What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffraction grating has 700 lines per mm? ​

Answer:

I] alpha = wf/t = (45*2pi/60)/1 = 4.712 rad/s

ii] W = alpha*t = 2.356 rad/s

iii] v = Wr =2.356*0.10 = 0.2356 m/s

iv] atan = alpha*r = 0.4712 m/s^2

v] arad= v^2/r = 0.2356^2/0.10 = 0.555 m/s^2

vi] a = sqrt(0.4712^2+0.555^2) = 0.728 m/s^2

vii] Fnet = ma = 1.6e-3*0.728 = 0.00116 N

Explanation:

Answer:

This structure is determined by the combined effect of deterministic forces and stochastic forces.

Explanation:

Genetic drift refers to random fluctuations in allele frequencies due to chance events , a stochastic (random) force which scrambles the predictable effects of selection, mutation, and gene flow.

Genetic drift is not a potent evolutionary force in very large randomly mating populations. to illustrate the consequences of genetic drift we will consider what happens when drift alone is altering the frequencies of alleles among many small populations. To prove this, we need to understand Population structure, which describes how individuals (or allele frequencies) in breeding populations vary in time and space.

Answer:

please read the answer below

Explanation:

The angular momentum is given by

[tex]|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)[/tex]

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

[tex]L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}[/tex]

b)

r=(0,-1)

angle = 90°

[tex]L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}[/tex]

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

[tex]L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}[/tex]

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

[tex]L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}[/tex]

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

The determinant method allows to find the z component of the angular momentum for the particle moving in the x axis are:

     a) L_z = - 5.0 kg m / s

     b) L_z = 5 kg m/s

     c)  L_z = 0

     d)  L_z = 0

     e) L_z = -5.0 kg m / s

     f)  L_z = 5.0 kg / ms

     g)  L_z = 5.0 kg m / s

The angular momentum is defined by the vector product of the position and the linear momentum, therefore it is a vector quantity.

      L = r x p

Where the bold letters represent vectors, L the angular momentum, r the position and p the linear moment

Momentum is defined by.

     p = m v

where m is the mass.

The modulus of angular momentum is:

    L = m r v sin θ

One way to find the angular momentum in various dimensions is to solve for the determinant method.

    [tex]L = m \ \left[\begin{array}{ccc}i&j&k\\x&y&z\\v_x&v_y&v_z\end{array}\right][/tex]  

The component z of the angular momentum is

     [tex]L_z = m ( x v_y - y v_x)[/tex]  

Indicate that the mass of the body is m = 2.5 kg and it moves in the direction x, v = 2 i m/s, let's calculate the angular momentum in direction z for each case:

The velocity is v = 2i

a) point r = (0,1)

    [tex]L_z[/tex] = - 2,5 1 2

    L_z = - 5.0 kg m / s

b) point r = (0, -1)

    L_z = -2.5 (-1) 2

    L_z = 5 kg m/s

c) point r = (1,0)

   L_z = 2,5 0 2

   L_z = 0

velocity is parallel to position

d) point r = (-1,0)

   L_z = 0

Velocity is parallel to position

e) point r = (1,1)

   L_z= 2.5 (1 0 - 2 1)

   L_z = -5.0 kg m / s

f) point r = (- 1, -1)

   L_z = 2.5 (1 0 - 2 (-1))

   L_z = 5.0 kg / ms

g) point r = (1, -1)

    L_z = 2,5 (1 0 - 2 (-1))

    L_z = 5.0 kg m / s

In conclusion using the determinant method we can find the z component of the angular momentum for the particle moving in the x axis are:

     a) L_z = - 5.0 kg m / s

     b) L_z = 5 kg m/s

     c)  L_z = 0

     d)  L_z = 0

     e) L_z = -5.0 kg m / s

     f)  L_z = 5.0 kg / ms

     g)  L_z = 5.0 kg m / s

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Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

a)

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

t = 0.0185 s = 18.5 ms

b)

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹

T = 874.8 N

First, we calculate the current through the cable using given Power and Voltage. Then, we use this current to find the resistance and subsequently, the diameter of the cable. For Part B, the electric field inside a cable is zero in steady state conditions.

To solve Part A of the question, we will use the formula Power = I2R, where I is current and R is resistance. First, we find the Current from the given power and voltage, Power = Voltage * Current. Second, we use this current to calculate the resistance, R = Power / I2. Lastly, we find the diameter using the formula for the Resistance of a cylindrical conductor, R = rho * L / A, where rho is resistivity, L is length and A is area. For Part B, the electric field inside a conductor under steady state conditions is zero.

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The diameter of the copper wire can be determined by solving an equation using its cross-sectional area. The electric field inside the copper wire is calculated by dividing the potential difference by the distance.

The question pertains to calculating the diameter of a cylindrical copper cable and the electric field inside it given certain conditions. This involves the use of formulas related to the principles of electromagnetism, and specifically electric current and electric field.

In order to determine the diameter of the cable (Part A), we would first need to deduce the cross-sectional area using data provided. The formula AD² = 4 could be used, where A represents the cross-sectional area and D the diameter of the cable. Substituting the value found for area, we can solve for the diameter.

For Part B, the electric field inside the cable can be calculated using the formula E = V/d, where V is the potential difference and d is the distance. Given that the potential difference is 220V and the length or distance is 3.3 km, the desired electric field can be computed.

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Answer:

The induced emf in the loop is 0.452 volts.

Explanation:

Given that,

Radius of the circular loop, r = 12 cm = 0.012 m

Magnetic  field, B = 0.8 T    

When released, the radius of the loop starts to shrink at an instantaneous rate of, [tex]\dfrac{dr}{dt}=75\ cm/s=0.75\ m/s[/tex]

The induced emf in the loop is equal to the rate of change of magnetic flux. It is given by :

[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{-dr}{dt}\\\\\epsilon=2\pi \times 0.12\times 0.8\times 0.75\\\\\epsilon=0.452\ V[/tex]

So, the induced emf in the loop is 0.452 volts.

Answer:

Check attachment for circuit diagram

And better understanding

Explanation:

Let find the equivalent inductance.

The 6mH in parallel to 6mH

Parallel connection will give

1 / Leq = 1 / L1 + 1 / L2

L1 = L2 = L = 6mH

1 / Leq = 1 / 6 + 1 / 6

1 / Leq = 2 / 6

1 / Leq = 1 / 3

Then, take reciprocal

Leq = 3 mH

Now the combInation of this parallel connection is connect in series with an 8mH inductor

Series connection is given as

Leq = L1 + L2

So, the equivalent of the parallel connection is 3mH and this will be connect in series with 8mH

Then, final inductance is

Leq = 3 + 8

Leq = 11 mH

Therefore, the repairman need to replace the element with one inductor of inductance 11 mH.

This question involves the concepts of inductance, series combination, and parallel combination.

The inductance to be used is found to be "11 mH".

First, we will find out the resultant inductance of the two inductors connected in a parallel combination. Since it is given that inductors behave like resistors in combination. Therefore, resultant of two inductances connected in parallel combination will be:

[tex]\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}[/tex]

where,

Therefore,

[tex]\frac{1}{L}=\frac{1}{6\ x\ 10^{-3}\ H}+\frac{1}{6\ x\ 10^{-3}\ H}\\\\[/tex]

L = 3 x 10⁻³ H = 3 mH

Now, we will calculate the total inductance of this combination in series combination with the third inductor as follows:

[tex]L_T=L+L_3=3\ mH+8\ mH\\L_T=11\ mH[/tex]

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Answer with Explanation:

We are given that

Tension=T=233 N

The displacement of  the rope is given by

[tex]y=(0.320 m)sin(\frac{\pi x}{3})sin(10\pi)t[/tex]

a.By comparing with

[tex]y=Asin(kx)sin(\omega t)[/tex]

We get

A=0.32

k=[tex]\frac{\pi}{3}[/tex]

[tex]\omega=10\pi[/tex]

[tex]k=\frac{2\pi}{\lambda}[/tex]

[tex]\frac{\pi}{3}=\frac{2\pi}{\lambda}[/tex]

[tex]\lambda=3\times 2=6m[/tex]

n=2

[tex]n\lambda=2L[/tex]

[tex]2\times 6=2L[/tex]

[tex]L=6[/tex]m

b.[tex]\omega=2\pi f[/tex]

[tex]2\pi f=10\pi[/tex]

[tex]f=\frac{10}{2}=5Hz[/tex]

Speed,[tex]v=f\lambda=5\times 6=30m/s[/tex]

c.Let

Mass of the rope=m

[tex]\mu=\frac{m}{L}=\frac{m}{6}[/tex]

[tex]v^2=\frac{T}{\mu}[/tex]

[tex](30)^2=\frac{233}{\frac{m}{6}}[/tex]

[tex]900\times \frac{m}{6}=233[/tex]

[tex]m=\frac{233\times 6}{900}=1.553 kg[/tex]

A parallel-plate capacitor consists of two plates with opposite charges, creating an electric field between them. The electric field can be measured using an electric field sensor.

In a parallel-plate capacitor, a second line of charges equal in size but opposite in charge is placed below the line of positive charges, creating an electric field (E-field) between the plates. The E-field is a vector quantity that points from the positive plate to the negative plate. Its magnitude can be determined using a sensor, such as an electric field sensor, which measures the strength of the electric field.

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A parallel-plate capacitor creates a uniform electric field between its plates, where the field's magnitude is proportional to the surface charge density and directly proportional to the amount of charge on the plates.

A parallel-plate capacitor consists of two identical, parallel conducting plates separated by a certain distance. When we create a system where one plate is charged with positive charges and the other with an equal magnitude of negative charges, by, for instance, connecting the plates to the opposite terminals of a battery, an electric field (often referred to as the E-field) is established between the plates. This electric field is extremely uniform because of the geometry of the plates and the even distribution of charges.

The magnitude of the electric field (E) between the plates is proportional to the surface charge density (σ), which is the charge per unit area on a plate. It is described mathematically as E = σ/ε0, where ε0 is the permittivity of free space. The strength of the electric field is also directly proportional to the amount of charge (Q) on the plates. Therefore, when more charge is present on the plates, the electric field will be stronger, illustrated by the increase in the density of field lines that begin on positive charges and end on negative ones.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is [tex]n =7[/tex]

Explanation:

    From the question we are told that

          The value of m = 2

            For every value of [tex]m, n = m+ 1, m+2,m+3,....[/tex]

           The modified version of  Balmer's formula is [tex]\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ][/tex]

             The Rydberg constant has a value of [tex]R = 1.097 *10^{7} m^{-1}[/tex]

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

    The wavelength is

                          [tex]\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ][/tex]

                          [tex]\lambda = \frac{1}{1523611.1112}[/tex]

                             [tex]\lambda = 656nm[/tex]

For m = 2 and n = 4

    The wavelength is

                          [tex]\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ][/tex]

                          [tex]\lambda = \frac{1}{2056875}[/tex]

                             [tex]\lambda = 486nm[/tex]

For m = 2 and n = 5

    The wavelength is

                          [tex]\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ][/tex]

                          [tex]\lambda = \frac{1}{2303700}[/tex]

                             [tex]\lambda = 434nm[/tex]

For m = 2 and n = 6

    The wavelength is

                          [tex]\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ][/tex]

                          [tex]\lambda = \frac{1}{2422222}[/tex]

                             [tex]\lambda = 410nm[/tex]

For m = 2 and n = 7

    The wavelength is

                          [tex]\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ][/tex]

                          [tex]\lambda = \frac{1}{2518622}[/tex]

                             [tex]\lambda = 397nm[/tex]

So the value of n is  7

Answer:

[tex]2.75\ rad/s^2[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 1.25 rad/s

[tex]\alpha[/tex] = Angular acceleration

[tex]\theta[/tex] = Angle of rotation = 8 rad

t = Time taken = 2 s

[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 8=1.25\times 2+\frac{1}{2}\times \alpha\times 2^2\\\Rightarrow \alpha=\dfrac{2}{2^2}(8-1.25\times 2)\\\Rightarrow \alpha=2.75\ rad/s^2[/tex]

The angular acceleration of the rotor is [tex]2.75\ rad/s^2[/tex]

Answer:

Explanation:

Force on Q due to q₁

= k Q.q₁ / r² , r is distance between two.

X component

= k Q.q₁ / r²  x cos θ ,  θ is angle r makes with x - axis.

= k Q.q₁ / r²  x cos θ

r = a / sinθ

X component

= k Q.q₁ sin²θ . cosθ/ a²  

= k Q.q₁ / 2 a² x sin2θ . sinθ

for maximum value of it ,

sin2θ = 1  , θ = 45°

a / d = tan45

a = d .

Answer:

the mass of steam at 100°C must be mixed is 150 g

Explanation:

given information:

ice's mass, [tex]m_{i}[/tex] = 490 g = 0.49 kg

steam temperature, T = 100°C

liquid water temperature, T = 89.0°C

specific heat of water, [tex]c_{w}[/tex] = 4186 J/kg.K = 4.186 kJ/kg.K

latent heat of fusion, [tex]L_{f}[/tex] = 333 kJ/kg

latent heat of vaporization, [tex]L_{v}[/tex] = 2256 kJ/kg

first, we calculate the heat of melted ice to water

Q₁ = [tex]m_{i} L_{f}[/tex]

where

Q = heat

[tex]m_{i}[/tex] = mass of the ice

[tex]L_{f}[/tex]  = latent heat of fusion

thus,

Q₁ = [tex]m_{i} L_{f}[/tex]

    = 0.49 x 333

    = 163.17 kJ

then, the heat needed to increase the temperature of water to 89.0°C

Q₂ = [tex]m_{i}[/tex] [tex]c_{w}[/tex] (89 - 0), the temperature of ice is 0°C

[tex]c_{w}[/tex] = specific heat of water

so,

Q₂ = [tex]m_{i}[/tex] [tex]c_{w}[/tex] (89 - 0)

     = 0.49 x 4.186 x 89

     = 182.55 kJ

so, the heat absorbed by the ice is

Q = Q₁ + Q₂

   = 163.17 + 182.55

   = 345.72 kJ

the temperature of the steam is 100°C, so the mass of the steam is

Q = [tex]m_{s}[/tex][tex]L_{v}[/tex]  +  [tex]m_{s}[/tex][tex]c_{w}[/tex] (100 - 89)

Q = [tex]m_{s}[/tex]([tex]L_{v}[/tex]  +  [tex]c_{w}[/tex] (11))

[tex]m_{s}[/tex] = Q/ [[tex]L_{v}[/tex]  +  [tex]c_{w}[/tex] (11)]

      = 345.72/ [2256 + (4.186 x 11)]

      = 0.15 kg

      = 150 g

Answer:

a )Decreases

b ) Increases

c.) It's likely to break

Explanation:

a )The torque is as a result of friction, and also the magnitude is dependent on the normal force that exist between the surface in contact. The torque decrease because there is decrease in frictional force as the weight of the roll decrease

b.)The angular velocity will increase as radius o f the roll decrease because it is operating at constant speed.

c)if the child suddenly hit the roll, the papper will break, because it's angular acceleration might not be able to move along with the papper. Situation like this occur when there is large radius

Answer:

A) R = 1019.11 Ω

B) V = 101.91 V

Explanation:

a) The resistance is given by the formula;

R = ρL/A

Where;

ρ is bulk density = 5.0 Ωm

L is length of cylinder = 1.6m

A is area = (π/4)D² = (π/4) x 0.1² = 0.007854 m²

Plugging in the relevant values, we obtain ;

R = (5 Ωm x 1.6m)/(0.007854 m²)

R = 1019.11 Ω

b) potential difference is given by;

V = RI

Thus, plugging in relevant values, we obtain;

V = 1019.11 Ω ∙ 0.1A

V = 101.91 V

The resistance between the hands if the skin resistance is negligible is approximately [tex]\( 1019 \, \Omega \)[/tex], the potential difference between the hands needed for a lethal shock current of 100 mA is approximately [tex]\( 101.9 \, \text{V} \).[/tex]

To solve this problem, we'll use the formulas for resistivity and Ohm's law.

Part (a): Resistance Between the Hands

The resistance R of a cylindrical conductor can be calculated using the formula:

[tex]\[ R = \rho \frac{L}{A} \][/tex]

where:

- [tex]\( \rho \)[/tex] is the resistivity of the material (in this case, the human body's resistivity is [tex](\( 5.0 \, \Omega \cdot \text{m} \))[/tex].

- L  is the length of the conductor (1.6 m).

- A is the cross-sectional area of the conductor.

First, we need to calculate the cross-sectional area A of the cylindrical conductor. The cross-sectional area of a cylinder is given by:

[tex]\[ A = \pi \left(\frac{d}{2}\right)^2 \][/tex]

where d is the diameter of the cylinder. Here, [tex]\( d = 0.10 \, \text{m} \)[/tex].

So,

[tex]\[ A = \pi \left(\frac{0.10 \, \text{m}}{2}\right)^2 = \pi \left(0.05 \, \text{m}\right)^2 \]\[ A = \pi \cdot (0.05)^2 \, \text{m}^2 \]\[ A \approx 3.14 \cdot 0.0025 \, \text{m}^2 \]\[ A \approx 0.00785 \, \text{m}^2 \][/tex]

Now we can calculate the resistance R:

[tex]\[ R = 5.0 \, \Omega \cdot \text{m} \frac{1.6 \, \text{m}}{0.00785 \, \text{m}^2} \]\[ R = 5.0 \cdot \frac{1.6}{0.00785} \, \Omega \]\[ R = 5.0 \cdot 203.82 \, \Omega \]\[ R \approx 1019.1 \, \Omega \][/tex]

Part (b): Potential Difference for Lethal Shock Current

Ohm's law states that the voltage V across a resistor is given by:

[tex]\[ V = I \cdot R \][/tex]

where I is the current and R is the resistance.

Given that the lethal shock current is [tex]\( 100 \, \text{mA} = 0.1 \, \text{A} \)[/tex] and the resistance [tex]\( R \) is \( 1019 \, \Omega \)[/tex]:

[tex]\[ V = 0.1 \, \text{A} \cdot 1019 \, \Omega \]\[ V = 101.9 \, \text{V} \][/tex]

Answer:

0.999958c

Explanation:

Remember that the trip involves traveling for six months at a constant velocity thus returning home at same speed . The time interval of this trip will therefore be one year.

Accurate time interval to be measured by the clock on the spaceship Δt0 = 1.0 years

Time interval as advanced on earth observed from the spacecraft Δt = 110 years

the formula for time dilation is

Δt = Δt0 /√(1-v2/c2)

or v = c*√1-(Δt0/Δt)2

=c*√[1-(1year/110years)2]

= 0.999958c

Answer:

Frad = 2IA/C

Explanation:

see attached file

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

[tex]\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}[/tex]

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

[tex]E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})[/tex]

a) for y=0.200m, r1=0.200m and r2=0.200m:

[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C[/tex]

b) for y=0.600m, r1=0.600m, r2=0.200m:

[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C[/tex]

a) The electric field strength for case 1 will be 228391.8 N/C

b) The electric field strength for case 2 will be 59345.91 N/C

The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity,

According to Gauss Law. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.

The given data in the problem is;

λ₁ is the charge per unit length for charge 1  = 4.80 μC/m

λ₂ is the charge per unit length for charge 2  =-2.26 μC/m

y₁ is the distance from charge 1 =0.400 m.

From the gauss law, the net electric field is produced by both lines of charges will be .;

[tex]\rm E= \frac{\lambda_}{2 \pi \epsilon_0 r_2} \\\\ \rm E_1= \frac{\lambda_1}{2 \pi \epsilon_0 r_1} \\\\ \rm E_2= \frac{\lambda_2}{2 \pi \epsilon_0 r_2}[/tex]

The total electric field will be the sum of the electric field for the charge 1 and2;

[tex]\rm E=E_1+E_2 \\\\ \rm E=\frac{\lambda_1}{2 \pi \epsilon_0 r_1}+\frac{\lambda_2}{2 \pi \epsilon_0 r_2} \\\\[/tex]

The net electric field for y=0.200m, r₁=0.200m, and r₂=0.200 m will be;

[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.200}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=228391.8 \ N/C[/tex]

The net electric field for  y=0.600m, r₁=0.600m, r₂=0.200 m will be;

[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.600}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=-59345.51 \ N/C[/tex]

Hence the electric field strength for case 1 and case 2 will be 228391.8 N/C and 59345.91 N/C respectively.

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Answer:

a. 3.86 m/s²

b. 332.64 N directed upwards towards the center of the ferris wheel.

c. 764.96 N directed downwards towards the center of the ferris wheel.

d. 589.84 N at 21.5° counterclockwise from the horizontal direction.

Explanation:

a. Since the ferris wheel rotates 4 times per minute, its period, T = 60 s/4 = 15 s.

We now find it angular speed ω = 2π/T = 2π/15 = 0.418 rad/s

We then calculate its centripetal acceleration from a = rω² where r = radius of ferris wheel = 22.0 m.

So, a = 22 m × (0.418 rad/s)² = 3.86 m/s²

b. At the lowest point, the normal force, N and the centripetal force, F both act in opposite directions to the weight, mg of the object. So,

N + F = mg

N = mg - F      

N = mg - ma where a is the centripetal acceleration

N = m(g - a)

N = 56 kg(9.8 m/s² - 3.86 m/s²)

N = 56 kg × 5.94 m/s²

N = 332.64 m/s²

The normal force the seat exerts on the child is thus 332.64 N directed upwards towards the center of the ferris wheel.

c. At the highest point, the weight, mg of the object and the centripetal force, F both act in opposite directions to the normal force, N. So,

N = mg + F      

N = mg + ma where a is the centripetal acceleration

N = m(g + a)

N = 56 kg(9.8 m/s² + 3.86 m/s²)

N = 56 kg × 13.66 m/s²

N = 764.96 N

The normal force the seat exerts on the child is thus 764.96 N directed downwards towards the center of the ferris wheel.

d. Half way between the top and the bottom of the ferris wheel, the normal force must balance the weight and the centripetal force so the child doesn't fall off. For this to happen, the normal force is thus the resultanf of the centripetal force and the weight of the child. Since these two forces are perpendicular at this instance,

N = √[(mg)² + (ma)²] = m√(g² + a²) = 56 kg√[(9.8 m/s²)² + (3.86 m/s²)²] = 56 kg√[(97.196 (m/s²)² + 14.8996(m/s²)²] = 56kg√110.9396 (m/s²)² = 56 kg × 10.53 m/s² = 589.84 N.

Since the centripetal force acts towards the center of the ferris wheel in the horizontal direction, it is equal to the horizontal component of the normal force, Also, the weight acts downwards and is equal to the vertical component of the normal force.

So, the direction of the normal force is gotten from

tanθ = ma/mg = a/g

θ = tan⁻¹(a/g) = tan⁻¹(3.86 m/s² / -9.8 m/s²) = tan⁻¹(-0.3939) = -21.5°. Since the angle is it shows a counter clockwise direction.

So, the normal force is 589.84 N at 21.5° counterclockwise from the horizontal direction.

(a) The centripetal acceleration of the child is 1.94 m/s².

(b) The force the seat exert on the child at the lowest point of the ride is 657.4 N.

(c) The force the seat exert on the child at the highest point of the ride is 440.2 N.

(d) The force the seat exert on the child when the child is halfway between the top and bottom is 108.64 N.

The given parameters;

The centripetal acceleration of the child is calculated as follows;

[tex]a_c = \omega ^2 r\\\\[/tex]

where;

ω  is angular speed of the wheel in rad/s

The angular speed of the wheel in rad/s is calculated as follows;

[tex]\omega = 4 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 0.419 \ rad/s[/tex]

The centripetal acceleration of the child is calculated as;

[tex]a_c = \omega^2 r\\\\a_c = (0.419)^2 \times 11\\\\a_c = 1.94 \ m/s^2[/tex]

The force the seat exert on the child at the lowest point of the ride is calculated as;

[tex]T_{bottom} = ma_c + mg\\\\T_{bottom} = m(a_c + g)\\\\T_{bottom} = 56(1.94 + 9.8)\\\\T_{bottom} = 657.4 \ N[/tex]

The force the seat exert on the child at the highest point of the ride is calculated as;

[tex]T_{top} = mg- ma_c\\\\T_{top} = m(g - a_c)\\\\T_{top} = 56(9.8 - 1.94)\\\\T_{top} = 440.2 \ N[/tex]

The force the seat exert on the child when the child is halfway between the top and bottom;

[tex]T = ma_c\\\\T = 56(1.94)\\\\T = 108.64 \ N[/tex]

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Answer:

Explanation:

question answered below

We described the diffusion process of a slowly evaporating hemispherical water droplet by adopting a spherical coordinate system. We made five assumptions: steady-state conditions, isotropic medium, uniform temperature, constant concentration of water vapor in the surrounding air, and no bulk motion of air. The differential form of Fick's law was derived, and two boundary conditions regarding the concentration of water were proposed.

To answer your question regarding the diffusion process of a hemispherical droplet of water slowly evaporating on a flat surface, we will start by selecting a spherical coordinate system. This is chosen because a spherical system best represents a hemispherical droplet with a shrinking radius R.

Now, let's list at least five reasonable assumptions for the mass-transfer aspects of the water-evaporation process:

To express the mass transfer in terms of the flux NA, we need to derive from Fick's first law, which states that the molar flux due to diffusion is proportional to the concentration gradient. In this case, it simplifies to: NA = -D(dCA/dx), where D is the diffusion coefficient and dCA/dx is the concentration gradient of species A (water).

As for the boundary conditions, we can consider two conditions:

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#SPJ3

Answer:

Noennt ied olod l qeu es ocmo  a balazuna de cozcna

in

Explanation:

Answer:

see detailed solution attached.

Explanation:

Final answer:

In the Heisenberg picture of quantum mechanics, the creation and annihilation operators for a one-dimensional harmonic oscillator evolve in time as â(t) = â(0)e-iwt and â¹(t) = â¹(0)eiwt, respectively, which are consistent with the time evolution of the position and momentum operators derived from the Heisenberg equations of motion.

Explanation:

Time-Dependent Creation and Annihilation Operators

In the Heisenberg picture of quantum mechanics, unlike the Schrödinger picture, the state vectors are stationary and the operators evolve with time. For the one-dimensional harmonic oscillator, the time-dependent creation (â¹(t)) and annihilation (â(t)) operators can be solved explicitly. The Heisenberg equations of motion for these operators are given by:

â(t)= â(0)e-iwt
â¹(t) = â¹(0)eiwt

where â(0) and â¹(0) are the initial operators at t=0, and w is the angular frequency of the oscillator. These time-dependent operators are then consistent with the previously derived time-dependent position (q(t)) and momentum (p(t)) operators:

q(t) = q(0)cos(wt) + (p(0)/mw)sin(wt)
p(t) = p(0)cos(wt) - mwq(0)sin(wt)

Here, m is the mass of the oscillator, and w is again the angular frequency. This consistency is because the position and momentum operators can be expressed in terms of the creation and annihilation operators. The expressions for â(t) and â¹(t) in the Heisenberg equations demonstrate that these operators oscillate with time in a manner that corresponds to the harmonic nature of the oscillator's motion in quantum mechanics.

The time evolution of any quantum mechanical operator in the Heisenberg picture can be described by the time evolution operator, U, which is unitary. This allows time propagation of operators while keeping the wavefunction unchanged.

Find the attached photo for the solution

The final charge on the 3.0-µF capacitor is 90 µC.

To find the final charge on the 3.0-µF capacitor, we need to use the concept of charge conservation in a series circuit. In a series circuit, the charge across capacitors connected in series is the same. Therefore, the final charge on the 3.0-µF capacitor will be the same as the charge on the 5.0-µF capacitor.

Using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can calculate the charge on the 5.0-µF capacitor.

Q = (5.0 µF)(18 V) = 90 µC.

Therefore, the final charge on the 3.0-µF capacitor is also 90 µC.

Complete question:

new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP

a) Find the p value.

b) Do you believe it is plausible that the blocks do not meet the specification, or are you convinced that they do? Expain your reasoning.

Answer:

a) p-value= 0.1949

b) It is possible the blocks do not meet the specifications

Explanation:

Given:

n = 100

Sample mean, X' = 1356

Standard deviation, s.d = 70

a) To find p- value.

Null hypothesis:

H0: u ≤ 1350

Alternative hypothesis:

H1 : u > 1350

The test statistic wll be:

[tex] z= \frac{X'-u_o}{s.d/ \sqrt{n}}[/tex]

[tex] = \frac{1356-1350}{70/\sqrt{100}}[/tex]

=0.86

The p value will be:

= P(z>0.86)

= 1-P(z≤0.86)

Using the normal distribution table, we now have:

1 - 0.8051

= 0.1949

P value = 0.1949

b) Since our p-value is 0.1949, we do not reject the null hypothesis, because the p-value, 0.1949, is not small. This means that it is possible the blocks do not meet the specifications.

The probability that the compressive strength is greater than 1350 kPa is 80.51%

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\Where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation, n=sample\ size[/tex]

Given that n = 100, μ = 1356, σ = 70. For x > 1350 kPa:

[tex]z=\frac{1350-1356}{70/\sqrt{100} } =-0.86[/tex]

From the normal distribution table, P(x > 1350) = P(z > -0.86) = 1 - P(z < -0.86) = 1 - 0.1949 = 80.51%

Hence the probability that the compressive strength is greater than 1350 kPa is 80.51%

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Answer:

10.16 degrees

Explanation:

Apply Snells Law for both wavelenghts

\(n_{1}sin\theta_{1} = n_{2}sin\theta_{2}\)

For red

(1.620)(sin 25.5) = (1)(sin r)

For red, the angle is 35.45degrees

For violet

(1.660)(sin 25.5) = (1)(sin v)

For violet, the angle is 45.6 degrees

The difference is 45.6- 35.45 = 10.16 degrees

Answer:

(a) Find the final velocity of the person and cart relative to the ground. = 1.367m/s

(b) Find the frictional force acting on the person while he is sliding across the top surface of the cart = 286.944N

(c) How long does the frictional force act on the person?  = 0.5809s

(d) Find the change in momentum of the person.  = 166.774N

(e) Determine the displacement of the person relative to the ground while he is sliding on the cart. = 1.5878m

(f) Determine the displacement of the cart relative to the ground while the person is sliding. = 0.3970m

(g) Find the change in kinetic energy of the person. = -455.71J

(h) Find the change in kinetic energy of the cart = 113.99j

(i) Explain why the answers to (g) and (h) differ.

the force  exerted on the cart by the person must be equal in magnitude and opposite in direction to the force exerted on the cart by the person. The changes in momentum  of the two cart must be equal in magnitude and must add up to zero.  The following represents two way thinking about "WAY". The distance the cart moves is different from the distance moved by the point of application of frictional force on the cart.

Explanation:

check attachment for explanation.

Answer:

1) tensile stress = 76.648 Mpa

2) extension = 0.0215 m

Explanation:

Detailed explanation and calculation is shown in the image below

Final answer:

Tensile stress and the force constant for the Achilles tendon can be calculated using Young's modulus, Hooke's Law, and the given dimensions. A 75 kg sprinter exerting a force 13 times his weight would cause a specific amount of stretch in the tendon, calculated by applying Hooke's Law with the determined spring constant.

Explanation:

The tensile stress required to stretch the Achilles tendon by 5.2% of its length can be calculated using Hooke's Law and the definition of stress and strain. Stress (σ) is the force applied per unit area (A) and strain (ε) is the relative change in length (ΔL/L). The Young's modulus (E) relates stress and strain through σ = Eε.

First, calculate the strain which is 5.2% or 0.052. Since the Young's modulus (E) is given as 1474 MPa and the strain (ε) is 0.052, we can now calculate the stress:

σ = Eε = 1474 MPa × 0.052 = 76.648 MPa

Next, we model the tendon as a spring to find the force constant (k). Hooke's Law states that F = kΔx, where F is the force applied and Δx is the change in length. To find k, we rearrange the formula to k = F/Δx. Here, F is the tensile force, which is σ × A. We substitute the stress we found and the area of the tendon to find k.

For a 75 kg sprinter exerting a force of 13.0 times his weight, the force F would be 13.0 × 75 kg × 9.8 m/s² (acceleration due to gravity). To find the stretch (Δx), we use Hooke's Law again with the calculated force constant k.

The Achilles tendon will stretch according to that force, proportionally to the spring constant k. The exact calculations require substitution of the calculated force into the Hooke's Law equation to solve for Δx.

Answer:

A. 4.40m/s²

B. 4.615N

C. 0.7511J

Explanation:

A. Ac= V²/L

= (2.15)²/1.05

= 4.50m/s²

B. T =Mv²/L + mg

0.325(4.40)+ 0.325(9.8)

= 4.615N

C. K.E = 1/2mv²

=0.7511J

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Answer:

(a) 4.4 m/s²

(b) 4.615 N.

(c) 0.751 J

Explanation:

(a)

Using,

a' = v²/r..................... Equation 1

Where a' = centripetal acceleration, v = speed of the pendulum bob, r = radius or length of the pendulum bob.

Given: v = 2.15 m/s, r = 105 cm = 1.05 m

Substitute into equation 1

a' = 2.15²/1.05

a' = 4.4 m/s².

(b) The force of tension in the string = Tangential force + weight of the bob.

T = ma'+mg..................... Equation 2

Where T = Force of tension in the string, m = mass of the bob, g = acceleration due to gravity.

Given: m = 325 g = 0.325 kg, a' = 4.4 m/s², g = 9.8 m/s²

Substitute into equation 2

T = 0.325×4.4+0.3325×9.8

T = 1.43+3.185

T = 4.615 N.

(c) Kinetic energy of he bob at that point = 1/2mv²

Ek = 1/2mv²...................... Equation 3

Where Ek = kinetic energy of the bob

Given: m = 0.325 kg, v = 2.15 m/s

Substitute into equation 3

Ek = 1/2(0.325)(2.15²)

Ek = 0.751 J