The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
Answer:
It represents the change in charge Q from time t = a to t = b
Explanation:
As given in the question the current is defined as the derivative of charge.
I(t) = dQ(t)/dt ..... (i)
But if we take the inegral of the equation (i) for the time interval from t=a to
t =b we get
Q =∫_a^b▒〖I(t) 〗 dt
which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.
The temperature scale which starts at absolute zero is the Kelvin scale. The correct option in respect to the given question is the last option. William Thompson was the British scientist and inventor that invented the Kelvin scale. William Thompson was also popularly known as Lord Kelvin.His discovery of the Kelvin scale is considered one among the three best scales in use for measuring temperatures.Each measuring unit of this scale is never called a degree but a Kelvin. This specialized scale gives the option of measuring temperature in both centigrade and Fahrenheit.
Answer:
The gauge pressure of air is 110 kpa
Explanation:
Atmospheric pressure, 
= 101 Kpa
where;
ρw is the density of water = 1000 kg/m³
ρo is the density of oil = 800 kg/m³
ρHg is the density of mercury = 13,600 kg/m³
g is acceleration due to gravity = 9.8 m/s²
Therefore, the gauge pressure of air is 110 kpa
Answer:
<h3>idl low presure but de 2.518 has a 6 divide 8 equal di. ko alam?</h3>
