Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, [tex]q_1=-3\ nC[/tex]

It is placed at a distance of 9 cm at x axis

Charge, [tex]q_2=+4\ nC[/tex]

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]

Here,

[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]

So,

[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]

Squaring both sides,

[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

According to the question,

Charge,

Now,

→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]

or,

→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]

→  [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]

here,

[tex]r_1 = \sqrt{y^2+81}[/tex]

[tex]r^2 = \sqrt{y^2+225}[/tex]

By substituting the values,

→      [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]

By applying cross-multiplication,

[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]

By squaring both sides, we get

→  [tex]9(y^2+225) = 16(y^2+81)[/tex]

    [tex]9y^2+2025 = 16 y^2+1296[/tex]

   [tex]2025-1296=7y^2[/tex]

               [tex]7y^2=729[/tex]

                  [tex]y = 10.2 \ m[/tex]

Thus the solution above is correct.

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Answer:

a. 24 m

Explanation:

Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.

Answer:

a)  A = 449526  J,  b) 449526 J

Explanation:

In this exercise they do not ask for the work of different elements.

Note that as the box rises at constant speed, the sum of forces is chorus, therefore

           T-W = 0

           T = W

           T = m g

           T = 1,390 9.8

           T = 13622 N

Now that we have the strength we can use the definition of work

           W = F .d

            W = f d cos tea

a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel

              A = A  x

              A = 13622  33

               A = 449526  J

b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180

               W = T x cos 180

               W = - 13622 33

               W = - 449526 J

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

Answer:

Trial 1: 2 Volts, 0 %

Trial 2: 2.8 Volts, 0%

Trial 3: 4 Volts, 0 %

Explanation:

Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:

V = IR

TRIAL 1:

V = IR

V = (0.1 A)(20 Ω)

V = 2 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2 - 2)/2| x 100%

% Difference = 0 %

TRIAL 2:

V = IR

V = (0.14 A)(20 Ω)

V = 2.8 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2.8 - 2.8)/2.8| x 100%

% Difference = 0 %

TRIAL 3:

V = IR

V = (0.2 A)(20 Ω)

V = 4 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(4 - 4)/4| x 100%

% Difference = 0 %

Answer:

m = 300668.9 kg

L₀ = 12.47 m

Explanation:

The relativistic mass of the space vehicle is given by the following formula:

[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]

where,

m = relativistic mass = ?

m₀ = rest mass = 150000 kg

v = relative speed = 2.6 x 10⁸ m/s

c = speed of light = 3 x 10⁸ m/s

Therefore

[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]

m = 300668.9 kg

Now, for rest length of vehicle:

L = L₀√(1 - v²/c²)

where,

L = Relative Length of Vehicle = 25 m

L₀ = Rest Length of Vehicle = ?

Therefore,

25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]

L₀ = (25 m)(0.499)

L₀ = 12.47 m

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = Δ[tex]V_{C}[/tex]

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

                  q = q₀ sin (wt)

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

[tex] M = log(\frac{I}{S}) [/tex]

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]

The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.

According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.

Mathematically, the net work done (W) on an object or physical body is given by the formula:

[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]

Where:

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Answer:

Your answer is( D) - Arago

Answer:

Explanation:

distance of third dark fringe

= 2.5 x λ D / d

where λ is wavelength of light , D is screen distance and d is slit separation

putting the given values

required distance = 2.5  x 739 x 10⁻⁹  x 6.3 / .49 x 10⁻³

= 23753.57 x 10⁻⁶

= 23.754 x 10⁻³ m

= 23.754 mm .

Answer:

0.9m/s²

Explanation:

See attached files

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Answer:

Explanation:

area of the coil  A = .08 x .08 = 64 x 10⁻⁴ m ²

flux through the coil Φ = area of coil x no of turns x magnetic field

= 64 x 10⁻⁴ x 50 x B where B is magnetic field

emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Answer:

35.6 m

Explanation:

The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.

The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.

Given data -

The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].

The initial height of the projection is, h = 1.50 m.

The horizontal displacement is, R = 3.00 m.

The mathematical expression for the horizontal displacement (Range) of the projectile is given as,

[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]

here,

u is the launch velocity.

g is the gravitational acceleration.

Solving as,

[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]

Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.

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Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r²           [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

The number of maxima of the standing wave pattern is two.

At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.

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Answer:

Explanation:

The sum of force along both components x and y is expressed as;

[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]

The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

[tex]a_x = \frac{d^2 x }{dt^2}[/tex]

[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]

Similarly,

[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]

[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]

[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

Answer:

5.33333... seconds

Explanation:

800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

Answer:

151.58 V

Explanation:

From the question,

The work done in a circuit in moving a charge is given as,

W = 1/2QV..................... Equation 1

Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.

make V the subject of the equation

V = 2W/Q.................. Equation 2

Given: W = 144 J. Q = 1.9 C

Substitute into equation 2

V = 2(144)/1.9

V = 151.58 V

Answer:

4.1 N/C

Explanation:

First of all, we know from maths that the surface area of a sphere = 4πr²

Charge on inner sphere ..

Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²

Q(i) = 5.03*10^-14 C

Charge on outer sphere

Q(o) = 60*10^-12 x 4π(0.03m)²

Q(o) = 6.79*10^-13 C

Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge

= +6.79*10^-13C + (+5.03*10^-14C)

= +7.29*10^-13 C

Now again, we have

E = kQ /r²

E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²

E = 6.561*10^-3 / 1.6*10^-3

E = 4.10 N/C

Thus, the magnitude of the electric field is 4.1 N/C

Answer:

Explanation:

Let the acceleration be a of the system

T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

for motion of 3.85 kg , applying newton's law

3.85g - T₁ = 3.85 a

for motion of 2.6 kg

T₂ - 2.6g = 2.6 a

T₂ - T₁ + 1.25 g = 6.45 a

T₁ - T₂ = 1.25 g -  6.45 a

for motion of pulley

(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

(T₁ - T₂ ) x R = 1 /2  m R² x a / R

(T₁ - T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

1.25 g -  6.45 a = .395 a

1.25 g = 6.845 a

a = 1.79 m /s²

B )

When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

Average deceleration = .2 / 6.4 = .03125 m /s²

Total deceleration created by frictional torque = 1.79 + .03125

= 1.82125 m /s²

If R be the average frictional torque  acting on the pulley

angular deceleration of pulley = a / R

= 1.82125 / .045

= 40.47 rad /s²

Now  R = I x 40.47 , I is moment of inertia of pulley

= 1 /2 x .79 x .045² x 40.47

= .0323 N.m

Torque created = .0323 Nm

The acceleration of the two masses hanging from ends of the pulley is 31 m/s².

The average frictional torque acting on the pulley is 0.55 Nm.

The given parameters;

The acceleration of the two masses is determined by taking net torque acting on the pulley;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_1R - T_2R = I \alpha\\\\[/tex]

where;

[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]

Substitute the given parameters, to solve for the acceleration of the masses;

[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]

The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.

[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]

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Answer:

Mb²/2

Explanation:

Pls see attached file

Answer:

Car 3

Explanation:

Answer:

[tex]I=1.19\times 10^{-2}\ W/m^2[/tex]

Explanation:

It is given that,

Maximum value of magnetic field strength, [tex]B=10^{-8}\ T[/tex]

We need to find the intensity of the light at that place.

The formula of the intensity of magnetic field is given by :

[tex]I=\dfrac{c}{2\mu _o}B^2[/tex]

c is speed of light

So,

[tex]I=\dfrac{3\times 10^8}{2\times 4\pi \times 10^{-7}}\times (10^{-8})^2\\\\I=1.19\times 10^{-2}\ W/m^2[/tex]

So, the intensity of the light is [tex]1.19\times 10^{-2}\ W/m^2[/tex].