How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

We need to find the ratio of the number of turns in the primary to the secondary  for step up transformer.

For a transformer, [tex]\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}[/tex]

So,

[tex]\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}[/tex]

So, the ratio of the number of turns in the primary to the secondary is 1:2.

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

Answer:

Explanation:

Let the original resistance be R and voltage be V

Applying ohm's law

V / R = 15

V = 15 R

In second case

V / (R+8 ) = 12

V = 12 R + 96

15 R = 12 R + 96

3R = 96

R = 32 ohm .

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]

Then the addition of these two velocities equals:

[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]

That time is the "period" of the wave.

(It's not one of the choices.)

The blank of a sine wave is the time it takes to complete one cycle of the wavelength, the correct answer is D.

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

where c is the speed of light

ν is the frequency of the wave

λ is the wavelength of the wave

The time taken by the sine wave to complete one cycle of the wavelength is called blank the correct answer is D.

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Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m

r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m

electric potential V = [tex]\frac{kq}{r}[/tex]

change in potential ΔV = [tex]V_{1} - V_{2}[/tex]

ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC

ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]

ΔV= 2.789×10⁵

[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃

[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

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Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

Answer:

The  velocity is [tex]v = 37 .46 \ m/s[/tex]

Explanation:

From the question we are told that

    The start distance above the ground is  [tex]h = 71.6 \ m[/tex]

Generally according to the  law of energy conservation we have that

     [tex]PE_{top} = KE_{bottom }[/tex]

Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as

      [tex]PE_{top} = m * g * h[/tex]

And  [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as

     [tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]

Therefore  

       [tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]

=>    [tex]v = \sqrt{2 * g * h }[/tex]

substituting value

     [tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]

     [tex]v = 37 .46 \ m/s[/tex]

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

Answer:

The major cause is "lack of high frequencies in a sound wave".

Explanation:

So that the above would be the correct solution.

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

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

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale [tex]F_2[/tex] reads  [tex]F_2 = 2.4225 \ N[/tex]

Explanation:

From the question we are told that

      The first force is  [tex]F_1 = 10.5 \ N[/tex]

      The acceleration by which the cart moves to the right is  [tex]a = 2.50 \ m/s^2[/tex]

      The mass of the cart is  m  = 3.231  kg

Generally the net force on the cart is  

       [tex]F_{net} = F_1 - F_2[/tex]

This net force is mathematically represented as

      [tex]F_{net} = m * a[/tex]

So  

        [tex]m* a = 10 - F_2[/tex]

        [tex]F_2 = 10.5 - 2.5 (3.231)[/tex]

        [tex]F_2 = 2.4225 \ N[/tex]

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

Answer:

Explanation:

Change in entropy is expressed as ΔS = ΔQ/T where;

ΔQ is the total heat change during conversion of ice to water.

T is the room temperature

First we need to calculate the total change in heat using the conversion formulae;

ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)

m is the mass of ice/water = 900g = 0.9kg

L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg

c is the specific heat capacity of water = 4200J/kgK

Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K

Substituting the given values into ΔQ;

ΔQ = 0.9(333000)+0.9(4200)(283)

ΔQ = 299700 + 1069740

ΔQ = 1,369,440 Joules

Since Change in entropy ΔS = ΔQ/T

ΔS =  1,369,440/30+273

ΔS = 1,369,440/303

ΔS = 4519.60 J/K

Hence, the change in entropy of the universe is 4519.60 J/K

Answer:

a. 0.849 micro farad

b. 2.89 s

Explanation:

a) V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B) time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

a. The capacitance is 0.849 micro farad

b. The  time constant of the circuit is 2.89 s

a)

We know that

V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B)

Now

time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

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Answer:

   K_A = 32.2 10⁶ J

Explanation:

In this exercise we must relate the quantities given to find the kinetic energy

Asteroid A data

              m_A = 3.5 m_B

               v_A = 2.0 v

they also give the value of the kinetic energy of asteroid A

              K_B = 2.3 10⁶ J

the expression for scientific energy is

               K = ½ m v²

let's replace

              K_A = ½ m_a V_a2

               K_A = ½ 3.5 m_B (2.0 v_B)^2

                K_A = 3.5 2² (½ m_B v_B²)

                K_A = 14 K_B

               K_A = 32.2 10⁶ J

Answer:

The intensity is [tex]I = 500 mW/m^2[/tex]

Explanation:

From the question we are told that

    The  intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]

Generally the intensity of the light emerging from the polarizer is  mathematically represented as

          [tex]I = \frac{I_o}{2}[/tex]

substituting values

         [tex]I = \frac{1000 *10^{-3}}{2}[/tex]

         [tex]I = 500 *10^{-3} W/m^2[/tex]

         [tex]I = 500 mW/m^2[/tex]

Answer:

The frequency is  [tex]f_n = 257.1 \ Hz[/tex]

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  [tex]f_3 = 540 \ Hz[/tex]

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         [tex]f = \frac{v_s}{ 2L}[/tex]

Now when it finger at 70% it original length is

      [tex]f_n = \frac{v}{2 * (0.7 L)}[/tex]

      [tex]f_n = \frac{v}{1.4 L}[/tex]

Here v  the velocity of sound

  So  

         [tex]\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

Also the fundamental frequency for the original length can also be represented as

       [tex]f = \frac{f_3}{3}[/tex]

substituting values

          [tex]f = \frac{540}{3}[/tex]

          [tex]f = 180 \ Hz[/tex]

So

       [tex]\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

=>  [tex]f_n =\frac{180}{0.7}[/tex]

=>   [tex]f_n = 257.1 \ Hz[/tex]

The fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

f is the frequency of tight guitar string = 540 Hz

Let's call the original length L.

The amount of time it is fingered is then 0.7 L.

In general, the fundamental frequency is expressed mathematically as;

[tex]\rm f = \frac{v_0}{2L} \\\\[/tex]

For the given conditions;

[tex]\rm f_n=\frac{v}{2 \times 0.7L} \\\\ \rm f_n=\frac{v}{1.4L}[/tex]

The ratio of the frequency is;

[tex]\rm \frac{f_n}{f} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }[/tex]

Also, the fundamental frequency for the original length can also be represented as;

[tex]\rm f= \frac{f'}{3} \\\\ f=\frac{540}{3} \\\\ \rm f=180\ Hz[/tex]

On putting the given data;

[tex]\rm \frac{f_n}{180} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }\\\\ \rm f_n=\frac{180}{0.7}\\\\\ \rm f_n=257.1\ Hz[/tex]

Hence the fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

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Answer:

Option 3 = both spheres are at the same potential.

Explanation:

So, let us complete or fill the missing gap in the question above;

" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"

The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:

=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.

Answer:

118.6 rad/s²

Explanation:

Δθ = 4100.0 rad

ω₀ = 159.0 rad/s

ω = 999.0 rad/s

Find: α

ω² = ω₀² + 2αΔθ

(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)

α = 118.6 rad/s²

Answer:

Net charge = E• b • L^3.

Explanation:

NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).

So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.

(1). Starting from -z and z faces which are the back and front faces of the cube:

Thus, We have that the flux,p = 0 for -z and z.

(2). Recall that we are given that E = =(a+bx)i^+cj^.

Thus, p_-y = (a + bx)i + cj (-j) (L^2)

Where y = 0

p_-y = -cL^2.

Obviously for p_j, we will have cL^2 and y = L

(3). For p_-x = =(a + bx)i + cj (-i) (L^2).

p_-x = -aL^2

Where x = 0.

When x = L and p_x = (a + bL)L^2.

This, adding all together gives Net charge = E • b • L^3.

Answer:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

Answer:

5.49 x 10^17 m  is the distance between the sun-like star to the earth

Explanation:

Radiation intensity on Earth = 1.0 x 10^-10 W/m^2

Power of radiation of the star = 3.8 x 10^26 W

Recall that the intensity of radiation is given as

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where

[tex]I[/tex] = intensity of radiation

P = power of radiation

A is the area through which the radiation spreads out in all three dimensional direction.

A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2

This area is spread out in the form of a sphere of area

A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]

3.8 x 10^36 = 12.568[tex]r^{2}[/tex]

[tex]r^{2}[/tex] =  (3.8 x 10^36)/12.568 = 3.02 x 10^35

r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m   this is the distance of the star to the Earth

Answer:

B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.

Explanation:

This is the Fleming's right hand rule, which was stated to explain the relationship or induction ability of the magnetic field, current or velocity of charged particles and magnetic force. These three variables are held mutually perpendicularly to one another.

The most suitable description of the right-hand rule is option B which clarifies the perpendicular mutual relationship of the thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.