pls help me solve dis​

Answer:

The electroscope will become discharged by loosing electrons in a phenomenon called photoelectric effect

Explanation:

The UV light knocks off loosely bounded electrons in the outer shell causing the metal electroscope to discharge

Answer:

The amount of energy delivered is 730.84 J

Explanation:

Given;

rms intensity of the electric field, [tex]E_{rms}[/tex] = 6.75 v/m

area of the wall, A = 2.0 cm² = 2.0 x 10⁻⁴ m²

The average intensity of the wave is given  by;

[tex]I_{avg} = c \epsilon_o E_{rms}^2[/tex]

where;

c is the speed of light

ε₀ is permittivity of free space

I = (3 x 10⁸)(8.85 x 10⁻¹²)(6.75)²

I = 0.121 W/m²

Average power delivered, P = I x A

P = 0.121 x 2 x 10⁻⁴

P = 2.42 x 10⁻⁵ W

The amount of energy delivered is calculated as;

E = P x t

E = 2.42 x 10⁻⁵ x 3.02 x 10⁷

E = 730.84 J

Therefore,  the amount of energy delivered is 730.84 J

Answer:

C. 0.25 m

Explanation:

Given;

current flow in the solenoid, I = 6.0 A

number of turns of the solenoid, N = 180 turns

the resistance of the circuit, R = 17 Ω

change in the magnetic field, dB/dt = 3.0 T/s

The emf of the circuit is given by;

V = IR

V = 6 x 17

V = 102 Volts

Magnitude of induced emf is given by;

[tex]E = N(\frac{dB}{dt} )A[/tex]

Where;

A is area of the solenoid

[tex]E = N(\frac{dB}{dt} )A\\\\A = \frac{E}{N(\frac{dB}{dt} )} \\\\A = \frac{102}{180 *3} \\\\A = 0.18889 \ m^2[/tex]

Area of the circular solenoid is given by;

A = πr²

where;

r is radius of the solenoid

[tex]r = \sqrt{\frac{A}{\pi} }\\\\r = \sqrt{\frac{0.18889}{\pi} }\\\\r = 0.25 \ m[/tex]

Therefore, the correct option is C. 0.25 m

Explanation:

It is given that,

A nearsighted woman has a far point of 180 cm.

The object distance for nearsightedness is infinity, u = ∞

The image distance is, v = -180 cm

Using lens formula,

[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-180)}-\dfrac{1}{\infty}\\\\f=-180\ cm\\\\f=-1.8\ m[/tex]

For nearsightedness, a diverging lens is used.

Power of the lens,

P = 1/f

[tex]P=\dfrac{1}{-1.8}\\\\P=-0.56\ D[/tex]

So, the power of the lens is -0.56 D.

The power of lens to be prescribed is 3.5 D.

A person that is far sighted can see far objects clearly but not nearby objects. The farpoint of a normal eye is infinity. Far sightedness is corrected by the use of a convex lens. The near point of the normal eye is 25 cm.

We have that;

u = 25 cm

v = -180 cm

1/f = 1/u - 1/v

1/f = 1/25 - 1/180

1/f = 0.04 - 0.0056

f = 29 cm

Power of lens = 100/f = 100/29 = 3.5 D

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Answer:

ΔS =  - k ln (3)

Explanation:

Using the Boltzmann's expression of entropy, we have;

S = k ln Ω

Where;

S = Entropy

Ω = Multiplicity

From the question, the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value. This also causes a change in the entropy of the gas as follows;

ΔS = k ln (ΔΩ)

ΔS = kln(Ω₂)  - kln(Ω₁)

ΔS = kln(Ω₂ / Ω₁)              -------------(i)

Where;

Ω₂ = Final/Current value of the multiplicity

Ω₁ = Initial/Previous value of the multiplicity

Ω₂ = [tex]\frac{1}{3}[/tex] Ω₁         [since the multiplicity is reduced to one-third of the previous value]

Substitute these values into equation (i) as follows;

ΔS = k ln ([tex]\frac{1}{3}[/tex] Ω₁ / Ω₁)

ΔS = k ln ([tex]\frac{1}{3}[/tex])

ΔS = k ln (3⁻¹)

ΔS =  - k ln (3)

Therefore, the entropy changes by - k ln (3)

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, [tex]\rho=2.44\times 10^{-8}\ \Omega-m[/tex]

Resistance in terms of resistivity is given by :

[tex]R=\dfrac{\rho l}{A}[/tex]

Also, V = IR

So,

[tex]\dfrac{V}{I}=\dfrac{\rho l}{A}[/tex]

A is area of wire,

[tex]\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}[/tex], r is radius, r = d/2 (diameter=d)

[tex]\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m[/tex]

Out of four option, near option is (C) 17 μm.

The diameter of the wire is 18μm and the closest answer is option C.

The resistance of a wire is proportional to its length and inversely proportional to its area.

Given that a  4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. Then,

R = (ρL)/A

where

ρ = resistivity = 2.44 × 10-8 Ω·m

L = length

A = Area = [tex]\pi ^{2}[/tex]D/2

D = diameter of the wire

From Ohm's law, V = IR

make resistance R the subject of the formula

R = V/I

R = 1.5/4 x [tex]10^{-3}[/tex]

R = 375 Ω

Substitute all the parameters into the formula

R = (ρL)/A

375 = (2.44 × [tex]10^{-8}[/tex] x 4)/A

A = (9.76 × [tex]10^{-8}[/tex])/ 375

A = 2.603 x [tex]10^{-10}[/tex] [tex]m^{2}[/tex]

but

A =  [tex]\pi (D/2)^{2}[/tex]

2.603 x [tex]10^{-10}[/tex] =  [tex]\pi (D/2)^{2}[/tex]

[tex]\pi[/tex][tex]D^{2}[/tex] /4 = 2.603 x [tex]10^{-10}[/tex]

[tex]\pi[/tex][tex]D^{2}[/tex] = 1.04 x [tex]10^{-9}[/tex]

[tex]D^{2}[/tex] = (1.04 x [tex]10^{-9}[/tex])/ [tex]\pi[/tex]

D = [tex]\sqrt{3.3 * 10^{-10} }[/tex]

D = 1.8 x [tex]10^{-5}[/tex] m

D = 18 μm

Therefore, the diameter of the wire is 18μm and the closest answer is option C.

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Answer:

Kindly check explanation

Explanation:

Given the following :

As train approaches ; frequency, f1 = 94Hz

As train recedes; frequency, f2 = 71Hz

Speed of sound in air ; v = 340m/s

A) speed of sound source (speed of train) = vs

From doppler effect :

As the train recedes ;

f2 = fs [v / (v + vs)] - - - - (1)

As train approaches :

f1 = fs [v / (v - vs)] ----- (2)

To find vs equate (1) and (2)

fs [v / (v - vs)] = fs [v / (v + vs)]

f1/f2 = v / (v - vs) ÷ v / (v + vs)

f1 / f2 = v / (v - vs) × (v + vs) / v

f1 / f2 = (v + vs) / (v - vs)

Let f1 / f2 = f

f = (v + vs) / (v - vs)

f (v - vs) = v + vs

fv - fvs = v + vs

fv - v = vs + fvs

v(f - 1) = vs(1 + f)

v(f - 1) / (1 + f) = vs

B)

v(f - 1) / (1 + f) = vs

f = f1 / f2 = 94/71 = 1.32 Hz

340(1.324 - 1) / (1 + 1.324) = vs

vs = 340(0.324) / 2.324

vs = 110.16 / 2.324

vs = 47.40 m/s

C.) To calculate fs, frequency of train, substitute vs into our equation.

f2 = fs [v / (v + vs)]

Following our substitikn we obtain:

fs = (2f / (f + 1))f2

D)

fs = (2f / (f + 1))f2

fs = 2(1.324) / (1.324 +1)) × 71

fs = (2.648 / 2.324) × 71

fs = 1.1394148 × 71

fs = 80.898450

fs = 80.90 Hz

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

Answer:

Explanation:

angle of reflection = angle of incidence i = 2θ

angle of refraction r = θ

sin i / sin r = 1.52

sin2θ / sin θ = 1.52

2 sinθ . cosθ / sin θ  = 1.52

2 cosθ = 1.52

cosθ = .76

θ= 41°

Hence angle of incidence = 41°

.

Answer:

D-The asthenosphere and outer core would solidify.

Explanation:

The athenosphere is a region of the earth mantle below the lithosphere.

If the Sun didn’t retain the heat formed during its formation then it will have an adverse effect on other parts such as the core and mantle.The inner core wouldn’t liquefy but solidify. The asthenosphere and outer core would also solidify.

The Seismic waves wont be able to move the crust and the Earth’s magnetic field wouldn’t disappear but would be more pronounced.

The Earth’s magnetic field would disappear as well as solidification of asthenosphere and outer core will occur.

The asthenosphere and outer core would solidify and the Earth’s magnetic field would disappear if the Earth did not retain the heat from its formation because this heat make all the materials present in the outer core and asthenosphere in liquid form. This liquid form in the outer core is responsible for the generation of magnetic field around the earth.

If this heat will not retain in asthenosphere and outer core, the heat will escape and cooldown the asthenosphere and outer core which change it into solid form so the Earth’s magnetic field will be disappear so we can conclude that the Earth’s magnetic field would disappear as well as solidification of asthenosphere and outer core will occur if the Earth did not retain the heat from its formation.

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Answer:

0.514 m/s

Explanation:

The knot is a unit of nautical speed used in maritime navigation.

1 knot is equal to one nautical mile per hour,

1 nautical mile per hour = 1.852 km/h

The basic SI system of units of speed is in 'm/s'

1.852 km/h = (1.852 x 1000)/3600 = 0.514 m/s

Answer:

0.135E-19kgm/s

Explanation:

Using the uncertainty principle, we find

Dp = h / (4π Dx)

= (6.63×10-34Js)/4π(3.90×10-15m) = 0.135×10-19kg m/s

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

             Q_c = m ce ([tex]T_{f}[/tex]-T₀)

Heat released by the largest container is

              Q_a = M ce (T_{i}-T_{f})

how

        Q_c = Q_a

       m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

Answer: A. similar chemical properties and characteristics

Answer:

Explanation:

1 km=1000 m

1 hour =60 min =60*60 sec=3600 sec

Now put 1000 m instead of km and 3600 sec instead of hour in the given expression.

=50 km/hour

=50*1000 m/3600 sec

=500 m/36 sec

=13.89 m/s

Answer:

= 13.89 m/s

Explanation:

     km     1000 m        1 hr

50 ----- x ----------  x ------------  

       hr        1 km      3600 secs

= 13.89 m/s

Answer:

To the fish the man appears to be 3.06m above its eyes

Explanation:

We know that refractive index n

n = real dept/ apparent depth

While apparent depth is the distance the fish appears to the man which is 2.3m so using the equation

Real dept= n x apparent depth

= 1.333* 2.3= 3.06m

Answer: v₂ = 5962 km

the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Explanation:

Given that;

Lunar surface is in an altitude h = 43.0 km =  43 × 10³ m

we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²

speed of the space craft when it crashes into the lunar surface , v

decreasing speed of the space craft = 23 m/s

Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r

but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²

HERE r = Rm + h

we substitute

r = 1.74 × 10⁶ m + 43 × 10³ m

= 1.783 × 10⁶ m

On equating these, we have

G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)

v²/r = GM/r²

v = √ ( GM/r)

v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )

v = √ (2750787.9978)

v = 1658.55 m/s

Now since speed is decreasing by 23 m/s

the speed of the space craft into the lunar face is,

v₁ = 1658.55 m/s - 23 m/s

v₁ = 1635.55 m/s

Now applying conversation of energy, we say

1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)

v₂ =  √ [ v₁² + GMe (1/Rm - 1/r)]

v₂ =   √ [ 1635.55²  + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]

v₂ =  √ (2675023.8025 + 67979.24)

v₂ = √(2743003.046)

v₂ = 1656.2 m/s

now convert

v₂ = 1656.2 × 1km/1000m × 3600s/1hrs

v₂ = 5962 km

Therefore the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Answer:

Lorenz force law

Explanation:

The equation given name after the physicist and the mathematician James Clerk  Maxwell. He published his first equation in 1861 and 1862 that included the Lorenz force law. These equation were used by Maxwell first of all for the light in electromagnetic phenomena.

In his equations Maxwell proposed that how the electric and magnetic field is fluctuate and constant speed in the vacuum which is called a electromagnetic radiation.

These are waves that occurs at different level of wavelength spectrum. These equation have two major variant such as universal applicability. The Maxwell equations are used for the Equivalent alternative formulation.

Answer:

A

  [tex]\epsilon_{max} = 0.774 \ V[/tex]

B

[tex]wt = 0^o[/tex]

Explanation:

From the question we are told that

   The  length of the side is  [tex]l = 16.0 \ cm = 0.16 \ m[/tex]

    The  number of turns is  [tex]N = 29 \ turns[/tex]

    The  magnetic field is  [tex]B = 83.0 mT = 83 *10^{-3} \ T[/tex]

    The  angular speed is  [tex]w = 1.20 * 10^2 rev/min = \frac{1.20 *10^2 * 2\pi}{60 } = 12.6 \ rad/s[/tex]

Generally  the area is  [tex]A = l^2[/tex]

Generally the induced emf is mathematically represented as

         [tex]\epsilon = N * w * B * A * cos(wt)[/tex]

At maximum  [tex]cos(wt) = 1[/tex]

    So

       [tex]\epsilon_{max} = N * w * B * A[/tex]

       [tex]\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* (l^2)[/tex]

=>   [tex]\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* ((0.16)^2)[/tex]

=>  [tex]\epsilon_{max} = 0.774 \ V[/tex]

At maximum emf

      [tex]cos (wt) = 1[/tex]

=>   [tex](wt) = cos^{-1} (1)[/tex]

=>   [tex]wt = 0^o[/tex]

Answer:

The amount of time for the whole journey is 8 hours.

Explanation:

A truck covered 2/7 of a journey at an average speed of 40  mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: [tex]1-\frac{2}{7} =\frac{5}{7}[/tex]

So if the truck covered the remaining 200 miles at [tex]\frac{2}{7}[/tex], this means that [tex]\frac{5}{7}[/tex] of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if [tex]\frac{5}{7}[/tex] of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?

[tex]miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles[/tex]

miles= 280

So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ([tex]speed=\frac{distance}{time}[/tex]), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:

[tex]time=\frac{distance}{speed}=\frac{280 miles}{35 mph}[/tex]

time= 8 h

The amount of time for the whole journey is 8 hours.

Answer:

[tex]f=1.98\times 10^5\ Hz[/tex]

Explanation:

The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by :

[tex]B=1.2\times 10^{-6}\sin [2\pi (\dfrac{z}{240}-\dfrac{10^7t}{8})][/tex] .....(1)

The general equation of the magnetic field wave is given by :

[tex]B=B_o\sin (kz-\omega t)[/tex] ....(2)

Equation (1) is in form of equation (2), if we compare equation (1) and (2) we find that,

[tex]\omega=\dfrac{10^7}{8}[/tex]

We need to find the frequency of the wave. It is given by :

[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{10^7}{8\times 2\pi}\\\\f=1.98\times 10^5\ Hz[/tex]

So, the frequency of the wave is [tex]1.98\times 10^5\ Hz[/tex]

Answer: B. A magnet pulls a nail towards it

Explanation: Out of the list of answers, this is the only answer where a force is acting on an object. the act of the magnet pulling on the nail is the force, and the nail is the object.

Answer:

a)143.8 decays/minute

b)0.41 decays/minute

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2=half-life of C-14= 5670 years

t= time taken to decay

Ao= activity of a living sample

A= activity of the sample under study

a)

0.693/5670 = 2.303/1000 log(162.5/A)

1.22×10^-4 = 2.303×10^-3 log(162.5/A)

1.22×10^-4/2.303×10^-3 = log(162.5/A)

0.53 × 10^-1 = log(162.5/A)

5.3 × 10^-2 = log(162.5/A)

162.5/A = Antilog (5.3 × 10^-2 )

A= 162.5/1.13

A= 143.8 decays/minute

b)

0.693/5670 = 2.303/50000 log(162.5/A)

1.22×10^-4 = 4.61×10^-5 log(162.5/A)

1.22×10^-4/4.61×10^-5 = log(162.5/A)

0.26 × 10^1 = log(162.5/A)

2.6= log(162.5/A)

162.5/A = Antilog (2.6 )

A= 162.5/398.1

A= 0.41 decays/minute

Answer:

The inductance of one conductor in the unstretched cord is 0.7849 μH

Explanation:

Given;

number of turns of the coil, N = 52 turns

diameter of the coil, d = 1.30 cm

radius of the coil, r = d/2 = 1.30 cm / 2 = 0.65 cm = 0.0065 m

length of the unstretched cord, l = 57.5 cm = 0.575 m

The inductance of one conductor in the unstretched cord is given by;

[tex]L = \frac{N^2 \mu_o A}{l}[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

A is area of coil, = πr² = π x (0.0065)² = 1.328 x 10⁻⁴ m²

[tex]L = \frac{(52)^2(4\pi *10^{-7})(1.328*10^{-4})}{0.575} \\\\L = 7.849*10^{-7} \ H[/tex]

L = 0.7849 μH

Therefore, the inductance of one conductor in the unstretched cord is 0.7849 μH

Answer:

The total force on the charge at the top of the triangle is approximately

[tex]1.56\,\,10^{-14}\,\,N[/tex]

Explanation:

Please look at the attached image to follow the explanation.

Since all charges are of the same positive value, the force exerted on the top charge by the other two are going to be on the line that joins the top charge with each of the other vertices of the triangle, pointing away from the top charge (illustrated by green vectors of the same length in the image).

We need to find the x and y components of these force vectors in order to find the resultant force by combining the x  components among themselves, and the y components among themselves. Notice that the angle needed is in all cases [tex]60^o[/tex].

The x- components include the cosine of [tex]60^o[/tex], while the y components include the sine of [tex]60^o[/tex].

Notice as well that the x-components cancel each other (they have the same magnitude but point in opposite directions), while the y-components are both of the same value but pointing in the same direction (pointing up).

Then we just need to multiply by two the y component of one of the forces to find this total force.

Now, the magnitude of the forces in question are given by Coulomb's Law:

[tex]F_C=k\,\frac{q_1\,\,q_2}{d^2} =9\,\,10^9\,\frac{2\,\,10^{-12}\,\,2\,\,10^{-12}}{2^2} =9\,\,10^{-15}\,\,N[/tex]

therefore we can calculate what the y-component of this is using:

[tex]F_y=F\,sin(60^o)=9\,\,10^{-15} \,\frac{\sqrt{3} }{2} \\[/tex]

and twice this value becomes:

[tex]2\, \,F_y=(2)\,\,9\,\,10^{-15} \,\frac{\sqrt{3} }{2}=9\,\,\sqrt{3} \,\,10^{-15}\approx 1.56\,\,10^{-14}\,\,N[/tex]

Answer:

34.3Hz

Explanation:

We know that

For a pipe open at one end, fundamental frequency (Fo)

is

Fo = v/4L

Where v= 340m/s

Fo = 340/4 x 2.48 = 34.3Hz

Answer:

The speed of the observer is 2.4 x 10^8 m/s

Explanation:

The standard length of a meter stick is 100 cm

we are to calculate at what speed parallel to the length the length reduces to 60 cm.

This is a relativistic effect question. We know that the length will contract to this 60 cm following the equation below

[tex]l = l_{0}\sqrt{1 - \beta ^{2} }[/tex]

where

[tex]l[/tex] is the new length of 60 cm

[tex]l_{0}[/tex] is the original length which is 100 cm

[tex]\beta[/tex] is the the ratio v/c

where

v is the speed of the observer

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

60 = [tex]100\sqrt{1 - \beta ^{2} }[/tex]

0.6 = [tex]\sqrt{1 - \beta ^{2} }[/tex]

we square both side

0.36 = 1 - [tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 1 - 0.36 = 0.64

β = [tex]\sqrt{0.64}[/tex] = 0.8

but β = v/c

v/c = 0.8

substituting value of c, we have

v = 0.8 x 3 x 10^8 = 2.4 x 10^8 m/s

Answer:

Resistance =330 Ω

Tolerance = 33 Ω

Explanation:

see attached resistor color code table

The first stripe is orange, which means the leftmost digit is a 3.

The second stripe is orange , which means the next digit is a 3.

The third stripe is brown.  Since brown is 1, it means add one zero to the right of the first two digits.

The resistance is:

orange-orange-brown=  330 Ω

The tolerance is:

The fourth color band indicates the resistor's tolerance.  Tolerance is the percentage of error in the resistor's resistance.

silver is 10%

A 330 Ω resistor has a silver tolerance band.  

Tolerance = value of resistor x value of tolerance band

= 330 Ω x 10% = 33 Ω

330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.

The resistance of the resistor is 330 Ω and the tolerance is within  363 Ω and 297 Ω

In physics, resistor's resistance is coded using colors.

Orange colors are coded as 3

The  brown color is coded as 0

The silver color determines the tolerance and silver means 10%

The resistor with four colored stripes in the following order: orange, orange, brown has a resistance value of 330 Ω

Tolerance = 330 × 10%

Tolerance = 33Ω

Resistor value = 330±33

Resistor value = (330+33) and (330-33)

Resistor value = 363 Ω and 297 Ω

Hence the resistance of the resistor is 330 Ω and the tolerance is within  363 Ω and 297 Ω

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Answer:

43200c

Explanation:

Answer:

The relationship is [tex]E=\frac{ kQ}{d^2}[/tex]

Explanation:

The electric field strength is denoted by the symbol E,

the test charge is denoted be q and the source charge be Q

distance is denoted by d

Then the equation can be rewritten in symbolic form as

Electric field strength is = Force/charge

[tex]E=\frac{F}{q}------1[/tex]

we know that the formula for force is given as

[tex]F= \frac{kQq}{d^2} ------2[/tex]

where [tex]k= 9*10^9 N.m^2/C^2[/tex]

and d is the separation distance between charges

We can insert the expression for Force in equation one

we have

[tex]E= \frac{kQ\sout{q}/d^2}{\xout{q}}--------3[/tex]

We can strike out both qs in the numerator and denominator we have

[tex]E=\frac{ kQ}{d^2}[/tex]