Identical 4.0-uC (microCoulomb) charges are placed on the y axis at y = +/-4.0 m. (plus or minus 4.0 m) Point A is on the x axis at x = +3.0 m. Determine the electric potential of point A (relative to zero at the origin).

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex]    ..............1

put here value and we get

F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]

F(r) =  11.45 N

so magnitude of the resultant of forces is 11.45 N

Answer:

no the only things that can travel at the speed of light are waves in the electromagnetic spectrum

Answer:

The largest  possible distance is [tex]x = 4.720 \ m[/tex]

Explanation:

From the question we are told that

    The distance of  separation is   [tex]d = 4.30 \ m[/tex]

      The  frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The  wavelength of the tone played by the speaker is  mathematically evaluated as

              [tex]\lambda = \frac{v}{f}[/tex]

substituting values

            [tex]\lambda = \frac{343}{103}[/tex]

            [tex]\lambda = 3.33 \ m[/tex]

Let the the position of the observer be O

Given that the line of sight between observer and speaker B is  perpendicular to the distance between A and B then

        The distance between A and the observer is  mathematically evaluated using Pythagoras theorem as follows

               [tex]L = \sqrt{d^2 + x^2}[/tex]

Where x is the distance between the observer and B

  For the observer to observe destructive interference

          [tex]L - x = \frac{\lambda}{2}[/tex]

So  

          [tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]

       [tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]

        [tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]

         [tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]

       [tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]

substituting values              

       [tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]

      [tex]x = 4.720 \ m[/tex]

Answer:

The  maximum error is  [tex]\Delta \eta = 2032.9[/tex]

Explanation:

From the question we are told that

     The length  is  [tex]l = 1\ m[/tex]

      The radius is  [tex]r = 0.002 \pm 0.0002 \ m[/tex]

        The pressure is  [tex]P = 4 *10^{5} \ \pm 1750[/tex]

        The  rate  is  [tex]v = 0.5*10^{-9} \ m^3 /t[/tex]

       The viscosity is  [tex]\eta = \frac{\pi}{8} * \frac{P * r^4}{v}[/tex]

The error in the viscosity is mathematically represented  as

       [tex]\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v[/tex]

   Where  [tex]\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}[/tex]

and         [tex]\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}[/tex]

and          [tex]\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}[/tex]

So  

             [tex]\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v][/tex]

substituting values

            [tex]\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ][/tex]

  [tex]\Delta \eta = \frac{\pi}{8} [56 + 5120 ][/tex]

   [tex]\Delta \eta = 647 \pi[/tex]

    [tex]\Delta \eta = 2032.9[/tex]

Answer:

0.8 m

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing towards the center.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the centripetal direction:

∑F = ma

Nμ = m v²/r

Substitute and simplify:

mgμ = m v²/r

gμ = v²/r

Write v in terms of ω and solve for r:

gμ = ω²r

r = gμ/ω²

Plug in values:

r = (10 m/s²) (0.5) / (2.5 rad/s)²

r = 0.8 m

The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.

Given the following data:

To determine how far (radius) from the axis of rotation can the man stand without sliding:

We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.

[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.

But, Normal force, [tex]F_n = mg[/tex]  

Substituting the normal force into eqn. 1, we have:

[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.

Also, Linear speed, [tex]v = r\omega[/tex]

Substituting Linear speed into eqn. 2, we have:

[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]

Radius, r = 0.784 meters

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Answer:

The center of mass velocity is  [tex]v = 32.25 \ m/s[/tex]

Explanation:

From the question we are told that

          The mass of the cylinder is  [tex]m = 2.50 \ kg[/tex]

            The radius  is  [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]

             The length is  [tex]l = 2.7 \ cm = 0.027 \ m[/tex]

              The height of the plane is  h  = 79.60  m

               and the distance covered is  [tex]d = 4.64 \ m[/tex]

The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as

              [tex]v = \sqrt{\frac{4gh}{3} }[/tex]

substituting values  

               [tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]

              [tex]v = 32.25 \ m/s[/tex]

Answer:

i

  [tex]v = 142.595 \ m/s[/tex]

ii

  [tex]f = 109.69 \ Hz[/tex]

iii1 )

  [tex]f_2 =219.4 Hz[/tex]

iii2)

   [tex]f_3 =329.1 Hz[/tex]

iii3)

    [tex]f_4 =438.8 Hz[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]l = 0.65 \ m[/tex]

     The tension on the string is  [tex]T = 61 \ N[/tex]

     The mass per unit length is  [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]

The speed of wave on the string is mathematically represented as

       [tex]v = \sqrt{\frac{T}{m} }[/tex]

substituting values

      [tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]

     [tex]v = 142.595 \ m/s[/tex]

generally the  string's  frequency is mathematically represented as

         [tex]f = \frac{nv}{2l}[/tex]

n = 1  given that the frequency we are to find is the fundamental frequency

So

      substituting values

       [tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]

       [tex]f = 109.69 \ Hz[/tex]

The  frequencies at which the string would vibrate include

1       [tex]f_2 = 2 * f[/tex]

Here [tex]f_2[/tex] is  know as the second harmonic and the value is  

      [tex]f_2 = 2 * 109.69[/tex]

      [tex]f_2 =219.4 Hz[/tex]

2

[tex]f_3 = 3 * f[/tex]

Here [tex]f_3[/tex] is  know as the third harmonic and the value is  

      [tex]f_3 = 3 * 109.69[/tex]

     [tex]f_3 =329.1 Hz[/tex]

3

     [tex]f_3 = 4 * f[/tex]

Here [tex]f_4[/tex] is  know as the fourth harmonic and the value is  

      [tex]f_3 = 4 * 109.69[/tex]

     [tex]f_4 =438.8 Hz[/tex]

Answer:

Required KVP = 86 KVP (Approx)

Explanation:

Given:

Current KVP = 75 KVP

Find:

KVP required to double exposure

Computation:

According to 15% rule of KVP,

15% change increse in KVP required to get double exposure.

So,

Required KVP = Current KVP + Current KVP(15%)

Required KVP = 75 KVP + 75 (15%)

Required KVP = 75 KVP + 11.25 KVP

Required KVP = 86.25 KVP

Required KVP = 86 KVP (Approx)

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

Explanation:

a. Mechanical advantage = force out / force in

MA = 830 / 100

MA = 8.3

b. Efficiency = work out / work in

0.80 = W / 600 J

W = 480 J

Answer:

The answer would be B ........m

Complete Question:

Given [tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex] at a point. What is the force per unit area at this point acting normal to the surface with[tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex]   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, [tex]\sigma_n = 28 MPa[/tex]

There are shear stresses acting on the surface since [tex]\tau \neq 0[/tex]

Explanation:

[tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex]

equation of the normal, [tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex]

[tex]\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]

Traction vector on n, [tex]T_n = \sigma \b n[/tex]

[tex]T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]

[tex]T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right][/tex]

[tex]T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z[/tex]

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

[tex]\sigma_n = T_n . \b n[/tex]

[tex]\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa[/tex]

If the shear stress, [tex]\tau[/tex], is calculated and it is not equal to zero, this means there are shear stresses.

[tex]\tau = T_n - \sigma_n \b n[/tex]

[tex]\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z[/tex]

[tex]\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa[/tex]

Since [tex]\tau \neq 0[/tex], there are shear stresses acting on the surface.

Answer:

A C

Explanation:

The statement of the exercise is a bit strange, but if the distance between the load increases.

The following phenomena must occur.

* If the charge has a spatial distribution, the electric field should reduce the electric field of a point charge at the same distance

* As the distance increases the value of the electric field decreases in quadratic form

therefore when reviewing the correct answers are

if the total load is q, answer A is correct

and answer C is always correc

Answer:

momentum of the coupled cars V =  1.77 m/s

kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J

Explanation:

given

m₁ =3.05 × 10⁴kg

u₁ =2.90m/s

m₂=6.10× 10⁴kg

u₂=1.20m/s

using law of conservation of momentum

m₁u₁ + m₂u₂ = (m₁ + m₂) V

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V

V =  1.617×10⁵/9.15×10⁴

V = 1.77m/s

K.E =1/2mV²

ΔK.E = K.E(final) - K.E(initial)

ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔK.E = -2.892×10⁴J

The final speed is 1.77 m/s

The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]

2.892×10⁴J of energy is converted.

Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.

Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg

The initial speed of the railroad boxcar is u₁ = 2.90m/s

Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg

And the initial speed be u₂ = 1.20m/s

The initial momentum of the first car is:

m₁u₁ = 3.05 × 10⁴ × 2.90 =  8.84 × 10⁴ kgm/s

The initial momentum of the coupled car is:

m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s

Let the final speed after all the boxcars are coupled be v

From the law of conservation of momentum, we get:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv

v =  1.617×10⁵/9.15×10⁴

v = 1.77m/s

The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:

ΔKE = K.E(final) - K.E(initial)

ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔKE = -2.892×10⁴J

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Answer:

The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.

Explanation:

The riders will experience a centripetal force from the cylinder

[tex]F_{C}[/tex] = mrω^2    .... equ 1

where

m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of of the rider

For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as

[tex]F_{f}[/tex] = μR       ....equ 2

where

μ = coefficient of friction = 0.87

R is the normal force from the rider = mg

where

m is the rider's mass

g is the acceleration due to gravity = 9.81 m/s

substitute mg for R in equ 2, we'll have

[tex]F_{f}[/tex] = μmg     ....equ 3

Equating centripetal force of equ 1 and frictional force of equ 3, we'll get

mrω^2 = μmg

the mass of the rider cancels out, and we are left with

rω^2 = μg

ω^2 = μg/r

ω = [tex]\sqrt{\frac{ug}{r} }[/tex]

ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]

ω = 1.58 rad/second

The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s

The riders will experience a  centripetal force from the cylinder

[tex]F = mrw^2[/tex]

where  m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of the rider

For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as

f = μN

where

μ = coefficient of friction = 0.87

N is the normal force = mg

f = μmg  

Equating centripetal force of and frictional force of we'll get

[tex]mrw^2 = umg[/tex]

[tex]rw^2 = ug[/tex]

[tex]w^2 = ug/r[/tex]

[tex]w= \sqrt{ug/r}[/tex]

[tex]w= \sqrt{0.87*9.8/3.4}[/tex]  

ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.

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Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

Explanation:

Answer:

33.749

Explanation:

According to the given situation, the solution of the minimum angle is shown below:-

We will apply the law to no and n1 medium  which is

[tex]1.8\times sin(\theta_2)=1\times sin90[/tex]

[tex]\theta_2 = sin^{-1} \frac{1}{1.8}[/tex]

After solving the above equation we will get

= 33.749

Therefore for computing the minimum angle we simply applied the above formula.

Hence, the correct answer is 33.749

Answer:

E = 1.50 × [tex]10^{8}[/tex] V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

[tex]\mu _b = \frac{B}{2\mu _o}[/tex]   ...............1

and

energy density due to electric filed is

[tex]\mu _e = \frac{\epsilon _o E^2}{2}[/tex]     ...............2

and here [tex]\mu _b = \mu _ e[/tex]

so that

E = [tex]\frac{B}{\sqrt{\mu _o \times \epsilon _o}}[/tex]      ...................3

put here value and we get

[tex]E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}[/tex]  

E = 3 × [tex]10^{8}[/tex]  × 0.50

E = 1.50 × [tex]10^{8}[/tex] V/m

Answer:

Drop to half of the previous value

Explanation:

Energy stored in capacitor is inversly propotional to the distance between the plates.

If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor drops to half its previous value.

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is  given by

C = ε₀ A/d

where, ε₀ is the permittivity of free space, A = area of cross section of plates and d is the distance between them.

Capacitance is inversely proportional to the distance between them. So, if distance is doubled, the capacitance decreases to half its original value.

Thus, the correct option is 8.

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The heat emitted from anything is carried in the form of infrared waves. (C)

Answer:

The speed of the electron will be 6x10^8m/s

Explanation:

See attached file

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

Answer:

Because closed magnetic field loops have to be formed between both ends of the magnet, a magnet will always have two poles.

Explanation:

Magnetic Monopoles do not exist in nature because a magnetic field always forms a loop that runs from one end of the magnet to the other.

Since this loop of the magnetic field has an origination and termination point which are at the two ends of the magnet (North and South poles).  A magnet will always be bipolar which is in this case, North and South; even at an atomic level.

Answer:

c. tension, gravity, and the centripetal force

Explanation:

The ball experiences a variety of force as explained below.

Gravity force acts on the body due to its mass and the acceleration due to gravity. The gravity force on every object on earth due to its mass keeps all object on the surface of the earth.

Although the car moves around in circle, centripetal towards the center of the radius of turn exists on the ball. This centripetal force is due to the constantly changing direction of the circular motion, resulting in a force away from the center. The centripetal force keeps the ball from swinging off away from the center of turn.

Tension force on the string holds the ball against falling towards the earth under its own weight, and also from swinging away from the center of turn of the car. Tension force holds the ball relatively fixed in its vertical position in the car.

Answer:

3

Explanation:

When two plane mirrors are placed side by side such that they make some angle, θ, with each other, the number of images, n, of an object placed close to and in front of these mirrors is given by;

n = (360 / θ) - 1         ------------(i)

From the question;

θ = 90°            [since they stood making a right angle with each other]

Substitute this value into equation (i) as follows;

n = (360 / 90) - 1

n = 4 - 1

n = 3

Therefore, the number of images formed is 3

Answer:

The magnitude of the momentum is 49145.19 kg.m/s

The direction of the two vehicles is 44.6° North West

Explanation:

Given;

speed of first vehicle, v₁ = 14 m/s (East to west)

mass of first vehicle, m₁ = 1500 kg

speed of second vehicle, v₂ = 23 m/s (South to North)

momentum of the first vehicle in x-direction (E to W is in negative x-direction)

[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]

momentum of the second vehicle in y-direction (S to N is in positive y-direction)

[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]

Magnitude of the momentum of the system;

[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]

Direction of the two vehicles;

[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West

Answer:

at the top of the 9 story building i think

Explanation:

When the ball starts to move, its kinetic energy increases and potential energy decreases. Thus the ball will experience its maximum potential energy at the top height before falling.

Potential energy of a massive body is the energy formed by virtue of its position and displacement. Potential energy is related to the mass, height and gravity as P = Mgh.

Where, g is gravity m is mass of the body and h is the height from the surface.  Potential energy is directly proportional to mass, gravity and height.

Thus, as the height from the surface increases, the body acquires its maximum potential energy. When the body starts moving its kinetic energy progresses and reaches to zero potential energy.

Therefore, at the sate where the ball is at the  top of the building it have maximum potential energy and then changes to zero.

To find more about potential energy, refer the link below:

https://brainly.com/question/24284560

#SPJ2

Answer:

18.75 rad/s

Explanation:

Moment of inertia of the disk;

I_d = ½ × m_disk × r²

I_d = ½ × 10 × 4²

I_d = 80 kg.m²

I_package = m_pack × r²

Now,it's at 2m from the centre, thus;

I_package = 5 × 2²

I_package = 20 Kg.m²

From conservation of momentum;

(I_disk + I_package)ω1 = I_disk × ω2

Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.

Thus;

Plugging in the relevant values, we obtain;

(80 + 20)15 = 80 × ω2

1500 = 80ω2

ω2 = 1500/80

ω2 = 18.75 rad/s

Answer:

radius of the loop =  7.9 mm

number of turns N ≅ 399 turns

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = [tex]4\pi *10^{-7} T-m/A[/tex] = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = [tex]2\pi r[/tex]

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = [tex]7.9*10^{-4} m[/tex] = 0.0079 m = 7.9 mm

Answer:

The speed of the ball at this distance is 9.15 m/s

Explanation:

Given;

initial speed of the baseball, U = 5.0 m/s.

distance traveled along the path way, h = 3 m

final speed of the baseball at this distance, V = ?

The baseball is falling under the influence of gravity.

Acceleration due to gravity, g is positive, since the baseball is falling towards its direction.

g = 9.8 m/s²

Apply the third kinematic equation;

V² = U² + 2gh

V² = 5² + 2 x 9.8 x 3

V² = 25 + 58.8

V² = 83.8

V = √83.8

V = 9.15 m/s

Therefore, the speed of the ball at this distance is 9.15 m/s