Answer:
Though you have not gave the choices, I do believe it is “testing”
Explanation:
B. truthfully it's the only one that makes sense
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:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
Answer:
120s^-1
Explanation:
v=12v
I=10A
and since rate is with time, therefore rate=energy/time.
H=IV
10×12=120/s
therefore the rate is 120s^-1
