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natulia [17]
3 years ago
5

Light waves travel in straight lines in all directions from the source of the light, unless

Physics
1 answer:
Ainat [17]3 years ago
6 0

A.  the medium through which the light travels changes.

Explanation:

Light waves will continue to travel in a straight line in all directions from their source unless the medium through which the light travels changes.

A change in medium causes light to exhibit different properties. Also, when light hits an obstacle, they can be diffracted.

  • The way light travels on crossing a boundary differs.
  • At the boundary between two medium, light can either be reflected back or refracted when they cross the medium
  • This will cause the light rays to bend towards or away from the normal depending on the properties of the medium.

Learn more:

Refraction brainly.com/question/12370040

#learnwithBrainly

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A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1 ° from the normal to the plane of the l
umka2103 [35]

Answer:

Magnetic flux through the loop is 1.03 T m²

Explanation:

Given:

Magnetic field, B = 4.35 T

Radius of the circular loop, r = 0.280 m

Angle between circular loop and magnetic field, θ = 15.1⁰

Magnetic flux is determine by the relation:

\Phi = BA\cos \theta   ....(1)

Here A represents area of the circular loop.

Area of circular loop, A = πr²

Hence, the equation (1) becomes:

\Phi=B\pi r^{2} \cos \theta

Substitute the suitable values in the above equation.

\Phi=4.35\times\pi (0.28)^{2} \cos 15.1

\Phi = 1.03 T m²

8 0
3 years ago
PLEASE HELP!!
leva [86]
I think the answer is B
8 0
2 years ago
Un atleta de 70 kg de masa que ha efectuado un salto de altura cae una vez que ha
Allushta [10]

Answer:

a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards

Explanation:

The forces on the athlete are

a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,

therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards

b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles

c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.

3 0
3 years ago
You walk exactly 250 steps North, turn around, and then walk exactly 400 steps South. How far are you from your starting
german

Answer:

150 steps south

Explanation:

250 north 250 back to start then continue south for remainder of 400 steps. 150 south

3 0
3 years ago
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
3 years ago
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