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:
   ....(1)
Here A represents area of the circular loop.
Area of circular loop, A = πr²
Hence, the equation (1) becomes:

Substitute the suitable values in the above equation.

 = 1.03 T m²
 
        
             
        
        
        
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.
 
        
             
        
        
        
Answer:
150 steps south
Explanation:
250 north 250 back to start then continue south for remainder of 400 steps. 150 south
 
        
             
        
        
        
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 = 
           Fx = 
  ( -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 = 
           F = 5.43 10¹¹  √2
           F = 7.68 10¹¹ N
in angle is
            θ = 45º