A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initially, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

[tex]\theta = \frac{1.22 \lambda }{D}[/tex]

Here,

D is diameter of the eye

[tex]D = \frac{1.22 (539nm)}{5.11 mm}[/tex]

[tex]D= 1.287*10^{-4}m[/tex]

The angle that relates the distance between the lights and the distance to the lamp is given by,

[tex]Sin\theta = \frac{d}{L}[/tex]

For small angle, [tex]sin\theta = \theta[/tex]

[tex]sin \theta = \frac{d}{L}[/tex]

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle [tex]sin \theta = \theta[/tex]

Therefore,

[tex]L = \frac{d}{sin\theta}[/tex]

[tex]L = \frac{0.691m}{1.287*10^{-4}}[/tex]

[tex]L = 5367m[/tex]

Therefore the distance is 5.367km.

Final answer:

Constructive interference at point P occurs when the path length difference between the two antennas is an integer multiple of the wavelength. The values of x where this occurs can be found using the equation path length difference = n * wavelength. Solving for n, we find that constructive interference occurs at x = 0.00906 m.

Explanation:

In order to have constructive interference at point P, the path length difference between the two antennas must be an integer multiple of the wavelength. Since the antennas are identical and radiating in phase, the path length difference is simply the distance between the two antennas, which is 9.05 mm.

The wavelength can be calculated using the formula: wavelength = speed of light / frequency. For a frequency of 115 MHz, the wavelength is approximately 2.6 m.

To find the values of x where constructive interference occurs, we can set up an equation: path length difference = n * wavelength, where n is an integer. So, 9.05 mm = n * 2.6 m.

Solving for n, we get:

n = (9.05 mm) / (2.6 m) = 0.00348

Therefore, constructive interference will occur at point P when x = n * wavelength. Substituting n = 0.00348 and wavelength = 2.6 m, we get:

x = (0.00348)(2.6 m) = 0.00906 m

Answer:

Explanation:

Given that the inputted angle of incidence is accepted

Angle of incidence θi = 76.5°

But angle of refraction is not acceptable

Angle of refraction θr = 76.5°

We are told that the value is too high

Then, θr < 76.5°

We want to calculate index of refraction n?

The acceptable value of refraction index of acrylic is 1.49

So the true value is 1.49

So, let calculate the measure value

Refractive index is given as

n = Sin(i) / Sin(r)

Then,

n = Sin(76.5) / Sin(76.5)

Then, n = 1

Now, percentage error of the refractive index,

Percentage error is

%error= |true value—measure value| / true value × 100

The true value is 1.49

The measure value is 1

Then,

%error = ( |1.49—1| / 1.49 ) × 100

% error = ( 0.49 / 1.49 ) × 100

%error = 0.3289 × 100

%error = 32.89%

Answer:

Explanation:

32.89%

Answer:

Explanation:

The pictures attached below shows the solution to the problem and i hope its explanatory enough. Thank you

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume ,  Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

Answer:

3.37x10^6J

Explanation:

The external energy is given as KQ1Q2/r

=(9*10^9)*(3*10^-3)*(25*10^-3)/(0.2)

=3.375 *10^6 J

The external energy required can be calculated using the electric potential energy formula. The external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.

The external energy required to bring a charge of 25µC from infinity to the center of the circle can be calculated using the electric potential energy formula, which is given by:

U = k * (q1 * q2) / r

Where U is the electric potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

In this case, the charge at the center of the circle is 25µC, and the charge distributed uniformly along the circumference is 3.0µC. The radius of the circle is given as 20 cm. Plugging these values into the formula, we have:

U = (9 * 10^9 Nm^2/C^2) * (25 * 10^-6 C) * (3 * 10^-6 C) / (0.20 m) = 337.5 J

Therefore, the external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.

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Answer:

a

The focal length of the lens in water is  [tex]f_{water} = 262.68 cm[/tex]

b

The focal length of the mirror in water is  [tex]f =79.0cm[/tex]

Explanation:

From the question we are told that

    The index of refraction of the lens material = [tex]n_2[/tex]

    The index of refraction of the medium surrounding the lens = [tex]n_1[/tex]

The lens maker's formula is mathematically represented as

            [tex]\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]

Where [tex]f[/tex] is the focal length

            [tex]n[/tex] is the index of refraction

            [tex]R_1 and R_2[/tex] are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           [tex]\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]

    When immersed in liquid the formula becomes

          [tex]\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]

The ratio of the focal length of the the two medium is mathematically evaluated as

           [tex]\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }[/tex]

From the question

      [tex]f_{air }[/tex]= 79.0 cm

       [tex]n_2 = 1.55[/tex]

and the refractive index of water(material surrounding the lens) has a constant value of  [tex]n_1 = 1.33[/tex]

         [tex]\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}[/tex]

           [tex]f_{water} = 262.68 cm[/tex]

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

Answer:

a) [tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex], b) [tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex], c) [tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex], d) [tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]

Explanation:

a) The acceleration of the object is:

[tex]a = - 9.8\,\frac{m}{s^{2}}[/tex]

The velocity function is found by integration:

[tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]

b) The position function is found by integrating the velocity function:

[tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]

c) The time when the object reaches its highest point ocurrs when speed is zero:

[tex]0\,m = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]

[tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex]

d) The time when the object hits the ground occurs when [tex]s = 0\,m[/tex]. The roots are found by solving the second-order polynomial:

[tex]t = \frac{-v_{o}\pm \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot (-9.8\,\frac{m}{s^{2}} )}[/tex]

[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)} \mp \frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]

Since time is a positive variable and [tex]v_{o} < \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex], the only possible solution is:

[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]

The physics of vertical motion under gravity deals with velocity, position, and time. Velocity over time can be calculated by integrating the acceleration function, while position over time is calculated by integrating the velocity function. The height and time at the peak of motion, as well as when the object hits the ground, are determined by setting these equations to zero at appropriate points.

Considering the physics of vertical motion under gravity, the concepts of velocity, position, and time are important. The acceleration needs to be considered for handling such scenarios, which in your case, has been described by the equation ​a(t)=v′(t)=g. The value of g=−9.8 m/s², where 'g' represents the acceleration due to gravity.

a. Velocity over time: Using the equation of acceleration a=v′=g, and because g is a constant, the velocity at any point can be determined by integrating the acceleration function. This gives v(t)=gt+c, where the constant c is the initial velocity. For a softball with an initial vertical velocity of 30 m/s, this becomes v(t)=-9.8*t+30.

b. Position over time: The position equation can be found by integrating the velocity function, that is: s(t)=½gt²+vt+s₀. The s₀ term is the initial height, which is 0 because the softball is popped up from the ground. Therefore, s(t)=-4.9*t²+30*t.

c. Time and height at the highest point: The motion reaches its highest point when the velocity is zero. So, set v(t) equal to zero and solve for t. Substituting this time into the position equation gives the maximum height.

d. Time when the softball strikes the ground: The softball will hit the ground when its position is again zero. Set s(t) equal to zero and solve for t to find this time.

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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

A pair of narrow slits that are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light?

Given Information:

Distance from the double slits to the screen = D = 4.8 m

Double slit separation distance = d = 1.8 mm = 0.0018 m

Number of fringes = m = 5

Distance between fringes = y = 1 cm = 0.01 m

Required Information:

Wavelength of the monochromatic light = λ = ?

Answer:

Wavelength of the monochromatic light = λ = 7.5x10⁻⁷ m

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

λ = yd/mD

where d is the double slit separation distance, D is the distance from the double slits to the screen, y is the distance between bright fringes and m is number of fringes.

λ = 0.01*0.0018/5*4.8

λ = 0.00000075 m

λ = 7.5x10⁻⁷ m

Therefore, the wavelength of the monochromatic light is 7.5x10⁻⁷ m

Answer:

Time = 77.63 s

Distance = 3673.3 ft

Explanation:

Using equation of motion

v = u + at'

50 = 6 * t'

t' = 50 / 6

t' = 8.33 s

v² = u² + 2a(s - s•)

50² = 0² + 2 * 6 * (s - 0)

2500 = 12 * s

s' = 2500 / 12

s' = 208.3 ft

At t' = 8.33 the distance traveled by the car is

s'' = v• * t'

s'' = 30 * 8.33

s'' = 250 ft

Now, the distance between the motorcycle and car is

6000 - 208.33 - 250 = 5541.67

For motorcycle, when passing occurs,

s = v• * t, x = 50 * t''

For the car, when passing occurs,

s = v• * t, 5541.67 - x = 30 * t''

Solving both simultaneously, we have

5541.67 - [50 * t''] = 30 * t''

5541.67 = 30t''+ 50t''

5541.67 = 80t''

t''= 5541.67 / 80

t''= 69.3 s, substituting t'' = 69.3 back, we have

x = 50 * 69.3

x = 3465 ft

Therefore for motorcycle,

t = 69.3 + 8.33

t = 77.63 s

S = 3465 + 208.3

s = 3673.3 ft

Final Answer:

The total time [tex]\( t \)[/tex] for the motorcycle to pass the car is calculated to be [tex]\( 77.63 \)[/tex] seconds, and the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3673.3 \)[/tex]feet.

Explanation:

The calculation begins by using the equation of motion ( v = u + at ), where (v ) represents final velocity, ( u ) is initial velocity, ( a ) is acceleration, and (t) is time. Substituting the given values, the time ( t' ) for the motorcycle to reach a speed of ( 50 ) ft/s with a constant acceleration of [tex]( 6\) ft/s\(^2\) )[/tex]is found to be approximately ( 8.33 ) seconds.

Next, the equation of motion [tex]\( v^2 = u^2 + 2a(s - s_0) \)[/tex] is used, where [tex]\( s \)[/tex] represents distance, [tex]\( s_0 \)[/tex] is initial distance, and other variables have the same meanings as before. Solving for [tex]\( s' \),[/tex] the distance traveled by the motorcycle during this time is approximately [tex]\( 208.3 \)[/tex] feet.

At [tex]\( t' = 8.33 \)[/tex] seconds, the distance traveled by the car is calculated to be [tex]\( 250 \)[/tex]feet.

The distance between the motorcycle and car when they pass each other is found to be approximately [tex]\( 5541.67 \)[/tex] feet.

To determine the time [tex]\( t'' \)[/tex] when the passing occurs, simultaneous equations for the motorcycle and car are set up using their respective distances. Solving these equations yields [tex]\( t'' = 69.3 \)[/tex] seconds. Substituting this value back, the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3465 \)[/tex]feet.

The compound (NH4)2O is known as Ammonium Oxide, consisting of two ammonium ions and one oxide ion. It's produced via an acid-base reaction, with the ammonium ion acting as a weak acid and the oxide ion as a strong base. Such compounds have broad applications.

The compound (NH4)2O is known as Ammonium Oxide. It consists of two ammonium ions (NH4+) and one oxide ion (O2-). In terms of acid-base reactions, the ammonium ion is considered a weak acid due to its ability to donate a proton to water, while the oxide ion as a strong base accepts protons from water.

The formation of Ammonium Oxide involves the transfer of H+ ions from water molecules to ammonia molecules, which then react with the oxide anions to create the final compound. Ammonium compounds, including Ammonium Oxide, have applications in areas like agriculture and commodity chemical synthesizing.

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Answer:

Explanation:

Hooke's Law is a principle of physics that states that the that the force needed to extend or compress a spring by some distance is proportional to that distance. ... In addition to governing the behavior of springs, Hooke's Law also applies in many other situations where an elastic body is deformed

Answer:

Yeah ,The extension of the spring is directly proportional to the force applied.

Answer:

51.3 Degrees

The presence of person above the center of mass of ladder will make the ladder more likely to slip.

Explanation:

Explanation in the attachment.

51.3 Degrees, The presence of person above the centre of mass of ladder will make the ladder more likely to slip.

The middle of a uniform ladder is where the center of mass is located. A smooth wall signifies that there is no friction felt on the wall, whereas rough ground indicates that there is friction there.

How to use static equilibrium to determine the coefficient of friction between the bottom of the ladder and the ground is shown in the Ladder Problem from Static Equilibrium. The torque, the x-direction forces, and the y-direction forces are added together and set to zero.

It follows that the force exerted on the ladder as a result of the resultant moment of the forces must also equal zero due to a random point.

Thus, it is 51.3 Degrees.

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NOTE: The diagram is attached to this solution

Answer:

The acceleration of point A = 14.64 ft/s²

Explanation:

By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)

[tex]a_{A} = \bar{a} + \frac{\alpha L}{2}[/tex]

Weight, W = mg

g = 32.2 ft/s²

m = W/g

[tex]p = m \bar{a}\\p = w \bar{a} /g[/tex]

[tex]\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66[/tex]

[tex]\frac{pL}{2} = I \alpha[/tex]

but [tex]I = \frac{wL^{2} }{12g}[/tex]

[tex]\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L[/tex]

[tex]a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}[/tex]

Complete Question:

the first attached file shows the schematic

Answer:

a. irreversible cycle

b. 0.182

c. 0.2646

Explanation:

From the saturated vapour pressure table

When the pressure is 1 MPa

Temperature, T₄₁ = 179.9 °C = 452.9 °k

When Pressure is 20 kPa

Temperature, T₃₂ = 60.06 °C = 333.06 °K

attached images 2, 3, 4 and 5 shows a comprehensive solution to the questions

A vapor power plant uses water as the working substance, which goes through a cycle of becoming vaporized and condensed in order to drive a turbine. The schematic provided includes four components: the boiler, condenser, turbine, and pump. Two specific processes mentioned include 4-1 and 2-3, both of which occur at constant pressures.

A vapor power plant operates by using water as the working substance, which goes through a cycle of becoming vaporized, driving a turbine, and then condensing back into a liquid state. In this specific schematic, the water flows through four components: the boiler, condenser, turbine, and pump. The process data provided includes two specific processes: 4-1 which occurs at a constant pressure of 8 MPa, and 2-3 which also occurs at a constant pressure of 8 kPa.

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Answer:

m * c1 * (9 - 6) = m * c2 * (19 - 9) → mass m cancels

c2 = c1 * 3 / 10 = 0.3*c1

Explanation:

However, for the second mix, there should be nothing from the first mix's answer except for 0.3c1. Essentially, the left side of the second mix equation should look like: m *0.3c1 * (31.1-19). Then, you can properly solve the equation.

Answer:

[tex]1.75\cdot 10^4 V/m[/tex]

Explanation:

For a totally reflecting surface, the radiation pressure is related to the intensity of radiation by

[tex]p=2\frac{I}{c}[/tex] (1)

where

I is the intensity of radiation

c is the speed of light

The radiation pressure is given by

[tex]p=\frac{F}{A}[/tex]

where in this case:

[tex]F=3.6\cdot 10^{-9}N[/tex] is the force exerted by the beam

A is the area on which the force is exerted

The beam has a diameter of d = 1.30 mm, so its area is

[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{0.0013 m}{2})^2=1.33\cdot 10^{-6} m^2[/tex]

Now we can write eq(1) as

[tex]\frac{F}{A}=\frac{2I}{c}[/tex]

From which we find the intensity of radiation, I:

[tex]I=\frac{Fc}{2A}=\frac{(3.6\cdot 10^{-9})(3.0\cdot 10^8)}{2(1.33\cdot 10^{-6})}=4.06\cdot 10^5 W/m^2[/tex]

Now we can find the amplitude of the electric field using the equation

[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]

where

[tex]\epsilon_0=8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity

E is the amplitude of the electric field

And solving for E,

[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(4.06\cdot 10^5)}{(3\cdot 10^8)(8.85\cdot 10^{-12})}}=1.75\cdot 10^4 V/m[/tex]

Answer with Explanation:

We are given that  the mathematical form of wave

[tex]y(x,t)=Asin(kx-\omega t)[/tex]

a.We have to find the speed of propagation of this wave

In given mathematical form

k=Wave number

[tex]\omega[/tex]=Angular frequency

A=Amplitude

We know that

Speed of propagation of the wave=[tex]v_p=\frac{\omega}{k}[/tex]

b.Differentiate w.r.r t

[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]

The velocity [tex]v_y(x,t)[/tex] of a point on the string as  a function of x and t is given  by

[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]

(a) The speed of propagation of the wave is [tex]\frac{\omega}{k}[/tex]

(b) (b) The velocity of a point on the string as a function of x and t is [tex]v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]

The given parameters;

[tex]y(x, t) = Asin(kx - \omega t)[/tex]

where;

(a) The speed of propagation of the wave is calculated as follows;

[tex]v_p = \frac{\omega }{k}[/tex]

(b) The velocity of a point on the string as a function of x and t is calculated as follows;

[tex]v = \frac{dy}{dt} \\\\v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]

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Answer:

Maximum wavelength will be [tex]3.96\times 10^{-7}m[/tex]

Explanation:

It is given wavelength [tex]\lambda =242nm=242\times 10^{-9}m[/tex]

Speed of light [tex]c=3\times 10^8m/sec[/tex]

Plank's constant [tex]h=6.6\times 10^{-4}Js[/tex]

So energy is equal to

[tex]E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{242\times 10^{-9}}=8.18\times 10^{-19}J[/tex]

Maximum kinetic energy is given

[tex]KE_{max}=1.99eV=1.99\times 1.6\times 10^{-19}=3.184\times 10^{-19}J[/tex]

Work function is equal to

[tex]w_0=E-KE_{max}=8.18\times 10^{-19}-3.184\times 10^{-19}=5\times 10^{-19}J[/tex]

[tex]\frac{hc}{\lambda _0}=5\times 10^{-19}[/tex]

[tex]\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda _0}=5\times 10^{-19}[/tex]

[tex]\lambda _0=3.96\times 10^{-7}m[/tex]

Answer:

The sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m²

Explanation:

The intensity of Sun located at a distance of R away is [tex]100\ W/m^2[/tex].

We need to find the sun's intensity for a planet located at a distance 5 R from the Sun. Intensity is given by :

[tex]I\propto \dfrac{1}{r^2}[/tex]

So,

[tex]\dfrac{I_1}{I_2}=(\dfrac{r_2}{r_1})^2[/tex]

Here, I₁ = 100 W/m² and r₁ = R and r₂ = 5R

[tex]I_2=\dfrac{I_1r_1^2}{r_2^2}[/tex]

[tex]I_2=\dfrac{100(R^2)}{(5R)^2}\\\\I_2=\dfrac{100}{25}\\\\I_2=4\ W/m^2[/tex]

So, the sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m².

A main sequence star relies on heat pressure to counteract gravity, a white dwarf uses electron degeneracy pressure, a neutron star uses neutron degeneracy pressure, and in a black hole, gravity ultimately triumphs.

In a main sequence star, it is the heat pressure, due to nuclear fusion, which pushes against gravity and maintains the star's stability. A white dwarf is stabilized against gravity by what is known as electron degeneracy pressure, a quantum mechanical effect that arises because no two electrons can be in the same place doing the same thing at the same time (the Pauli exclusion principle).

The neutron star is supported against gravity by neutron degeneracy pressure, the same quantum effect but explained by neutrons instead of electrons. However, if the mass of a star's core is high enough (more than about three times that of the Sun), no known force can stop it from collapsing into a black hole; gravity ultimately overcomes all other forces.

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Answer:

a. The maximum strain is 60 * 10^-6

b. Resistance change = 0.014395 ohms

Explanation:

The complete question is as follows;

Two strain gauges are mounted so that they sense axial strain on a steel member in uniaxial tension. The 120 V gauges form two legs of a Wheatstone bridge, and are mounted on opposite arms (e.g., arms 1 and 4). The gauge factor for each of the strain gauges is 2 and E m for this steel is 29 × 106 psi. For a bridge excitation voltage of 4 V and a bridge output voltage of 120 μV under load (i.e., Ei = 4 V and 0  E =120 μV ):

(a) Estimate the maximum strain.

(b) What is the resistance change experienced by each gauge?

solution;

Please check attachment for complete solution and step by step explanation

Answer:

A correlation coefficient represents the following:

1- The direction of the relationship

2- The strength of the relationship

A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship.

A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship. The correlation coefficient ranges from -1 to 1, where a value close to -1 indicates a strong negative correlation, a value close to 1 indicates a strong positive correlation, and a value close to 0 indicates no or weak correlation.

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Answer:

The angular velocity is

5.64rad/s

Explanation:

This problem bothers on curvilinear motion

The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s

We know that the velocity v is given as

v= ωr

Where ω is the angular velocity

r is 300mm to meter = 0.3m

the radius of the circle

described by the level

v=1.64m/s

Making ω subject of the formula and solving we have

ω=v/r

ω=1.64/0.3

ω=5.46 rad/s

Answer:

Average velocity = 1.835 m/s

Explanation:

Detailed explanation and calculation is shown in the image below

The average velocity in the 0.3 m diameter pipe. The result is 1.833 m/s.

To find the average velocity in the 0.3 m diameter pipe, we'll use the principle of continuity for incompressible fluid flow. The flow rate (Q) must be the same at all points in the system.

First, we calculate the flow rates in the two initial pipes:

For the 0.15 m diameter pipe: Q1 = A₁ v₁ = π (0.075)² 2 m/s = π × 0.005625 m² × 2 m/s = 0.01125π m³/s

For the 0.2 m diameter pipe: Q2 = A2 v2 = π (0.1)² 3 m/s = π × 0.01 m² × 3 m/s = 0.03π m³/s

The combined flow rate entering the 0.3 m diameter pipe is:

[tex]Q_t = Q_1 + Q_2[/tex] = 0.01125π m³/s + 0.03π m³/s = 0.04125π m³/s

Next, we find the area of the 0.3 m diameter pipe:

A₃ = π (0.15)² = π 0.0225 m²

The average velocity in the 0.3 m diameter pipe is then:

v₃ = Qtotal / A₃ = (0.04125π m³/s) / (π 0.0225 m²) ≈ 1.833 m/s

Thus, the average velocity in the 0.3 m diameter pipe is 1.833 m/s.

Answer:

100 newtons

Explanation:

Recall force is the product of mass and acceleration from Newton's 2nd law given base units kgms^-2 which is equivalent to a newton( N)

Answer:

(a) 1.298 * 10^(-4) J

(b) 5.82 * 10^6 m/s

Explanation:

Parameters given:

Electric field, E = 640 N/C

Distance traveled by electron, r = 15 cm = 0.15 m

Mass of electron, m = 9.11 * 10^(-31) kg

Electric charge of electron, q = 1.602 * 10^(-19) C

(a) The kinetic energy of the electron in terms of Electric field is given as:

K = (q² * E² * r²) / 2m

Therefore, Kinetic energy, K, is:

K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]

K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]

K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J

(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.

Force is generally given as:

F = ma

Electric force is given as:

F = qE

Therefore, equating both, we have:

ma = qE

a = (qE) / m

a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))

a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²

Using one of the equations of motion, we have that:

v² = u² + 2as

Since the electron started from rest, u = 0 m/s

Therefore:

v² = 2 * 1.13 * 10^(14) * 0.15

v² = 3.39 * 10^(13)

v = 5.82 * 10^6 m/s

The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.

(a) The kinetic energy of the electron is 1.53×10⁻¹⁷ J

(b) The final speed of the electron is 5.8×10⁶ m/s

The electron is under motion due to the electrostatic force F produced by the electric field E = 640 N/C.

The equation of motion of the electron can be written as:

F = qE

⇒ ma = qE

where q = 1.6×10⁻¹⁹ C is the charge on the electron

m = 9.1×10⁻³¹ kg is the mass of the electron

and a is the acceleration.

So,

a = qE/m

a =  1.6×10⁻¹⁹×640 /  9.1×10⁻³¹

a = 1.12×10¹⁴ m/s²

The initial speed of the electron is zero, so u = 0

From the third equation of motion:

v² = u² + 2as

where v is the final speed,

and s is the distance traveled  = 15cm = 0.15m

v² = 0 + 2× 1.12×10¹⁴×0.15

v = 5.8×10⁶ m/s

So the Kinetic energy of the electron is:

KE = ¹/₂mv²

KE = 0.5×9.1×10⁻³¹×(5.8×10⁶)²

KE = 1.53×10⁻¹⁷ J

Learn more about kinetic energy:

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Answer:

a) n2=(n1sin1)(sin2)

b) 1.18

c) 201081632.7m/s

d) 254237288.1m/s

Explanation:

a) We can calculate the index of refraction of the second material by using the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}[/tex]

b) By replacing in the equation of a) we obtain:

[tex]n_2=\frac{(1,47)sin49\°}{sin69.5\°}=1.18[/tex]

c) light velocity in the medium is given by:

[tex]v=\frac{c}{n_1}=\frac{3*10^{8}m/s}{1.47}=204081632.7\frac{m}{s}[/tex]

d)

[tex]v=\frac{c}{n_2}=\frac{3*10^{8}m/s}{1.18}=254237288.1\frac{m}{s}[/tex]

hope this helps!!

The index of refraction for the second material can be found using Snell's Law, which is given by:  n₁sin(θ₁) = n₂sin(θ₂). The refractive index of the second material is approximately 1.19, The light's velocity in medium 1 is 2.04 x 10⁸ m/s. The light's velocity in medium 2 is 2.52 x 10⁸ m/s.

Using the provided angles and refractive index, we calculate the index of the second material to be approximately 1.19, with the corresponding light velocities as 2.04 x 10⁸ m/s in medium 1 and 2.52 x 10⁸ m/s in medium 2.

This question involves the concept of light refraction and the application of Snell's Law. Let's go through each part step-by-step.

The index of refraction for the second material can be found using Snell's Law, which is given by:

where:

Rearranging for n₂ gives:

Using the values:

Let's calculate it step by step:

So, the refractive index of the second material is approximately 1.19.

The velocity of light in a medium is given by:

where c is the speed of light in vacuum (approximately 3.00 x 10⁸ m/s), and n is the refractive index of the medium. For medium 1:

Similarly, for medium 2:

Explanation:

Given:

Distance between the transmitter and receiver = 160 km

Distance between Earth and Sun (D) = 150000000 km

Distance between Earth and Moon (D') = 380000 km

Distance between Earth and space ship (D") = 360000000 km

Distance between Earth and Supernova (d) = 9300 LY

We know that,

Speed of Light = 300000 km/sec

Lets us solve each problem now:

a) We know that the radio waves travle with speed of light. So the time(T) taken by the signal from transmitter to receiving antenna will be,

[tex]Time = \frac{Distance}{Speed}[/tex]

[tex]T = \frac{160}{300000}[/tex]

T = 0.53 millisecond

b) Time taken by sunlight to reach Earth is,

[tex]T_{1} = \frac{150000000}{300000} seconds[/tex]

T₁ = 500 seconds = 08 min 20 seconds

Time taken by light from Moon to reach Earth will be,

[tex]T_{2} = \frac{380000}{300000} seconds[/tex]

T₂ = 1.26 seconds

Let us assume that in the given situation Earth and Moon are at the same distance from the Sun. This means that Sun light will take the same time to reach Moon as it will take to reach Earth. So Sun light reaches the Moon in 500 seconds. After that it will be reflected towards Earth and will take 1.26 seconds more to reach Earth.

Thus the light left the Sun 501.26 seconds earlier.

c) Time taken by the light from space ship to reach Earth will be,

[tex]T = \frac{360000000}{300000} seconds[/tex]

T = 1200 seconds.

Thus round trip time will be = 2T = 2400 seconds

d) Lightyear is a unit of distance and is equal to the distance travelled by light in a year. If a supernova is at a distance of 9300 LY it means that light has taken 9300 years to reach Earth from that supernova. It means that the light we see today was actually generated from the supernova 9300 years ago.

It is evident that the explosion actually occurred 9300 years back.

We have calculated the times light and radio signals take to travel given distances at the speed of light; these times depend directly on the distances. For the observed supernova, we use the reverse calculation, determining the time light has traveled given the distance in light-years.

(a) The radio signal is essentially light, so it travels at the speed of light, which is about 300,000 kilometers per second (3 × 10⁸ m/s). To find the time it takes to travel 160 km, we simply divide the distance by the speed: 160 km / 300000 km/s = 0.00053 seconds, or about 0.53 milliseconds.

(b) The light we see from the Moon has indeed traveled from the Sun, bounced off the Moon, and then hit our eyes. Therefore, it has traveled a distance of the Earth–Sun distance plus the Earth–Moon distance. Adding these distances, the light has traveled a total of 1.5 × 10⁸ km + 3.8 × 10⁵ km = 1.50038 × 10⁸ km. Dividing this distance by the speed of light gives the time this light has traveled: 1.50038 × 10⁸ km / 300000 km/s = 500.13 seconds, or about 8 minutes and 20 seconds.

(c) The round-trip travel time of light between Earth and a spaceship would be twice the time it would take light to make a one-way trip. The one-way trip time is calculated as the distance divided by the speed of light, which is 3.6 × 10⁸ km / 300000 km/s = 1200 seconds. Therefore, the round-trip travel time would be 2400 seconds or 40 minutes.

(d) A light-year is the distance that light travels in one year, so if a supernova is observed to be 9300 light-years away, this means the light from the supernova has been traveling for 9300 years before reaching Earth. Therefore, the supernova actually happened 9300 years ago.

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Answer:

The current is  [tex]I =0.2mA[/tex]

Explanation:

From the question we are told that

   The first radius is [tex]R_1 = 5cm = \frac{5}{100} = 0.05cm[/tex]

    The number of turns is [tex]N = 17 \ turn/cm[/tex]

    The current rate is  [tex]\frac{dI}{dt} = 5 A/s[/tex]

    The second radius is  [tex]R_2 = 7cm = \frac{7}{100} = 0.07m[/tex]

     The resistance is [tex]r = 5 \Omega[/tex]

Generally the magnetic flux induced in the solenoid is mathematically represented as

      [tex]\O = B A[/tex]

 Where  is the magnetic field mathematically represented as

            [tex]B = N \mu_o I[/tex]

Where [tex]\mu_o[/tex] is the permeability of free space with a value of [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]

     and A is the area mathematically represented as

             [tex]A = \pi (R_2 - R_1)^2[/tex]

So

         [tex]\O = N \mu I * \pi R^2[/tex]

            Substituting values

        [tex]\O = 17 * 4\pi *10^{-7} * \pi (7-5)^2I[/tex]

           [tex]\O = 2.68*10^{-4}I[/tex]

The induced emf is mathematically represented as

              [tex]\epsilon =- |\frac{d\O}{dt}|[/tex]

                  [tex]\epsilon = 2.68*10^{-4 } \frac{dI}{dt}[/tex]

substituting values

               [tex]\epsilon =2.68 *10^{-4} * 5[/tex]

                  [tex]=1.3 *10^{-3} V[/tex]

From Ohm law

      [tex]I = \frac{\epsilon }{r}[/tex]

Substituting values

     [tex]I = \frac{1.3*0^{-3}}{5}[/tex]

        [tex]I =0.2mA[/tex]