Answer:
6.07 N
Explanation:
Given that,
Force, F = 35 N
It makes 10 degree angle with the positive x-axis.
We need to find the magnitude of the vertical component of the force. It can be given by :
So, the magnitude of the vertical component of the force is 6.07 N.
The way to do this is very easy so do 4125 x 2 = ? then the ? will be times by 2 again after the answer to both of those is your answer!!!
Answer:
d. According to the previous theorem:
(kh^2)/2-mgh-(m(2v)^2)/2=0,
126h^2-2.577h-8.236=0,
h=0.266 m.
Explanation:
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
