex]

[tex]a_e=\dfrac{5.46\times10^{-10}}{9.109\times10^{-31}} [/tex]

[tex]a_e=5.992\times10^{20}[/tex]

[tex]a_p=\dfrac{F}{m_p} [/tex]

[tex]a_p=\dfrac{5.46\times10^{-10}}{1.673\times10^{-27}} [/tex]

[tex]a_e=3.625\times10^{17}[/tex]

Thus the acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

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Answer:

0.407

Explanation:

5.5÷13.5

i believe it is this as

density =mass÷volume

At a speed of 58.8mi/hr must he drive for the remainder of the trip to complete the trip in the usual amount of time.

The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The time it takes to go 120 miles at that speed in normal conditions is;

[tex]\rm t = \frac{d}{v} \\\\ t =\frac{120}{55} \\\\ t= 2 hour \ 11 minute[/tex]

However, the driver was 30 minutes behind schedule at this distance, so the time the has to spend is;

T== 2hr 11 min +30 min

T= 2hr 41 mins

He has to fulfill the initial planned moment of 10 hours. The time to cover the remaining distance;

T'=(10-2.41)

T' = 7hr19mins

The remaining distance will be ;

S={550-120}

S=430mile.

The speed is to be maintained the following distance on the time;

[tex]\rm V'=\frac{430}{(\frac{439}{60} )} \\\\V'=58.7699 \ miles / hour[/tex]

The speed must he drive for the remainder of the trip to complete the trip in the usual amount of time will be 58.5 mile/hr.

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To complete his remaining 430 miles in 8 hours, the truck driver needs to drive at a speed of 53.750 mph for the rest of the trip.

The driver has covered 120 miles during which he used up 30.0 minutes more than what he would usually take.

We can calculate the remaining distance he needs to cover which is 550 total miles minus the 120 miles he's already driven, which equals to 430 miles.

The remaining time he has is the total normal trip time of 10.0 hours minus the 2.0 hours he's already spent (previous 1.5 hours plus the extra 0.5 hours), which equals 8 hours.

To cover the remaining 430 miles in the 8 hours left, he must drive at a speed of 430 miles divided by 8 hours, which equals to 53.750 mph.

This is the speed the driver needs to maintain for the remainder of the trip to complete it in the usual amount of time.

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Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

Answer

A)   Positively charged two insulating rod are brought closed to an object  they repel each other. It means the Object is positively charged. Because similar charge repel each other.

   The correct answer is Option A.

B) we know force between to charges is calculated using Formula

         [tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex].......(1)

   form the given condition in the question

          [tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}[/tex]

          [tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}[/tex]

          [tex]F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}[/tex]

from equation (1)

          [tex]F' = \dfrac{F}{4}[/tex]

hence, the correct answer is Option C.

The object is positively charged if repelled by a positively charged rod. The force between two charged objects reduces to F/16 when both charges are one-fourth and the distance is halved.

Charged Objects and Electrostatic Forces

If the object suspended by a string is repelled away from a positively charged insulating rod, we can conclude that the object is positively charged (A). This is because like charges repel each other, according to electrostatic principles.

Regarding the force between two charged objects, Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2, the new force is calculated as:

New charge = Original charge / 4

New distance = d / 2

New force = (1/4 × 1/4) / (1/2 × 1/2)^2 = 1/16 / 1/4 = F/16 (A)

Answer:

There is a force that has the same magnitude as that of the hammer applied on the astronaut and with direction away from the asteroid, movement is given by

                F_hammer - F_Gravitation = m a

Explanation:

For this exercise we will propose its solution from Newton's third law, which states that every action has a reaction of equal magnitude, but felt different.

As it is in space, we must assume that it is not subject to the gravitational attraction of nearby bodies, except the asteroid that attracts it. When he extends his hand and hits the asteroid, he exerts a force on him, by Newton's third law he responds with a force of equal magnitude applied to the astronaut, therefore without the two they are not united they could separate if this force is greater than the force of universal attraction between the two.

In summary There is a force that has the same magnitude as that of the hammer applied on the astronaut and with direction away from the asteroid, movement is given by

                F_hammer - F_Gravitation = m a

The magnitude of the force of friction is 6.1 N

Explanation:

The work done by a force when moving an object is given by the equation:

[tex]W=Fd cos \theta[/tex]

where :

F is the magnitude of the force

d is the displacement

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

In this problem, we have:

W = -55.2 J (work done by the force of friction)

d = 8.98 m (displacement of the sled)

[tex]\theta=180^{\circ}[/tex] (because the force of friction acts opposite to the direction of motion)

By solving the equation for F, we find the magnitude of the force of friction on the sled:

[tex]F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N[/tex]

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Answer:

6.1

Explanation:

Acellus

Explanation:

Given that,

Charge 1, [tex]q_1=6.75\ nC=6.75 \times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.46\ nC=4.46\times 10^{-9}\ C[/tex]

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

[tex]F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}[/tex]

[tex]F=6.84\times 10^{-8}\ N[/tex]

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

(a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N.

(b) The force is repulsive.

(a) To calculate the magnitude of the electrostatic force that one charge exerts on the other, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these  values into equation 1

(b) The force is repulsive because both charges a the same (positive).

Hence, (a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N (b) The force is repulsive.

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Incomplete question.The complete question is here

A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10m wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s.What is her acceleration on the rough ice?

Answer:

acceleration = -4.103 m/s²

Explanation:

Given data

Initial velocity Vi=8.30 m/s

Final velocity Vf=5.20 m/s

Initial distance xi=0 m.......(We choose xi=0 the start point of acceleration motion )

Final distance xf=5.10 m

To find

Acceleration

Solution

From the kinetic equation

[tex](v_{f})^{2}=(v_{i} )^{2}+2a(x_{f} -x_{i} )\\  (5.20 m/s)^{2}=(8.30m/s)^{2}+2a(5.10m-0m)\\a=\frac{(5.20m/s)^{2}-(8.30m/s)^{2}  }{2*(5.10m)}\\a=-4.103 m/s^{2}[/tex]

To find the line charge density (λ) on a long wire, the formula E = λ/ (2πε0r) can be used. The known values are plugged into the formula and solved to give λ = 1.31 x 10^-8 C/m.

To calculate the line charge density (λ) on a long wire, we use the formula that relates electric field and line charge density: E = λ/(2πε0r), where E is the magnitude of electric field, λ is the line charge density, ε0 is the permittivity of free space, and r is the distance from the wire.

In this given scenario, the magnitude of the electric field (E) is 260 kN/C (or 260,000 N/C). The distance from the wire (r) is 45 cm (or 0.45 m). The permittivity of free space (ε0) is a constant value of approximately 8.85 x 10^-12 C²/Nm².

By substituting these known values into the formula, we solve for λ: λ = E* (2πε0r) = 260,000 N/C * 2π * 8.85 x 10^-12 C²/Nm² * 0.45 m = 1.31 x 10^-8 C/m.

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Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by [tex] tanθ = y/D [/tex]

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            [tex]V^{2} = U^{2} + 2gh[/tex]

Substituting the values,

                             [tex]V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)[/tex]

                             [tex]V^{2} = 24.5 m/s[/tex]

                             [tex]V = \sqrt{24.5} \ m/s[/tex]

                             [tex]V = 4.95 \ m/s[/tex]

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            [tex]V^{2} = U^{2} + 2gh[/tex]

Substituting the values,

                            [tex]0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)[/tex]

                            [tex]0 = U^{2} - 16.072 m/s[/tex]

                            [tex]U^{2} = 16.072 m/s[/tex]

                            [tex]U = \sqrt{16.072} \ m/s[/tex]

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            [tex]F = \frac{mv - mu}{t}[/tex]

                          [tex]F.t = m(v - u)[/tex]

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

Answer:

They will become aligned according to the charges on the metal plate.

Explanation:

When Polar molecules are placed between oppositely charged metal plates the molecules will tend to be attracted by their corresponding oppositely charged plates that is the positive and negative plates,

A polar molecule is one which has opposite charges on its ends. Non-polar molecules however do not have charges on their end

When polar molecules are placed between oppositely charged metal plates, the poles orient towards the oppositely charged plate.

These are the molecules that have positive and negative in opposite poles. For example- water molecules.

Therefore, when polar molecules are placed between oppositely charged metal plates, the poles orient towards the oppositely charged plate.

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The question doesn't have any particular requirement, but we'll compute the displacement of the plane from its initial and final landing point in the pasture

Answer:

[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]

[tex]\displaystyle \theta =-19.395^o[/tex]

Explanation:

Displacement

The vector displacement [tex]\vec r[/tex] is a measure of the change of position of a moving object. The displacement doesn't depend on the path followed, only on the final and initial positions. Its scalar counterpart, the distance, does measure the total space traveled and considers all the changes in the direction taken by the object. To find the displacement, we must add all the particular displacements by using vectors.

The plane first flies 160 km at 66° east of north. To find the vector expression of this displacement, we must know the angle with respect to the East direction or North of East. Knowing the angle East of North is 66°, the required angle is 90°-66°=34°

The first vector is expressed as

[tex]\displaystyle \vec{r_1}=\left \langle 160^o\ cos34^o, 160\ sin34^o \right \rangle[/tex]

[tex]\displaystyle \vec{r_1}=\left \langle 132.646, 89.471 \right \rangle[/tex]

The second displacement is 260 km at 49° South of East. This angle is below the horizontal respect to the reference, thus we use -49°.  

The second vector is expressed as:

[tex]\displaystyle \vec{r_2}=\left \langle 260\ cos(-49^o), 260\ sin(-49^o)\right \rangle[/tex]

[tex]\displaystyle \vec{r_2}=\left \langle 170.575,-196.224\right \rangle[/tex]

The total displacement is computed as the vectorial sum of both vectors

[tex]\displaystyle \vec{r}=\vec{r_1}+\vec{r_2}=\left \langle 132.646+170.575\right \rangle+ \left \langle89.471-196.224\right \rangle[/tex]

[tex]\displaystyle \vec{r}=\left \langle 303.221,-106.753\right \rangle\ km[/tex]

The magnitude of the total displacement is

[tex]\displaystyle |\vec{r}|=\sqrt{303.221^2+(-106,753)^2}[/tex]

[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]

And the direction is

[tex]\displaystyle tan\ \theta =\frac{-106.753}{303.221}=-0.352[/tex]

[tex]\displaystyle \theta =-19.395^o[/tex]