Which wave diagram BEST represents a dim red sunset on the right side to the light from an intense ultraviolet bug light on the left side?


Answer 1

Diagram-A satisfies. High amplitude (bright) and long wavelengths are present on the left (red). & The right side has a short wavelength and low amplitude (dim) (violet).

What are light waves?

Light comes from a source as waves. Each wave has an electric and a magnetic component. Light is hence sometimes referred to as electromagneticradiation.

A large portion of the light in the universetravels with wavelengths that are too short or too long for the human eye to detect, yet our brains interpret light waves by giving distinct colours to the various wavelengths.

The infrared, microwave, and radio spectrum bands have the longest wavelengths. The ultraviolet, x-ray, and gammaradiation have the shortest wavelengths in the electromagnetic spectrum.

Diagram A is therefore satisfactory. On the left, there are long wavelengths with high amplitude (bright) (red). & The right side is dark and has a short wavelength (violet).

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Answer 2
Answer: The wave that best represents it is c.

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A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.



The acceleration expressed in the new units is 114.048 Km/h^(2)


To convert from m/s^(2) to Km/h^(2) it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:



Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

114.048 Km/h^2

So the acceleration expressed in the new units is 114.048 Km/h^2.

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.



vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)


Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)


h: displacement in meters (m)  

v₀: initial speed in m/s

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Free fall of the stone


v₀ =  10 m/s

vf =  30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:


(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone


v₀x =  10 m/s

v₀y =  0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)

v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, v_y_0 = 10 m/s

final vertical velocity of the stone, v_y_f = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s

The horizontal velocity doesn't change.

the final horizontal velocity, v_x_f = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)



a)  W₁ = - 127 J, b) W₂ = 148.18 J, c)   v_(f)= 3.43 m/s  and d)  v_(f) = 3.43 m / s


The work is given the equation

         W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

        W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

       W-T = m a

       T = W -m a

       T = mg -mg/7

       T = mg 6/7

       T = 3.6 9.8 6/7

       T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

      W₁ = F d cos θ

      W₁ = 30.24 4.2 cos 180

      W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

      W₂ = (mg) d cos 0º

      W₂ = 3.6 9.8 4.2

      W₂ = 148.18 J

c) kinetic energy

      K = ½ m v²

Let's calculate speed with kinematics

    v_(f)² = vo² + 2 a y

    v₀ = 0

    a = g / 7

     v_(f)² = 2g / 7 y

      v_(f) = √ (2 9.8 4.2 / 7)

      v_(f)= 3.43 m/s

We calculate

     K = ½  3.6  3.43²

     K = 21.18 J

d) the speed of the block and we calculate it in the previous part

       v_(f) = 3.43 m / s

A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?



The current flows in the second wire is 1.3*10^(10)\ A


Given that,

Upward current = 24 A

Force per unit length(F)/(l) =88*10^(4)\ N/m

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force


(F)/(l)=(\mu I_(1)I_(2))/(2\pi r)


(F)/(l)=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

88*10^(4)=(4\pi*10^(-7)*24* I_(2))/(2\pi *7*10^(-2))


I_(2)=1.3*10^(10)\ A

Hence, The current flows in the second wire is 1.3*10^(10)\ A

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)


Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d


F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg


m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J

PART B) Using Newton's Second law we have that,

F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN

The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?


(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

  • F force at α
  • 5.7 N force at 50⁰
  • 6.2 N force at 44⁰
  • 6.7 N force at 43⁰

The net vertical force on the knot is calculated as follows;

F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\n\nF_y = F sin(\alpha)  -4.51\n\nFsin(\alpha) = 4.51

The net horizontal force on the knot is calculated as follows;

F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\n\nF_x = -Fcos(\alpha) + 3.22\n\nFcos(\alpha) = 3.22

From the trig identity;

sin^2 \theta + cos^ 2 \theta = 1\n\n

(Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\n\nF^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\n\nF^2(1) = 30.71\n\nF = √(30.71) \n\nF = 5.54 \ N

The angle α of the force F is calculated as follows;

Fsin(\alpha) = 4.51\n\nsin(\alpha) = (4.51)/(F) \n\nsin(\alpha ) = (4.51)/(5.54) \n\nsin(\alpha ) = 0.814\n\n\alpha = sin^(-1)(0.814)\n\n\alpha = 54.5 \ ^0

Find the image uploaded for the complete question.

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The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N²   →   F5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399   →   α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.