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
Answer:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.
acceleration a=qE/m=
=
m/s
now we find the horizontal distance traveled by electrons hit the plates
horizontal distance
![X=u[2y/a]^{1/2}](https://tex.z-dn.net/?f=X%3Du%5B2y%2Fa%5D%5E%7B1%2F2%7D)
=![4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}](https://tex.z-dn.net/?f=4%2A10%5E6%5B2%2A2%2A10%5E%7B-2%7D%2F7%2A10%5E%7B14%7D%5D%5E%7B1%2F2%7D)
=
= 3 cm
Answer:
11760 joules
Explanation:
Given
Mass (m) = 75kg
Height (h) = 16m
Required
Determine the increment in potential energy (PE)
This is calculated as thus:
PE = mgh
Where g = 9.8m/s²
Substitute values for m, g and h.
P.E = 75 * 9.8 * 16
P.E = 11760 joules
