I believe it’s (D. Any object)
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
6.0 is the answer. hope this helps ya
Answer:
the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².
Explanation:
Given;
applied force, F = 5000 N
cross sectional area of the steel wire, A = 0.2 m²
The tensile stress of the steel wire, is calculated as;
Therefore, the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².
This problem is pretty straight forward since we are given
the half time and we simply have to find for the rate constant. The equation
relating the two variable is:
t1/2 = ln(2) / k
where k is the rate constant, therefore:
k = ln(2) / t1/2
k = ln(2) / 14.7 h
<span>k = 0.047 / hour</span>
