Answer:
A) electric field strength between the plates;E = 2 x 10^(6) N/C
B) exit velocity;v = 8.39 x 10^(7) m/s
Explanation:
We are given;
Potential difference; V = 20 kV = 20000 V
Distance between the 2 parallel plates; d = 1cm = 0.01 m
A) The electric field strength will be gotten from;
E = V/d
E = 20000/0.01
E = 2000000
E = 2 x 10^(6) N/C
B) For exit speed, we'll use the formula for Kinetic energy; KE = (1/2)mv²
KE is also expressed as; V•q_e
Thus,
(1/2)mv² = V•q_e
Where;
V is potential difference = 20000 V
Q_e is charge of electron which has a constant value of; (1.6 x 10^(-19))C
m is mass of electron with a constant value of (9.1 x 10^(-31)) kg
v is the velocity
Thus, making v the subject, we have;
v = √((2V•q_e)/m)
v = √((2 x 20000•(1.6 x 10^(-19)))/(9.1 x 10^(-31)))
v = 83862786 m/s or
v = 8.39 x 10^(7) m/s
The answer is A. 5 n up deals with balanced and unbalanced forces.
Answer: According to Newton's third law of motion, whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. These two forces are called action and reaction forces.
Explanation:
No cap i know
By definition we have to:
Where,
Pabs: absolute pressure
Patm: atmospheric pressure
Pg: gage pressure
The atmospheric pressure is constant and its value is:
Then, by clearing gage pressure we have:
Substituting values we have:
Answer:
If the absolute pressure of a gas is 550.280 kPa, its gage pressure is:
D. 448.955 kPa
