Answer:
a
The value at a point inside is Zero
b
The electric field is 
Explanation:
We know from the problem that
The charge magnitude is 
The radius of the spherical ball is 
According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero
On the outside, the electric field around the spherical ball is mathematically expressed as
Here a denotes a point outside the spherical ball with its value of 
and k represents Coulomb's constant, valued at
=> 
=>
Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
Answer:
a = 18.28 ft/s²
Explanation:
the values provided are:
duration of force application, t= 10 s
Work done = 10 Btu
mass of the object = 15 lb
acceleration, a =? ft/s²
1 Btu = 778.15 ft.lbf
thus, 10 Btu = 7781.5 ft.lbf
m = 0.466 slug
So,
the work is equivalent to the change in kinetic energy
The acceleration of the object is therefore
a = 18.28 ft/s²
the constant acceleration of the object is calculated to be 18.28 ft/s²
Respuesta:
Explicación:
Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!
