Answer:
The magnitude of the electric field in the second case will be 
of the electric field in the first case
Explanation:
The electric field at a distance 
< R from the center of the sphere is given by
the formula 
= 
where R is the radius of the sphere, and, Q is the charge uniformly distributed on the sphere.
It is given that the same charge Q is distributed uniformly throughout a sphere of radius 2R and we have to find the electric field at same distance from the center.
In the second case the electric field will be given by
= 
Therefore the magnitude of the electric field in the second case will be of the electric field in the first case.
Answer:
53.36 m.
Explanation:
Initial velocity of the toolbox u = 0
acceleration due to gravity g = 9.8 m s⁻²
time t = 3.3 s
height of roof h = ?
h = ut + 1/2 g t²
= 0 + .5 x 9.8 x 3.3²
= 53.36 m.
Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - 
= 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
The answer is uniformitarianism.
