Answer: a = 1.32 * 10^18m/s² due north
Explanation: The magnitude of the force required to move the electron is given as
F = ma
The force exerted on the charge by the electric field of intensity (E) is given by
F = Eq
Thus
Eq = ma
a = E * q/ m
Where a = acceleration of charge
E = strength of electric field = 7400N/c
q = magnitude of electronic charge = 1.609 * 10^-6c
m = mass of an electronic charge = 9.109 * 10^-31kg
a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31
a = 11906.6 * 10^-16 / 9.019 * 10^-31
a = 1.19 * 10^-12 / 9.019 * 10^-31
a = 0.132 * 10^19
a = 1.32 * 10^18m/s²
As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)
Answer:
<u>Drag force</u> is the frictional force needed to slow an object in motion
Explanation:
Answer:
276.135 J
Explanation:
Given that:
mass of Fe = 30.0 g
initial temperature = 24.5°C
final temperature = 45.0°C
specific heat of Fe = 0.449 J/g°C
We can determine the thermal energy added by using the formula;
Q = mcΔT
Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C
Q = 276.135 J
Answer:
new atmospheric pressure is 0.9838 × 
Pa
Explanation:
given data
height = 21.6 mm = 0.0216 m
Normal atmospheric pressure = 1.013 ✕ 10^5 Pa
density of mercury = 13.6 g/cm³
to find out
atmospheric pressure
solution
we find first height of mercury when normal pressure that is
pressure p = ρ×g×h
put here value
1.013 × = 13.6 × 10³ × 9.81 × h
h = 0.759 m
so change in height Δh = 0.759 - 0.0216
new height H = 0.7374 m
so new pressure = ρ×g×H
put here value
new pressure = 13.6 × 10³ × 9.81 × 0.7374
atmospheric pressure = 98380.9584
so new atmospheric pressure is 0.9838 × Pa
