In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.

Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

Answer:

It will take. the same distance up as before, but take a longer time

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]

where;

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

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

Answer: Search Results

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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.

Explanation:

Surface waves can exhibit characteristics of both longitudinal waves and transverse waves.

Surface waves are a type of mechanical wave that propagate along the interface between two different mediums, such as the ground and air or the surface of water. These waves combine properties of both longitudinal and transverse waves

Similar to longitudinal waves, surface waves involve particles oscillating in the same direction as the wave propagation. This creates compressions and rarefactions, leading to variations in density or pressure. These compressions and rarefactions are characteristic of longitudinal waves.

However, surface waves also exhibit transverse motion. As the wave propagates along the surface, particles move in a perpendicular direction to the wave's motion. This transverse motion causes particles to displace vertically or horizontally, similar to transverse waves.

By combining both longitudinal and transverse characteristics, surface waves possess a complex motion that allows them to travel along the surface while simultaneously causing particles to oscillate both parallel and perpendicular to the direction of wave propagation.

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Answer:

The cross-sectional area of the larger piston is 392 cm²

Explanation:

Given;

output mass of the piston, m₀ = 2000 kg

input force of the piston, F₁ = 500 N

input area of the piston, A₁ = 10 cm² = 0.001 m²

The output force is given by;

F₀ = m₀g

F₀ = 2000 x 9.8

F₀ = 19600 N

The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]

Therefore, the cross-sectional area of the larger piston is 392 cm²

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Answer:

Following are the answer to this question:

Explanation:

Formula:

[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]

Calculating point A:

when the value is [tex]0.38[/tex]

[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]

                   [tex]=2.632[/tex]

[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]

               [tex]=542,802.6[/tex]

Calculating point B:

when the value is [tex]0.75[/tex]

[tex]\to D(PC)=\frac{1}{0.75}[/tex]

                [tex]=1.33[/tex]

[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]

             [tex]=275,020[/tex]

Calculating point C:

when the value is [tex]0.28[/tex]

[tex]\to D(PC)=\frac{1}{0.28}[/tex]

                [tex]=3.571[/tex]

[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]

               [tex]=736660.7[/tex]

Calculating point D:

when the value is [tex]0.42[/tex]

[tex]\to D(PC)=\frac{1}{0.42}[/tex]

                [tex]=2.38[/tex]

[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]

               [tex]=490910.7[/tex]

Calculating point E:

when the value is [tex]0.31[/tex]

[tex]\to D(PC)=\frac{1}{0.31}[/tex]

                [tex]=3.226[/tex]

[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]

               [tex]=665370.97[/tex]

Answer:

Well, yes.

We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.

Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.

An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.

Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.

Answer:

i) 545.2 N  upwards

ii) 828.2 N  downwards

Explanation:

mass of the roller = 70 kg

force exerted = 200 N

angle the force makes with the ground ∅ = 45°

weight of the roller W = mg

where

m is the mass of the roller

g is the acceleration due to gravity = 9.81 m/s^2

weight of the roller = 70 x 9.81 = 686.7 N

The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°

==> F = 200 x 0.707 = 141.5 N

i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards

Resultant vertical force = 686.7 N - 141.5 N = 545.2 N  upwards

ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards

Resultant vertical force = 686.7 N + 141.5 N = 828.2 N  downwards

Answer:

A) Length of an edge = 3.38 × 10^(-9) m

B) 34 times the diameter of a molecule.

C) 11 times the atomic spacing in solids.

Explanation:

A) We will use Avogadro's hypothesis to solve this. It states that 1 mole of gas occupies 22.4 L at STP.

We want to find the volume occupied by 1 mole of gas at 1.06 atm pressure and temperature of 28 °C (= 301 K).

Thus, by the ideal gas equation, we have;

V_mole = (1 × 22.4/273) × (301/1.06) = 23.3 L = 0.0233 m³

Now, since from avogadros number, 1 mole of gas contains 6.02 x 10^(23) molecules, then volume occupied by a molecule is given by;

V_molecule = 0.0233/(6.02 × 10^(23)) m³ = 3.87 x 10^(-26) m³

Thus, length of an edge of the cube = ∛(3.87 × 10^(-26)) = 3.38 × 10^(-9) m

B) We are told that The diameter of a typical molecule is about 10^(-10) m.

Thus, the distance is about;

(3.38 × 10^(-9))/(10^(-10)) ≈ 34 times the diameter of a molecule.

C) We are told that the spacing of atoms is typically are about 0.3 nm apart

Thus;

The separation will be about;

(3.38 × 10^(-9))/(0.3 × 10^(-9)) ≈ 11 times the atomic spacing in solids.

Answer:

a) 0.988 m/s

b) 199512 Pa

c) 57.52 s

Explanation:

given that

A

A1 v1 = A2 v2

d1² v1 = d2² v2

v2 = [d1/d2]² v1

v2 = (1.2/5.4)² * 20

v2 = 0.049 * 20

v2 = 0.988 m/s

B

P + 1/2 ρ v² = K.

[p2 - p1] = 1/2 ρ [v1² - v2²]

[p2 - p1] = 1/2 * 1000 [20² - 0.988²]

[p2 - p1] = 500 * (400 - 0.976)

[p2 - p1] = 500 * 399

[p2 - p1] = 199512 Pa

C

Flow rate = AV = π [d²/ 4 ] * v

= π [0.012² / 4 ] * 20 = 0.00226 m³ /s

= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s

130 liters = 0.13 m³

t = 0.13/ 0.00226 = 57.52 s

Answer:

Not slow down or speed up.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

Answer:

The frequency does not change, but the wavelength does

Explanation:

Here are the options

A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.

B. The frequency does change, but the wavelength remains unchanged.

C. Both the frequency and wavelength change.

D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.

E. The frequency does not change, but its wavelength does.

When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.

[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]

where [tex]\lambda_o[/tex] indicates wavelength in vacuum

[tex]\lambda_m[/tex] indicates wavelength in medium

n indicates refractive index

v indicates velocity of light wave

c indicates velocity of light

And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.

Therefore the correct option is E

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]

So it would normally be written with a negative signs in front indicating a divergent lens.

Answer:

Hey there!

This can be a confusing topic, so it's totally fine if you get confused...

First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.

Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.

Let me know if this helps :)

Answer:

v₀(1 + B²L²t/mR)

Explanation:

We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Since F = ma where a = acceleration of loop and m = mass of loop

a = F/m = B²L²v₀/mR

Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0

Substituting the value of a and u into v, we have

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)

The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).

As we know that

Force on the loop

F = BIL

here

B = magnetic field strength,

I = current

and L = length of side of loop.

Now

the current in the loop I = ε/R

where

ε = induced e.m.f in the loop = BLv₀

where v₀ = velocity of loop

and r = resistance of loop

So,

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Also, F = ma where a = acceleration of loop and m = mass of loop

Now

a = F/m = B²L²v₀/mR

We have to use

v = u + at

where

u = initial velocity of loop = v₀,

t = time after t = 0

and v = velocity of loop after time t = 0

So, it be like

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

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Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

1. Electrons transferred from sphere to rod.

Option B is correct.

2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.

Given that initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let us consider that the charge absorbed by the plastic rod  is  q

[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Therefore, electrons transferred from sphere to rod

The charge on one electron is, 1.602 x 10⁻¹⁹ C .

Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]

Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.

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Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

​

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

Answer:

TOTAL INTERNAL REFLECTION

Explanation:

Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.

For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.

Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION

Answer:

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Answer:

D.

Inverted and reduced

If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.

A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

Hence option B is correct.

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Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km