Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Answer:
See explanation
Explanation:
The question is incomplete because the images were not attached but I will try to help you as much as possible.
Constant acceleration implies that the velocity increases uniformly with time.
The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.
The image attached shows a velocity-time graph depicting constant acceleration.
Answer:
33,02 lb
Explanation:
g_m ≈ 1,62 m/s2
g ≈ 9,81 m/s2
m = 200 lb
m_m = m * g_m / m = 200 * 1,62 / 9,81 = 33,02 lb
