HELPP ?Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volume of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?

Answer:

The average upward force exerted by the water is 988.2 N

Explanation:

Given;

mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free  fall.

V² = U² + 2gh

where;

V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]

Therefore, the average upward force exerted by the water is 988.2 N

Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]

So, its initial position was 471 m.

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]

distancia= 25 metros

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo

Usando la tasa de conversión;

1000 m = 1 km

1 hora = 3600 segundos

[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]

Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.

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Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

E = 389 MeV

Answer:

Energy is stored by a 50 ampere-minute 12

volt battery is approximately = 36,000 J = 36 kJ

Explanation:

Power in electrical circuits is given as

Power = IV

But power generally is defined as energy expended per unit time

Power = (Energy/time)

Energy = Power × Time

Energy = IV × Time

Energy = (I.t × V)

I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds

V = 12 V

Energy = 3,000 × 12 = 36,000 J = 36 kJ

Hope this Helps!!!

ANSWER: num 1 is black hole

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

Answer:

The five kinds of stressors are:

Acute time-limited

Brief naturalistic

Stressful events sequences

Chronic

Distant

Explanation:

yeah

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

Answer:

16.9 m

Explanation:

Answer:

(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

[tex]I(t)=I_{max}sin(\omega t)[/tex]   (1)

(a) The induced emf is given by the following formula

[tex]emf_L=-L\frac{dI}{dt}[/tex]    (2)

You derivative the expression (1) in the expression (2):

[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

[tex]emf_L=-\omega LI_{max}[/tex]

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

At t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

The current through an inductor of inductance L can be calculated by

[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1  

(a) The induced emf can be calculated by  

[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2  

Derivative the equation (1) in the equation (2)

[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]

(b) At t=0 the current is zero,

(c) At t = 0 the emf is:

[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]  

Therefore, at t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

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Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

Explanation:

The first equation of motion in kinematics is given by :

[tex]v=u+at[/tex] .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

Answer:

663N

Explanation:

We need to find the force that will overcome the frictional force.

The angle of the normal force is 15°.

The mass of the crate is 32 kg

The coefficient of static friction is 0.49

Frictional force is given in terms of Normal force as:

F = μNcosθ

where μ = coefficient of static friction

N = normal force

θ = angle of normal force

Frictional force is given as:

F = mg

=>mg = μNcosθ

=> N = mg/(μcosθ)

N = (32 * 9.8) / (0.49 * cos15)

N= 313.6 / 0.473

N = 663 N

The force needed to cause the box to move must be 663N or greater.

Answer:

65 m/h

Explanation:

Let the distance of the car moving south be y.

Let the distance of the car moving west be x.

Let the distance between the two cars be a.

These three distances can be represented as a right angled triangle. So we can say:

[tex]a^2 = x^2 + y ^2[/tex]

Let us differentiate with respect to time, since the distances are changing with respect to time:

[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)

da/dt = rate of change of distance between two cars

The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.

Therefore:

dy/dt = 60 m/h and dx/dt  = 25 m/h

After two hours, the distance of the two cars will be:

y = 2 * 60 = 120 miles

x = 2 * 25 = 50 miles

Therefore:

[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]

From (1):

130(da/dt) = 50(25) + 120(60)

130(da/dt) = 1250 + 7200 = 8450

da/dt = 8450/130 = 65 m/h

Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.

Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

[tex]Density = \frac{Mass}{Volume}[/tex]

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

Answer:

Spring Constant = 279.58 N/m

Explanation:

We are given;

Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg

Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m

Maximum height; h = 4.46 m

The formula for the energy in the spring is given by;

E = ½kx²

where:

k is the spring constant

x is the distance the spring is compressed.

Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;

½kx² = mgh

Plugging in the relevant values, we have;

½ * k * 0.0801² = 0.0205 * 9.81 * 4.46

0.003208005k = 0.8969

k = 0.8969/0.003208005

k = 279.58 N/m

Answer:

a) 5.143 m/s^2

b) T = 15.43 N

c) Fr = 10.29 N

d) 5.143 m/s^2 , T = 15.43 N

Explanation:

Given:-

- The mass of left most block, m1 = 1.0 kg

- The mass of center block, m2 = 2.0 kg

- The mass of right most block, m3 = 4.0 kg

- A force that acts on the right most block, F = 36 N

Solution:-

a)

- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.

- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:

                       [tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]

Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2

b)

- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).

- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.

- We will again apply the Newton's 2nd law of motion for the block m3 as follows:

                       [tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]

Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )

c)

- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).

- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:

                         [tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]

- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )

Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.

d)

- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2

- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).

- We apply the Newton's 2nd law of motion for the block m3 as follows:

                            [tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]

Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.

Answer:

Answer D is correct

Explanation:

[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]

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Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

Explanation:

The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.

The given parameters;

The volume of the bobber is calculated as follows;

[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]

The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;

[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]

The volume of the lead is calculated as follows;

[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]

The buoyant force experienced by the lead due to the volume submerged is calculated as follows

[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]

The total buoyant force is calculated as;

[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]

The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;

[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]

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Answer:

The width of the slit will be ".946 mm".

Explanation:

The given values are:

Wavelength = 610 × 10⁻⁹

Length, L = 3 m

As we know,

⇒  [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]

On putting the estimated values, we get

⇒  [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]

On applying cross-multiplication, we get

⇒  [tex]a=9.46\times 10^{-4}[/tex]

⇒  [tex]a = .946 mm[/tex]

Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.

Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.

Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.

Thus, the correct option is C, as the D image is showing the absorption.

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Answer: C.D

Explanation:...

Answer:

Light is less energetic than X-rays.

Explanation:

The electromagnetic spectrum refers to the range of wavelengths or frequencies over which electromagnetic radiation extends. It is the entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light. In the electromagnetic spectrum, the entire distribution of electromagnetic radiation is done according to their frequency or wavelength.

The energy of an electromagnetic wave depends on its frequency and wavelength. The shorter the wavelength, the greater the energy of the electromagnetic wave but the larger frequency, the greater the energy of the electromagnetic wave.

X-rays has a frequency of about 1×10^20 Hz compared to visible light of frequency of about 1×10^15 Hz. Hence X-rays, having a larger frequency, is more energetic than visible light.

Answer:

The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]

Explanation:

From the question we are told that

     The angle of incidence is  [tex]\theta _i = 26.50 ^o[/tex]

      The angle of refraction angle for  sheet 1 is  [tex]\theta _{r_1}} = 31.70 ^o[/tex]

       The angle of refraction for sheet 3 is  [tex]\theta _{r_3}} = 36.70 ^o[/tex]

According to Snell's  law  

       [tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]

Where  [tex]n_1 \ and \ n_2[/tex]  are refractive index of sheet 1  and  sheet 2  

       =>   [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]

Also  when sheet 3 in on top of sheet 2

       [tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

substituting for  [tex]n_2[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

=>    [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]

when sheet 1 in on top of sheet 3

        [tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3

substituting for  [tex]n_3[/tex]

         [tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

=>   [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]

substituting values

      [tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]

=>     [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]

=>   [tex]\theta_{r_s } = 23.1 ^o[/tex]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

Answer:

D. Presynaptic terminal button

explanation:

Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron

hope this helped!