Answer:
### What are light waves?

**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**).

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

A large portion of the light in the **universe****travels** 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 **gamma****radiation** 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:
The wave that best represents it is c.

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

**Answer:**

The acceleration expressed in the new units is

**Explanation:**

To convert from to 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:

So the acceleration expressed in the new units is .

**Answer:**

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

**Explanation:**

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

vf²=v₀²+2*g*h Formula (1)

Where:

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**

**Data**

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

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

vf²=v₀²+2*g*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**

**Data**

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)

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, ** = 10 m/s*

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

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

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;

The **horizontal velocity** doesn't change.

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

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

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

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**Answer:**

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

**Explanation:**

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

² = vo² + 2 a y

v₀ = 0

a = g / 7

² = 2g / 7 y

= √ (2 9.8 4.2 / 7)

= 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

= 3.43 m / s

**Answer:**

The current flows in the second wire is

**Explanation:**

**Given that,**

Upward current = 24 A

Force per unit length

Distance = 7.0 cm

**We need to calculate the current in second wire**

**Using formula of magnetic force**

Where,

=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

**Hence, The current flows in the second wire is **

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

Where,

F = Force

d = Distance

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

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

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

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;

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

From the **trig identity**;

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

*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 * F*₁,

* F*₁ +

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 → *F* ≈ **5.5 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.