Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless. they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.



Required:


a. What is the energy, in joules, of the photon?


b. What is the energy, in electron volts. of the photon?

Answer:

8.167

Explanation:

Answer:

c. 4.2 m/s

Explanation:

The definition of the average velocity, measured in meters per second, is given by the following expression:

[tex]\bar v = \frac{x_{f}-x_{o}}{t_{f}-t_{o}}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final positions, measured in meters.

[tex]t_{o}[/tex], [tex]t_{f}[/tex] - Initial and final instants, measured in seconds.

Positions and instants must be written in meters and seconds, respectively:

[tex]x_{o} = 0\,m[/tex], [tex]x_{f} = 1000\,m[/tex].

[tex]t_{o} = 0\,s[/tex], [tex]t_{f} = 240\,s[/tex].

Finally, the average velocity of the athlete that runs a total length of 1.0 kilometer in a time interval of 4 minutes is:

[tex]\bar v = \frac{1000\,m-0\,m}{240\,s-0\,s}[/tex]

[tex]\bar v = 4.167\,\frac{m}{s}[/tex]

Hence, the best option is C.

Answer:

3Nm

Explanation:

work = 0.5 x 12 x 0.5 = 3

The work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.

To calculate the work done in stretching a spring, we can use the formula for work done by a spring:

Work = (1/2) * k *[tex]x^2[/tex]

where:

k = spring constant

x = distance the spring is stretched

Given that the restoring force (F) acting on the spring is 24 N, and the distance the spring is stretched (x) is 0.5 m, we can find the spring constant (k) using Hooke's law:

F = k * x

k = F / x

k = 24 N / 0.5 m

k = 48 N/m

Now, we can calculate the work:

Work = (1/2) * 48 N/m * [tex](0.5 m)^2[/tex]

Work = (1/2) * 48 N/m * [tex]0.25 m^2[/tex]

Work = 6 joules

Therefore, the work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.

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Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]

Where

[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.

[tex]F(x)[/tex] - Force as a function of position, measured in newtons.

Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]

[tex]k = 250\,\frac{N}{m}[/tex]

Now, the work function is obtained:

[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]

[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]

[tex]W = 0.313\,J[/tex]

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:

[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]

[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]

By using trigonometrical identities, the integral is further simplified:

[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]

[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]

[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]

[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]

[tex]A = 4\pi[/tex]

The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

[tex]W_{g}=mgd\sin\theta[/tex]

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]

[tex]W_{g}=-158.8\ J[/tex]

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

[tex]\Delta U=-W[/tex]

Put the value into the formula

[tex]\Delta U=-(-158.8)\ J[/tex]

[tex]\Delta U=158.8\ J[/tex]

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

[tex]W=F\times d[/tex]

Put the value into the formula

[tex]W=100\times5.10[/tex]

[tex]W=510\ J[/tex]

We need to calculate the work done by frictional force

Using formula of work done

[tex]W=-f\times d[/tex]

[tex]W=-\mu mg\cos\theta\times d[/tex]

Put the value into the formula

[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]

[tex]W=-172.5\ J[/tex]

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]

Put the value into the formula

[tex]\Delta k=-158.8-172.5+510[/tex]

[tex]\Delta k=178.7\ J[/tex]

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]

[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]

Put the value into the formula

[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]

[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]

[tex]v_{2}=6.35\ m/s[/tex]

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

Answer:

a) a = 2.383 m / s², b)   T₂ = 120,617 N , c)   T₃ = 72,957 N

Explanation:

This is an exercise of Newton's second law let's fix a horizontal frame of reference

in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.

a) request the acceleration of the system

we can take the sledges together and write Newton's second law

     T = (m₁ + m₂ + m₃) a

    a = T / (m₁ + m₂ + m₃)

     a = 143 / (10 +20 +30)

     a = 2.383 m / s²

b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg

on the sled of m₁ = 10 kg

          T - T₂ = m₁ a

in this case T₂ is the cable tension

           T₂ = T - m₁ a

            T₂ = 143 - 10 2,383

            T₂ = 120,617 N

c) The cable tension between the masses of 20 and 30 kg

            T₂ - T₃ = m₂ a

             T₃ = T₂ -m₂ a

             T₃ = 120,617 - 20 2,383

             T₃ = 72,957 N

Answer:

The permittivity of rubber is  [tex]\epsilon = 8.703 *10^{-11}[/tex]

Explanation:

From the question we are told that

     The  magnitude of the point charge is  [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]

      The diameter of the rubber shell is  [tex]d = 32 \ cm = 0.32 \ m[/tex]

       The Electric field inside the rubber shell is  [tex]E = 2500 \ N/ C[/tex]

The radius of the rubber is  mathematically evaluated as

              [tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         [tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]

Where [tex]\epsilon[/tex] is the permittivity of rubber

    =>     [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]

   =>      [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]

substituting values

            [tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]

            [tex]\epsilon = 8.703 *10^{-11}[/tex]

Answer:

v= 4.823 × 10⁻⁹ cm/s

Explanation:

given

flow rate = 9.6 x10-5 cm³/s, d = 0.080mm

r = d/2= 0.080/2= 0.0040 cm

speed = rate of blood flow × area

v = (9.6 x 10⁻⁵ cm³/s) × (πr²)

v = (9.6 x 10⁻⁵ cm³/s) × π(0.0040 × cm)²

v= 1.536 × 10⁻⁹π cm/s

v= 4.823 × 10⁻⁹ cm/s

Answer:

Increase of volume (F)  = 8.01%

Explanation:

Given:

Density of fish = 1,080 kg/m³

Density of water = 1,000 kg/m³

density of air = 1.28 kg/m³

Find:

Increase of volume (F)

Computation:

1,080 kg/m³  + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³  

1,080 + 1.28 F =1,000 F + 1,000

80 = 998.72 F

F = 0.0801 (Approx)

F = 8.01%  (Approx)

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

Answer:

Explanation:

Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.

q = 1.6 x 10-19 C

v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s

the force acting on electron is

F= qvBsinΦ

F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)

F = 1.793x 10⁻¹⁸ N

The net force acting on electron is

F = e ( E+ ( vXB)

= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)

= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)

a= F/m

1.60 × 10¹² i =  ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹

9.11 i = - ( E- 6.7 k + 9.0 j)

E = -9.11i + 6.7k - 9.0j

Answer:

the coefficient of static friction is larger than kinetic coefficients of friction

Explanation:

The coefficient of static friction is usually larger than the kinetic coefficients of friction because an object at rest has increasingly settled agreements with the surface it's resting on at the molecular level, so it takes more force to break these agreement.

Until this force is been overcome, kinetic coefficient of friction is not going to surface.

Note: coefficient of static friction is the friction between two bodies when the bodies aren't moving. Meanwhile, kinetic coefficient is the ratio of frictional force of a moving body to the normal reaction.

[tex]F_{s}[/tex] ≤μ[tex]_{s}[/tex]N(coefficient of static friction)

where [tex]F_{s}[/tex]  is the static friction, μ[tex]_{s}[/tex] is the coefficient of static friction and N is the normal reaction

μ = [tex]\frac{F}{N}[/tex](kinetic coefficient of friction)

attached is diagram illustrating the explanation

Answer:

The angular momentum of the solid sphere is 0.667 kgm²/s

Explanation:

Given;

radius of the solid sphere, r = 0.15 m

mass of the sphere, m = 13 kg

angular speed of the sphere, ω = 5.70 rad/s

The angular momentum of the solid sphere is given;

L = Iω

Where;

I is the moment of inertia of the solid sphere

ω is the angular speed of the solid sphere

The moment of inertia of solid sphere is given by;

I = ²/₅mr²

I = ²/₅ x (13 x 0.15²)

I = 0.117 kg.m²

The angular momentum of the solid sphere is calculated as;

L = Iω

L = 0.117 x 5.7

L = 0.667 kgm²/s

Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s

Answer:

It always contains certain elements

Explanation:

Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.

Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.

The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

A mineral is a naturally occurring chemical compound, usually of a crystalline form.

Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

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Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:    

8 payments of $24,225:           $193,800    

Par value at maturity:                 $570,000    

Total repaid:                                   $763800 (193,800 + 570,000)  

Less amount borrowed:         $508050    

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period­ End; Unamortized Discount; Carrying Value

01/01/2019                                      61,950                           508,050  

06/30/2019                                      54,206                          515,794  

12/31/2019                                       46,462                         523,538  

06/30/2020                                       38,718                        531,282  

12/31/2020                                         30,974                          539,026

3) Record the interest payment and amortization on June 30:

June 30            Bond interest expense, dr                         31969  

                       Discount on bonds payable, Cr     (61950/8)  7743.75

                                        Cash, Cr                     ( 570000*8.5%/2)  24225  

4) Record the interest payment and amortization on December 31:

Dec 31                 Bond interest expense, Dr               31969  

                           Discount on bonds payable, Cr  7744  

                                    Cash, Cr                                24225

Terminal Velocity is the constant speed that a falling thing reaches when the resistence of a medium prevents the thing to reach any further speed.

Best of Luck!

Answer:

The discharging current is [tex]I_d = 36.8 \ A[/tex]

Explanation:

From the question we are told that  

     The radius of each circular plates is  R

     The displacement current is  [tex]I = 9.2 \ A[/tex]

      The radius of the central circular area is  [tex]\frac{R}{2}[/tex]

The discharging current is mathematically represented as

       [tex]I_d = \frac{A}{k} * I[/tex]

where A is the area of each plate which is mathematically represented as

       [tex]A = \pi R ^2[/tex]

and   k is central circular area which is mathematically represented as

     [tex]k = \pi [\frac{R}{2} ]^2[/tex]

So  

     [tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]

     [tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]

     [tex]I_d = 4 * I[/tex]

substituting values

     [tex]I_d = 4 * 9.2[/tex]

     [tex]I_d = 36.8 \ A[/tex]

Answer: The density in kilograms per cubic meter is 1025

Explanation:

Density is defined as mass contained per unit volume.

Given : Density of sea water = [tex]1.025g/cm^3[/tex]

Conversion : [tex]1.025g/cm^3=?kg/m^3[/tex]

As 1 g = 0.001 kg

Thus 1.025 g =[tex]\frac{0.001}{1}\times 1.025=0.001025kg[/tex]

Also [tex]1cm^3=10^{-6}m^3[/tex]

Thus  [tex]1.025g/cm^3=\frac{0.001025}{10^{-6}kg/m^3}=1025kg/m^3[/tex]

Thus density in kilograms per cubic meter is 1025

The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object

Answer:

Explanation:

The index of refraction of the liquid can be computed from ...

  [tex]n_i\sin{(\theta_t)}=n_t[/tex]

where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.

For the given incidence angles, the computed indices of refraction are ...

  A: n = 1.52sin(52.0°) = 1.198

  B: n = 1.52sin(44.3°) = 1.062

  C: n = 1.52sin(36.3°) = 0.900

Answer:

The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V

Explanation:

The equation for electric power is

power P = IV

also,  I = V/R,

substituting into the equation, we have

[tex]P = \frac{V^{2} }{R}[/tex]

[tex]R = \frac{V^{2} }{P}[/tex]

a)  [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω

b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω

c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω

d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω

from the calculations, one can see that the lightbulb with te greates resistance is

C. 4 W, 4.5 V

Explanation:

The Earth has a solid inner core

Answer:

The final velocity of the two players is 0.69 m/s in the direction of the opposing player.

Explanation:

Since the players are moving in opposite directions, from the principle of conservation of linear momentum;

[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v

Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.

[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;

103 × 2.0 - 110 × 3.2 = (103 + 110)v

206 - 352 = 213 v

-146 = 213 v

v = [tex]\frac{-146}{213}[/tex]

v = -0.69 m/s

The final velocity of the two players is 0.69 m/s in the direction of the opposing player.

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0  m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]

So, the plane slide a distance of 229.5 m.  

Answer:

1.363×10^15 seconds

Explanation:

The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.

To calculate the time taken from Kepler's third Law :

T^2 = ( 4π^2/GMe ) r^3

Where Me is the mass of the earth

r is the average distance travel

G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2

π = 3.14

Me = mass of earth = 5.972×10^24kg

r =( r minimum + r maximum)/2 ......1

rmin = 6590km

rmax = 385000km

From equation 1

r = (6590+385000)/2

r = 391590/2

r = 195795km

From T^2 = ( 4π^2/GMe ) r^3

T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3

= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15

= 39.4384/ 3.983×10^14 ) × 7.5059×10^15

= (9.901×10^14) × 7.5059×10^15

T^2 = 7.4321× 10^30

T =√7.4321× 10^30

T = 2.726×10^15 seconds

The time for one way trip from Earth to the moon is :

∆T = T/2

= 2.726×10^15 /2

= 1.363×10^15 secs

Answer:

8.7 m/s^2

82.15°

Explanation:

Given:-

- The initial speed of the car, vi = 25 m/s

- The radius of track, r = 129 m

- Car makes a circular " quarter turn "

- The constant tangential acceleration, at = 1.2 m/s^2

Solution:-

- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:

                          [tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]

- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:

                           at = r*α

                           α = ( 1.2 / 129 )

                           α = 0.00930 rad/s^2

- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).

- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:

                            [tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]

- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:

                            [tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]

- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:

                            [tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]

- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:

                             [tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]

- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:

                          [tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex] 

Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .

Answer:

Explanation:

Given;

Resistance of the coil, R = 20 W

Inductance of the coil, L = 0.35 H

Frequency of the alternating current, F = 25 cycle/s

Reactance of the coil is calculated as;

[tex]X_L=[/tex] 2πFL

Substitute in the given values and calculate the reactance [tex](X_L)[/tex]

[tex]X_L =[/tex] 2π(25)(0.35)

[tex]X_L[/tex] = 55 W

Impedance of the coil is calculated as;

[tex]Z = \sqrt{R^2 + X_L^2} \\\\Z = \sqrt{20^2 + 55^2} \\\\Z = 59 \ W[/tex]

Therefore, the reactance of the coil is 55 W and Impedance of the coil is 59 W

Answer:

7.74 x 10⁶ Joules

Explanation:

recall that "Watts" is the SI unit used for "energy per unit time"

Hence "Watts" may also be expressed as Joules / Second (or J/s)

We are given that the washer is rated at 350W (i.e. 350 Joules / s) and the dryer is rated at 1800W (i.e. 1800 Joules / s).

We are also given that the appliances are each run for 1 hour

1 hour = 60 min = (60 x 60) seconds = 3600 seconds

Hence the total energy used,

= Energy used by Washer in 1 hour + Energy used by dryer in 1 hour

= (350 J/s x 3600 s)  + (1800 J/s x 3600 s)

= 3600 ( 350 + 1800)

= 3600 (2150)

= 7,740,000 Joules

= 7.74 x 10⁶ Joules

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

[tex]\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})[/tex]

Where:

[tex]\omega_{o}[/tex], [tex]\omega[/tex] - Initial and final angular velocities, measured in radians per second.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

[tex]\theta_{o}[/tex], [tex]\theta[/tex] - Initial and final angular position, measured in radians.

Then,

[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}[/tex]

Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\theta-\theta_{o} = 4.9\,rad[/tex], the angular acceleration is:

[tex]\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}[/tex]

[tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex]

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

[tex]\omega = \omega_{o} + \alpha \cdot t[/tex]

Where [tex]t[/tex] is the time measured in seconds.

The time is cleared and obtain after replacing every value:

[tex]t = \frac{\omega-\omega_{o}}{\alpha}[/tex]

If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex],  [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex], the required time is:

[tex]t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }[/tex]

[tex]t = 14\,s[/tex]

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

[tex]\bar \alpha = \frac{\omega-\omega_{o}}{t}[/tex]

If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex],  [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]t = 14\,s[/tex], the average angular acceleration is:

[tex]\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}[/tex]

[tex]\bar \alpha = 0.05\,\frac{rad}{s^{2}}[/tex]

The average angular acceleration is 0.05 radians per square second.