A disk of radius 1.6 m and mass 3.8 kg rotates about an axis through its center. A force of 18.4 N is applied tangentially to the edge of the hoop, causing it to rotate counterclockwise. Assuming the disk is initially at rest, after 3 s... a) what is the radial acceleration of a point halfway between the axis and the edge of the disk

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.

Complete Question

The diagram for this question is shown on the first  uploaded image

Answer:

The initial angular momentum is  [tex]L_i= 0.134 Kg .m^2/s[/tex]

Explanation:

From the question we are told that

        The mass of the rod is [tex]M = 3.5 kg[/tex]

         The mass of the ball is [tex]m =45g = \frac{45}{1000} = 0.045kg[/tex]

          The speed is [tex]v = 5.02 m/s[/tex]

          The angle the ball strikes the rod is [tex]\theta = 42^o[/tex]

           The distance from the center of the rod is [tex]D = \frac{2}{3} L[/tex]

          The length L is  [tex]= 1.2m[/tex]

          The initial angular momentum of the ball is mathematically represented as

                  [tex]L_i = m\ v\ D\ cos \theta[/tex]

Substituting the value

                 [tex]L_i = 45*10^{-3} * 5.02 * \frac{2}{3} * 1.2 * cos (42)[/tex]

                      [tex]= 0.134 Kg .m^2/s[/tex]

Following are the calculation of the initial angular momentum of the ball:

Given:

[tex]M = 3.5 \ kg \\\\m = 45\ g = 0.045\ kg\\\\ v = 5.02\ \frac{m}{s}\\\\ \theta = 42^{\circ}\\\\D = \frac{2}{3}\ L\\\\ L = 1.2\ m\\\\[/tex]

To find:

Li=?

Solution:

Using formula:

[tex]Li= m v D \cos \theta=mv\times \frac{2L}{3}\cos \theta\\\\[/tex]

      [tex]=0.045 \times 5.02 \times \frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.2259 \times\frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.18072\times \cos 42^{\circ}\\\\=0.13430113286\\\\=13.43 \times 10^{-2} \ \frac{kg\ m^2}{s}\\\\[/tex]

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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]

Final answer:

The minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).

Explanation:

In order to suspend the bead in equilibrium, the radiation pressure from the laser beam must be equal to the gravitational force on the bead.

The radiation pressure is given by P = 2I/c, where P is the pressure, I is the laser intensity, and c is the speed of light. The gravitational force is given by F = (4/3)πr³ρg, where F is the force, r is the radius of the bead, ρ is the density, and g is the acceleration due to gravity.

Equating the radiation pressure and gravitational force, we have 2I/c = (4/3)πr³ρg. Rearranging this equation, we get I = (3c/8πr³ρg).

Therefore, the minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).

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 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

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:

Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot

Explanation:

Given

Pressure applied [tex]= 1700[/tex] pounds per square foot

Initial Volume [tex]= 5[/tex] cubic feet

Final Volume [tex]= 9[/tex] cubic feet

As we know pressure is inversely proportional to V

[tex]P = \frac{k}{V}[/tex]

where k is the proportionality constant

V is the volume and

P is the pressure

Work done

[tex]\int\limits^{V_2}_{V_1} {P} \, dV[/tex]

 [tex]\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV\\= \int\limits^{V_2}_{V_1} {\frac{1700 * 5}{V} } \, dV\\= 8500* \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV[/tex]

Integrating the above equation, we get-

[tex]8500 ln \frac{V_2}{V_1} \\8500 * 2.303 * \frac{9}{5} \\= 4996.18[/tex]

Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot

Answer:

The magnitude of the acceleration  is [tex]a = 0.33 m/s^2[/tex]

The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]

         The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]

         The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]

        The the magnetic field is  [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]

The charge experienced  a force which is mathematically represented as

                    [tex]F = q (\= v * \= B)[/tex]

    Substituting value

         [tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]

            [tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]

            [tex]= (-5.966*10^4 N) \r k[/tex]

Note :

           [tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]

Now force is also mathematically represented as

        [tex]F = ma[/tex]

Making a the subject

      [tex]a = \frac{F}{m}[/tex]

   Substituting values

     [tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]

        [tex]= (-0.33m/s^2)\r k[/tex]

        [tex]= 0.33m/s^2 * (- \r k)[/tex]

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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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.

Answer:

v = 1.95*10^5 m/s

3.13 x 10^16 proton

Explanation:

Identify the unknown:  

The speed of the protons  

List the Knowns:  

Energy of protons KE = 2 keV = 2*10^3 eV

Current produced by the generator: I= 5.05 mA = 5 x 10^-3 A  

1 eV = 1.6 x 10-19 Joule

Mass of proton: m = 1.67 x 10^-27 kg

Charge of proton: q_p = 1.6 x 10^-19 C  

Set Up the Problem:

Kinetic energy is given by:

KE= 1/2mv^2

v = √2KE/m

Solve the Problem:  

v = √2 x 2*10^3 x 1.6 x 10^-19/ 1.67 x 10^-27

v = 1.95*10^5 m/s

b.  Identify the unknown:  

Number of protons produced each second  

Set Up the Problem:  

Current is given by:

I =ΔQ/Δt

So, the total charge in one second:  

ΔQ =I*Δt = 5 x 10^-3 x 1 = 5 x 10^-3 C  

Number of protons in this charge:

n =   ΔQ/q_p

Solve the Problem:  

n =   5 x 10^-3/1.6 x 10^-19  

  =3.13 x 10^16 proton

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:

[tex]{\rm K} = 2.4\times 10^{-19}~J[/tex]

Explanation:

The electric field inside a parallel plate capacitor is

[tex]E = \frac{Q}{2\epsilon_0 A}[/tex]

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

[tex]F = qE = \frac{qQ}{2\epsilon_0 A}[/tex]

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

[tex]C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60[/tex]

Plugging this identity into the force equation above gives

[tex]F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q[/tex]

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is [tex]1.6 \times 10^{-19}[/tex]

Therefore, the kinetic energy is [tex]2.4\times 10^{-19}[/tex]

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%

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]

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:

The amount of charge on the plates increases during this process

Explanation:

In order to maintain a constant voltage across the plates, pushing the plates closer will increase the charge on the plate

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.

Answer:

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Explanation:

The magnitude of the magnetic field 'B' can be determined by;

             B = [tex]\frac{mV}{qR}[/tex]

where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.

Given that: speed 'V' of the proton = 3.5 × [tex]10^{6} ms^{-2}[/tex], distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × [tex]10^{-19}[/tex] C and mass of proton = 1.67 × [tex]10^{-27}[/tex]Kg.

Thus,

  B = (1.67 × [tex]10^{-27}[/tex] × 3.5 ×[tex]10^{6}[/tex]) ÷ (1.6 × [tex]10^{-19}[/tex] ×  0.23)

      = [tex]\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }[/tex]

      = 0.15883

 B = 0.16 Tesla

The magnitude of the magnetic field 'B' is 0.16 Tesla.

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:

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:

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:

It would raise further south

Explanation:

To those of us who live on earth, the most important astronomical object by far is the sun. It provides light and warmth. Its motions through our sky cause day and night, the passage of the seasons, and earth's varied climates.

On any given day, the sun moves through our sky in the same way as a star. It rises somewhere along the eastern horizon and sets somewhere in the west. If you live at a mid-northern latitude (most of North America, Europe, Asia, and northern Africa), you always see the noon sun somewhere in the southern sky.

The sun's path through the rest of the sky is similarly farther north in June and farther south in December. In summary:

In late March and late September (at the "equinoxes"), the sun's path follows the celestial equator. It then rises directly east and sets directly west. The exact dates of the equinoxes vary from year to year but are always near March 20 and September 22.

After the March equinox, the sun's path gradually drifts northward. By the June solstice (usually June 21), the sun rises considerably north of due east and sets considerably north of due west. For mid-northern observers, the noon sun is still toward the south, but much higher in the sky than at the equinoxes.

After the June solstice, the sun's path gradually drifts southward. By the September equinox, its path is again along the celestial equator. The southward drift then continues until the December solstice (usually December 21), when the sun rises considerably south of due east and sets considerably south of due west.

Answer:

It would rise further south

Explanation:

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

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Given Information:  

Initial speed = v₁ = 13 m/s

time = t = 4.50 sec

Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m

Altitude = h = 30,000 km = 3×10⁷ m

Required Information:  

a) mass of Mongo = M = ?

b) time in hours = t = ?

Answer:  

a) mass of Mongo = M =  8.778×10²⁵ kg

b) time in hours = t = 11.08 h

Explanation:  

We know from the equations of kinematics,

v₂ = v₁t - ½gt²

0 = 13*4.50 - ½g(4.50)²

58.5 = 10.125g

g = 58.5/10.125

g = 5.78 m/s²

Newton's law of gravitation is given by

M = gC²/4π²G

Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.

M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)

M = 8.778×10²⁵ kg

Therefore, the mass of planet Mongo is 8.778×10²⁵ kg

b) From the Kepler's third law,

T = 2π*(R + h)^3/2/(G*M)^1/2

Where R = C/2π

T = 2π*(C/2π + h)^3/2/(G*M)^1/2

T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2

T = 39917.5 sec

Convert to hours

T = 39917.5/60*60

T = 11.08 hours

Therefore, it will take 11.08 hours for the ship to complete one orbit.

To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The mass of Mongo can then be found using Newton's law of universal gravitation. To determine the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we can use Kepler's third law of planetary motion.

a. What is the mass of Mongo?

To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The equation for the motion of the stone is given by:

h = ut + 0.5gt^2

Where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. From the given information, we can find that the stone reaches a maximum height and returns to the ground in 4.50 s. Plugging in the values:

0.0 = (13.0 m/s)(4.50 s) + 0.5g(4.50 s)^2

Simplifying, we find:

g = 13.0 m/s / 4.50 s

Then, we can use Newton's law of universal gravitation to find the mass of Mongo:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant, m1 is the mass of the stone, m2 is the mass of Mongo, and r is the radius of Mongo. Since the stone is on the surface of Mongo, the radius is equal to the circumference divided by 2*pi:

r = 2.00×10^5 km / (2 * pi)

Substituting the known values, we can solve for m2:

m2 = (F * r^2) / (G * m1)

Calculating the gravitational force between the stone and Mongo using the known values and substituting them in the formula, the mass of Mongo is determined.

b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

To find the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we need to use Kepler's third law of planetary motion. This law states that the square of the orbital period (T) is proportional to the cube of the orbital radius (r):

T^2 = (4 * pi^2 * r^3) / (G * m)

Where T is the orbital period, r is the orbital radius, G is the gravitational constant, and m is the mass of Mongo. Plugging in the known values, we can solve for T and convert it to hours.

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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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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:

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.