HELP ASAP PLZ
What is the most likely reason for using the following model? A diagram that shows the different layers inside the sun Image © 2010 NASA The object is too small. The object represents a set of data. The object cannot be directly observed. The object represents a prediction about the future.

Explanation:

One idea would be to investigate the correlation between your pulse pressure and your pulse rate.  To do this, you'll need a blood pressure monitor.

First, measure your resting pressure and rate.  Then exercise for 30 seconds.  Measure your new blood pressure and pulse rate.  Wait for your pressure and rate to return to normal, then repeat the trial for 1 minute, 1.5 minutes, 2 minutes, etc.

List the results in a table.  This should include the amount of exercise time, your pulse rate, your systolic pressure (the high number, which is your blood pressure during contraction of your heart muscle), and your diastolic pressure (the low number, which is your blood pressure between heartbeats).  Calculate your pulse pressure (systolic minus diastolic) for each trial.  Graph the pulse pressure on the x-axis, and your pulse rate (beats per minute) on the y-axis.

What do you hypothesize will be the shape of the graph?  Consider Bernoulli's formula, which relates fluid pressure and flow.  How close do the results match your hypothesis?  What might explain any differences?

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  [tex]I = 2.50 \ kg \cdot m^2[/tex]

    The final  angular speed is [tex]w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s[/tex]

     The time taken is  [tex]t = 8.0 s[/tex]

      The initial angular speed is  [tex]w_i = 0\ rad/s[/tex]

Generally the average angular acceleration is mathematically represented as

        [tex]\alpha = \frac{w_f - w_i }{t}[/tex]

=>     [tex]\alpha = \frac{41.89}{8}[/tex]

=>      [tex]\alpha = 5.24 \ rad/s^2[/tex]

Generally the torque is mathematically represented as

   [tex]\tau = I * \alpha[/tex]

=>    [tex]\tau = 5.24 * 2.50[/tex]

=>     [tex]\tau = 13.09 \ N \cdot m[/tex]

Answer:

The stress is  [tex]stress = 2688000 \ N[/tex]

Explanation:

From the question we are told that

    The first temperature is  [tex]T_1 = 10 ^o \ C[/tex]

    The  young modulus is  [tex]Y = 7.00 *10^9\ N/m^2[/tex]

    The compressive strength is  [tex]\sigma = 2.00 *10^{9} \ N/m^2[/tex]

     The coefficient of  linear expansion is  [tex]\alpha = 1.2 *10^{-5} \ ^o C ^{-1}[/tex]

     The  second temperature is  [tex]T_2 = 42.0^o \ C[/tex]

Generally the change in length of the concrete is mathematically represented as

      [tex]\Delta L = \alpha * L * [T_2 - T_1 ][/tex]

=>  [tex]\frac{\Delta L}{L} = \alpha * [T_2 - T_1 ][/tex]

=> [tex]strain = \alpha * [T_2 - T_1 ][/tex]

Now  the young modulus is  mathematically represented as

        [tex]Y = \frac{stress}{strain}[/tex]

=>     [tex]7.00 *10^9 = \frac{stress}{\alpha(T_2 - T_1 ) }[/tex]

=>   [tex]stress = \alpha (T_2 - T_1 ) * 7.00 *10^9[/tex]

=>   [tex]stress = 1.2* 10^{-5} (42 - 10 ) * 7.00 *10^9[/tex]

=>   [tex]stress = 2688000 \ N[/tex]

Answer:

a) 607.95 J

b) 0 J

Explanation:

a) Initial volume = 1 L = 0.001 m^3

final volume = 4 L = 0.004 m^3

pressure = 2 atm = 202650 Pa     (1 atm = 101325 Pa)

work done by the gas on the environment = PΔV

P is the pressure = 101325 Pa

ΔV is the change in volume from the initial volume to the final volume

ΔV = 0.004 m^3 - 0.001 m^3 = 0.003 m^3

work done by the gas = 202650 x 0.003 = 607.95 J

b) If the gas is cooled at constant volume, then the gas does no work. For a gas to do work, there must be a change in its volume.

Therefore the work done in cooling at constant volume until pressure falls to 1.5 atm = 0 J

Answer:

[tex]n=6.25\times 10^{11}[/tex]

Explanation:

We need to find the number of electrons that would have to be removed from a coin to leave it with a charge of [tex]+10^{-7}\ C[/tex]. Then the number of electrons be n. Using quantization of electric charge as :

q = ne

e is charge on an electron

[tex]n=\dfrac{q}{e}\\\\n=\dfrac{10^{-7}}{1.6\times 10^{-19}}\\\\n=6.25\times 10^{11}[/tex]

So, the number of electrons are [tex]6.25\times 10^{11}[/tex].

Answer:

The mass is  [tex]m_w = 0.599 \ kg[/tex]

Explanation:

From the question we are told that    

     The mass of ice is  [tex]m_c = 0.20 \ kg[/tex]

     The  initial temperature of the ice is  [tex]T_i = -40.0 ^oC[/tex]

     The  initial temperature of the water is  [tex]T_{iw} = 80^o C[/tex]

     The  final temperature of the system is [tex]T_f = 20^oC[/tex]

Generally according to the law of energy conservation,

   The  total heat loss is  =  total heat gained

 Now the total heat gain is mathematically represented as

      [tex]H = H_1 + H_2 + H_3[/tex]

Here  [tex]H_1[/tex] is the energy required to move the ice from [tex]-40^oC \to 0^oC[/tex]

And it mathematically evaluated as

     [tex]H_1 = m_c * c_c * \Delta T[/tex]

Here the specific heat of ice is  [tex]c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]

So  

    [tex]H_1 = 0.20 * 2100 * (0-(-40))[/tex]  

     [tex]H_1 = 16800\ J[/tex]

[tex]H_2[/tex] is the energy to melt the ice

And it mathematically evaluated as

       [tex]H_2 = m * H_L[/tex]

The  latent heat of fusion of ice is  [tex]H_L = 334 J/g = 334 *10^{3} J /kg[/tex]

So  

    [tex]H_2 = 0.20 * 334 *10^{3}[/tex]

    [tex]H_2 = 66800 \ J[/tex]

[tex]H_3[/tex] is the energy to raise the melted ice to [tex]20^oC[/tex]

And it mathematically evaluated as

    [tex]H_3 = m_c * c_w * \Delta T[/tex]

Here the specific heat of water  is  [tex]c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]

    [tex]H_3 = 0.20 * 4190* (20-0))[/tex]  

     [tex]H_3 = 16744 \ J[/tex]  

So

  [tex]H = 16800 + 66800 + 16744[/tex]

   [tex]H = 100344\ J[/tex]

The  heat loss is mathematically evaluated as

     [tex]H_d = m * c_h ( 80 - 20 )[/tex]

     [tex]H_d = m_w * 4190 * ( 80 - 20 )[/tex]

     [tex]H_d = 167600 m_w[/tex]

So

      [tex]167600 m_w = 100344[/tex]

=> [tex]m_w = 0.599 \ kg[/tex]

Answer:

The velocity is [tex]v = 4.76 \ m/s[/tex]

Explanation:

From the question we are told that

   The first distance is   [tex]d_1 = 4.0 \ km = 4000 \ m[/tex]

   The  first speed  is  [tex]v_1 = 5.0 \ m/s[/tex]

    The  second distance is  [tex]d_2 = 1.0 \ km = 1000 \ m[/tex]

    The  second speed  is  [tex]v_2 = 4.0 \ m/s[/tex]

Generally the time taken for first distance is  

      [tex]t_1 = \frac{d_1 }{v_1 }[/tex]

        [tex]t_1 = \frac{4000}{5}[/tex]

       [tex]t_1 = 800 \ s[/tex]

The time taken for second  distance is

           [tex]t_1 = \frac{d_2 }{v_2 }[/tex]

        [tex]t_1 = \frac{1000}{4}[/tex]

       [tex]t_1 = 250 \ s[/tex]

The total time is mathematically represented as

     [tex]t = t_1 + t_2[/tex]

=>   [tex]t = 800 + 250[/tex]

=>    [tex]t = 1050 \ s[/tex]

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      [tex]v = \frac{d_1 + d_2}{t }[/tex]

=>    [tex]v = \frac{4000 + 1000}{1050 }[/tex]

=>    [tex]v = 4.76 \ m/s[/tex]

The constant speed that will let her finish in the same time as in the first race is 4.76 m/s

Time 1 = distance / speed

Time 1 = 4000 / 5

Time 1 = 800 s

Time 2 = distance / speed

Time 2 = 1000 / 4

Time 2 = 250 s

Constant speed = Total distance / total time

Constant speed = 5000 / 1050

Constant speed = 4.76 m/s

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Answer:

F = 2009.64 N

Explanation:

It is given that,

Mass of a baseball, m = 0.145 kg

Initial speed if the baseball, u = 33 m/s

It hit on a horizontal line drive straight back at the pitcher at 46.0 m/s, final velocity, v = -46 m/s

Time of contact between the bat and the ball is [tex]t=5.7\times 10^{-3}\ s[/tex]

We need to find the magnitude of the force exerted by the ball on the bat. It can be calculated using impulse-momentum theorem. So,

[tex]Ft=m(v-u)\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.145\times (-46-33)}{5.7\times 10^{-3}}\\\\F=-2009.64\ N[/tex]

So, the magnitude of force exerted on the ball by the bat is 2009.64 N.

Answer:

[tex] \boxed{\sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )} [/tex]

Given:

Refractive index of air ( [tex] \sf \mu_{air} [/tex] )= 1

Refractive index of glass slab ( [tex] \sf \mu_{glass} [/tex]) = 1.40

Angle of incidence (i) = 30.0°

To Find:

Angle of refraction (r)

Explanation:

From Snell's Law:

[tex] \boxed{ \bold{ \sf \mu_{air}sin \ i = \mu_{glass}sin \: r}}[/tex]

[tex] \sf \implies 1 \times sin \: 30 ^ \circ = 1.4sin \:r[/tex]

[tex] \sf sin \:30^ \circ = \frac{1}{2} : [/tex]

[tex] \sf \implies \frac{1}{2} = 1.4 sin \: r[/tex]

[tex] \sf \frac{1}{2} = 1.4 sin \: r \: is \: equivalent \: to \: 1.4 sin \: r = \frac{1}{2} : [/tex]

[tex] \sf \implies 1.4 sin \: r = \frac{1}{2} [/tex]

Dividing both sides by 1.4:

[tex] \sf \implies \frac{\cancel{1.4} sin \: r}{\cancel{1.4}} = \frac{1}{2 \times 1.4} [/tex]

[tex] \sf \implies sin \: r = \frac{1}{2 \times 1.4} [/tex]

[tex] \sf \implies sin \: r = \frac{1}{2.8} [/tex]

[tex] \sf \implies r = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]

[tex] \therefore[/tex]

[tex] \sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]

Answer:

The initial speed of the block is 1.09 m/s

Explanation:

Given;

mass of block, m = 1.7 kg

force constant of the spring, k = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

From principle of conservation of energy

kinetic energy of the block = elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv²  = kx²

[tex]v = \sqrt{\frac{kx^2}{m} }[/tex]

where;

v is the initial speed of the block

x is the compression of the spring

[tex]v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s[/tex]

Therefore, the initial speed of the block is 1.09 m/s

Answer:

(a) 5.0 kg

(b) 10 kg

Explanation:

Draw a free body diagram for each block.  There are 4 forces on block A:

Weight force mAg pulling down,

Normal force N pushing up,

Tension force T pulling right,

and friction force Nμ pushing left.

There are 2 forces on block B:

Weight force mBg pulling down,

and tension force T pulling up.

Whether the system is just starting to move, or moving at constant speed, the acceleration is 0.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = 0

mBg = T

Sum of forces on A in the +y direction:

∑F = ma

N − mAg = 0

N = mAg

Sum of forces on A in the +x direction:

∑F = ma

T − Nμ = 0

T = Nμ

Substitute:

mBg = mAg μ

mA = mB / μ

(a) When the system is just starting to move, μ = 0.40.

mA = 2.0 kg / 0.40

mA = 5.0 kg

(b) When the system is moving at constant speed, μ = 0.20.

mA = 2.0 kg / 0.20

mA = 10 kg

m_1=5kg

m=10kg

From the question we are told

the coefficient of static friction between mass mA  and the table is 0.40, where as the coefficient of kinetic friction  is 0.20.

a)  

Generally the equation for the Tension  is mathematically given as

T=mg

Where

[tex]m_1g=m_2g[/tex]

Therefore

[tex]m_1=\frac{2.0}{0.4}\\\\m_1=5kg[/tex]

b

Generally the equation for the Tension  is mathematically given as

[tex]T=f\\\\T=u_km_1g\\\\\m_1=\frac{m_2}{u}\\\\m_1=\frac{2}{0.2}[/tex]

m=10kg

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Answer:

d) y(x,t) = (10.5 mm)cos(172rad?m?1 x ?2730rad?s?1t)

Answer:

t = 5 s

Explanation:

In uniform rectilinear movement, the equation for final speed is:

vf = v₀ + a*t

In this case we need that the car stops just before 60 m after applied the brakes, then

vf = v₀ - a*t

vf = final speed = 0

v₀ = initial speed =  15 m/s

And negative acceleration is 3 m/s²

0 = 15 (m/s) - 3 ( m/s²)*t

t = 15 / 3   (m/s /m/s²)

t = 5 s

The point is that with that value ranger will hit the deer so in order to not to hit the deer that time should be smaller than 5 seconds

Answer:B

Explanation: I just did it on a p e x

273 K gives the object's temperature in the SI unit therefore the correct answer is option B

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

The International System of Units, sometimes known as the SI system of units, is the most frequently used and acknowledged system of units in use nowadays. There are three additional units and 7 SI basic units in this system of SI units.

The three supplemental SI units are radian, steradian, and becquerel, whereas the base SI units are meter, kilogram, second, kelvin, ampere, candela, and mole. These base units can be used to create all other SI units.

Thus,273 K gives the object's temperature in the SI unit therefore the correct answer is option B

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Answer:

Infrared light

Explanation:

Infrared light is the spectrum of electromagnetic wave given off by a body possessing thermal energy. Infrared light is preferred over visible light in this region of space because visible light is easily scattered in the presence of fine particles. Infrared ray makes it easy for us to observe Cold, dark molecular clouds of gas and dust in our galaxy that glows when irradiated by the stars . Infrared can also be used to detect young forming stars, even before they begin to emit visible light. Stars emit a smaller portion of their energy in the infrared spectrum, so nearby cool objects such as planets can be more readily detected with infrared light which won't be possible with an ultraviolet or visible light.

Answer:

The correct option is C

Explanation:

From the question we are told that

   The wavelength is  [tex]\lambda = 575 *10^{-9} \ m[/tex]

    The  angle is  [tex]\theta = 6.5^o[/tex]

    The order of maxima  is  n =  3

Generally for  constructive interference

       [tex]dsin \theta = n * \lambda[/tex]

=>   [tex]d = \frac{n * \lambda }{ sin \theta }[/tex]

=>   [tex]d = \frac{3 * 575 *10^{-9} }{ sin 6.5 }[/tex]

=>   [tex]d = 15.24 *10^{-6} \ m[/tex]

=>  [tex]d = 15 \mu m[/tex]

Explanation:

time taken fir 50 oscillations is 6 seconds

time taken for 1 oscillation is 6/50

convert it into a decimal

Answer:

The angular separation is  [tex]k = 0.8594^o[/tex]

Explanation:

From the question we are told that

   The  wavelength of the light is [tex]\lambda = 600 \ nm = 600*10^{-9} \ m[/tex]

   The  distance of separation between the slit is  [tex]d = 0.080 \ mm = 0.080 *10^{-3} \ m[/tex]

    The distance from the screen is

Generally the condition for  constructive interference is mathematically represented as

        [tex]d \ sin(\theta) = n \lambda[/tex]

=>    [tex]\theta = sin ^{-1} [ \frac{n * \lambda }{ d } ][/tex]

    here [tex]\theta[/tex] is the angular separation between the central maxima and one side of the first order maxima

given that we are considering the first order of maxima n =  1  

        =>   [tex]\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ][/tex]

        =>    [tex]\theta = sin ^{-1} [ 0.0075 ][/tex]

        =>   [tex]\theta = 0.4297^o[/tex]

So the angular separation of the two first order maxima  is  

     [tex]k = 2 * \theta[/tex]

     [tex]k = 2 * 0.4297[/tex]

      [tex]k = 0.8594^o[/tex]

Answer:

The  depth of water at the point is  [tex]d_A = 6.55 \ m[/tex]

Explanation:

From the question we are told that

   The height of the person above water   is  [tex]d = 3.00 \ m[/tex]

   The distance  of the coin as seen by the person  is [tex]d' = 8.00 \ m[/tex]

Generally the apparent depth is mathematically represented as

      [tex]d_a = \frac{d_A}{n}[/tex]

Here [tex]d_A[/tex] is the actual depth of water while  n is the refractive index of water with a constant value [tex]n = 1.33[/tex]

Now from the point the person is the apparent depth is evaluated as

     [tex]d_a = d'-d[/tex]

=>   [tex]d_a = 8 - 3[/tex]

=>  [tex]d_a = 5 \ m[/tex]

So

     [tex]5 = \frac{d_A}{1.33}[/tex]

=>   [tex]d_A = 5 * 1.33[/tex]

=>   [tex]d_A = 6.55 \ m[/tex]

Answer:

n₁ > n₂.

prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50

Explanation:

Total internal reflection occurs when the refractive index of the incident medium the light is greater than the medium to which the light is refracted, let's use the refraction equation

                 n₁ sin θ₁ = n₂ sin  θ₂

the incident medium is 1, at the limit point where refraction occurs is when the angle in the refracted medium is 90º, so sin θ₂ = 1

                 n₁ sin θ₁ = n₂

                 sin θ₁ = n₂ / n₁

We mean that this equation is defined only for n₁ > n₂.

In our case, for the total internal reflection to occur, the refractive incidence of the medium must be greater than the index of refraction of the prism.

In general, prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50

Answer:

its 45 over 6

Explanation:the answer is in  the question

Answer: Only the melted cube's shape changed.

Explanation:

Answer:

1.44*10^-3m

Explanation:

Given that distance BTW two bright fringes is

DetaY = lambda* L/d

So for second wavelength

Deta Y2= Lambda 2* L/d

=lambda 2 x deta y1/ lambda1

So substituting

= 360 x 10^-9 x (1.6*10^-3/640*10^-9)

1.44*10^ -3m

Answer:

   k = -3.1450 10⁻⁴  s⁻¹

Explanation:

In this exercise we are given the equation that describes the cooling process

        dT / dt = k (T -)

Let's solve is this equation,

        dT / (T-T_ {e}) = k dt

change of variable for integration

       T -T_{e} = T ’

       dT = dT '

       ∫ dT ’/ T’ = k  ∫ dt

we integrate

        ln T ’= k t

we change to the initial variables

        ln (T - T_{e}) = k t

Let's evaluate from the lower limit T = T for t = 0 to the upper limit T = T₀ for time t

       ln (T₀ -T_{e}) - ln (T -T_{e}) = k (t-0)

we simplify

       ln (T₀ -T_{e} / T -T_{e}) = k t

       k = ln (T₀ -T_{e})  / (T-Te) / t

In the exercise they indicate that the temperature T = 205 F, the ambient temperature is T_{e} = 70F, the temperature to which T₀ = 200 F falls in a time t = 2 min = 120 s

Let's calculate

           k = ln [(200- 70) / (205 -70)] / 120

           k = -0.0377403 / 120

           k = -3.1450 10⁻⁴  s⁻¹

Explanation:

If m₁ is at the origin, then the center of mass is at:

x = (m₁ × 0 m + m₂ × d m) / (m₁ + m₂)

x = m₂ d / (m₁ + m₂)

If m₁ = 1 kg, m₂ = 3 kg, and d = 2 m:

x = (3 kg) (2 m) / (1 kg + 3 kg)

x = 1.5 m

Sum of forces in the x direction:

∑F = ma

16 N = (1 kg + 3 kg) aₓ

aₓ = 4 m/s²

Sum of forces in the y direction:

∑F = ma

20 N = (1 kg + 3 kg) aᵧ

aᵧ = 5 m/s²

Answer:

A very high metabolism and a very small size.

Explanation:

The pygmy shrew is a very small mammal, that forages day and night. The metabolism of the Pygmy shrew is so high that it must eat at least every 30 minutes or it might die. The best explanation for what happens to the food's mass and energy is that most of the food mass is rapidly used fro building up of the shrew due to its very high metabolism, and a bigger portion of the food is lost from the surface of the body of the shrew, due to its very small size. The combination of these two factors; a very high metabolism (rapidly uses up food material, and generates a large amount of heat in a very short time) and the very small size (makes heat loss due to surface area to volume ratio high) explains what happens to the food mass and energy.

Answer:

6.624 x 10^-21 J

Explanation:

The temperature of the ideal gas = 320 K

The average translational energy of an ideal gas is gotten as

[tex]K_{ave}[/tex] = [tex]\frac{3}{2}K_{b}T[/tex]

where

[tex]K_{ave}[/tex]  is the average translational energy of the molecules

[tex]K_{b}[/tex] = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1

T is the temperature of the gas = 320 K

substituting value, we have

[tex]K_{ave}[/tex] = [tex]\frac{3}{2} * 1.38*10^{-23} * 320[/tex] = 6.624 x 10^-21 J

Answer:

Explanation:

If the mass of the probe is 500kg, its weight W = mass  acceleration due to gravity.

Weight of the probe = 500*9.81

Weight of the probe = 4,905N

If its average density =  1400kg/m³

Volume = Mass/Density

Volume = 500/1400

Volume = 0.3571m³

According to the floatation principle, the volume of the probe is equal to the volume of liquid displaced. Hence the volume of water displaced is 0.357m³.

Since density of water is 1000kg/m³, we can find the mass of the water using the formula;

Mass of water = Density of water * Volume of water

Mass of water = 1000*0.3571

Mass of water = 357.1kg

Weight of water displaced = 3571 * 9.81 = 3503.15N

The tension in the cable will be the difference between the weight of the probe and weight of the displaced fluid.

Tension in the cable = 4,905N -  3503.15N

Tension in the cable = 1,401.85N

Hence the tension in the cable is 1,401.85N

Answer:

a

 [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]

b

The  direction of the electric field is opposite that of the current              

Explanation:

From the question we are told that

   The current is  [tex]I = 17\ A[/tex]

   The diameter of the ring is  [tex]d = 3.0 \ cm = 0.03 \ m[/tex]

Generally the  radius is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

       [tex]r = \frac{0.03}{2}[/tex]

       [tex]r = 0.015 \ m[/tex]

The  cross-sectional area is mathematically represented as

       [tex]A = \pi r^2[/tex]

=>     [tex]A = 3.142 * (0.015^2)[/tex]

=>    [tex]A = 7.07 *10^{-4 } \ m^ 2[/tex]

Generally  according to ampere -Maxwell equation we have that

      [tex]\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }[/tex]

Now given that [tex]\= B = 0[/tex] it implies that

     [tex]\oint \= B \cdot \= ds = 0[/tex]

So

    [tex]\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0[/tex]

Where  [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

            [tex]\mu_o[/tex] is the permeability of free space with value  

[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

      [tex]\phi[/tex] is magnetic flux which is mathematically represented as

       [tex]\phi = E * A[/tex]

Where E is the electric field strength

  So  

       [tex]\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0[/tex]

=>   [tex]\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }[/tex]

=>   [tex]\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }[/tex]

=>   [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]

The  negative  sign shows that the  direction  of  the electric field is opposite that of the current

This question is incomplete, the complete question is;

In a simple model of a potassium iodide (KI) molecule, we assume the K and I atoms bond ionically by the transfer of one electron from K to I.

(a) The ionization energy of K is 4.34 eV, and the electron affinity of I is 3.06 eV. What energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms? This quantity is sometimes called the activation energy Ea.eV

(b) A model potential energy function for the KI molecule is the Lennard - jones potential:

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

where r is the internuclear separation distance and α and ∈ are adjustable parameters (constants) . The Ea term is added to ensure the correct asymptotic behavior at large r and is activation energy calculated in a. At the equilibrium separation distance, r=r₀=0.305 nm, U(r) is a minimum, and dU/dr=0. In addition, U(r₀)=-3.37 eV.

Us the experimental values for the equilibrium sepeartion and dissociation energy of KI to determine/find 'α' and '∈'.

(c) calculate the force needed to break the KI molecule in nN

Answer:

a) energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms is 1.28 eV

b) α = 0.272, ∈ = 4.65 eV

c) the force needed to break the KI molecule in nN 65.6 nN

Explanation:

a) The ionization energy of K is 4.34 ev ( energy needed to remove the outer most electrons)

And the electron affinity of I is 3.06 ev ( which is energy released when electron is added)

Now the energy that is need to transfer an electron from K to I,

i.e the ionization energy of K(4.34 ev) and the electron affinity of I (3.06 ev)

RE = 4.34 - 3.06 = 1.28 eV

b)

from the question we have

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

now taking d/drU(r₀)=0  (at r = r₀)

= 4∈d/dr [ (α/r)¹² - (α/r)⁶ ] = 0

= ( -12(α¹²/r¹³)) - (-6 (α⁶/r⁷)) = 0

12(α¹²/r¹³) = 6 (α⁶/r⁷)

α⁶ = r⁶/2

α = r/(2)^1/6

at equilibrium r = r₀ = 0.305 nm

α = 0.305 nm / (2)^1/6

C = 0.0305/1.1246

α = 0.272

Now substituting the values of U(r₀), α, Eₐ in the initial expression

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

we have

- 3.37eV = 4∈ [ (0.272 nm / 0.305 nm)¹² - (0.272 nm / 0.305 nm )⁶ ] + 1.28

- 1.65 eV = ∈(0.25 - 0.5)

∈ = 4.65 eV

c)

Now to break the molecule then the potential energy should be zero(0)

and we know r = 0.272 nm

therefore force needed to break the molecule is

F = -dU/dR_r-α

F = -4∈ (-12/α  + 6/α)

F = -4(4.65eV) ( -12/0.272nm + 6/0.272nm)

F = 65.6 nN