Well most of the particles did pass through and a few were deflected. but i think the answer is A
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;

Therefore, the magnitude of the electric field is 0.1108 N/C
Answer:
Approximately 0.0898 W/m².
Explanation:
The intensity of light measures the power that the light delivers per unit area.
The source in this question delivers a constant power of
. If the source here is a point source, that
of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.
The surface area of a sphere of radius
is equal to
. For the imaginary 9.6-meter sphere here, the surface area will be:
.
That
power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:
.
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed
Vf = 27.5 m/s Final speed
W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2 -24.0 ^2)
160kj = 4680 x m
convert kilo joules to jeoules 160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg