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
By definition, Ampere is a unit of current which is a measure of the amount of charge passing through a point in a circuit per unit of time, with an equivalent charge of 1.602 x 10^(-19) Coulomb per electron. To determine the number of electrons passing through the heater, we use the definition of the current. We calculate as follows:
13.5 A = 13.5 C per second
Charge = 13.5 C/s (10 min) ( 60 s / 1 min)
Charge = 8100 C
Number of electrons = 8100 C / 1.602 x 10^(-19) C per electron
Number of electrons = 5.1 x 10^22 electrons
Therefore, there are 5.1 x10^22 electrons that assed through the heater for 10 minutes.
Answer:
plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)
Explanation:
<span>The Gravitational Force of an object is a measure of the amount of matter it contains. on the other hand __Matter__ is a measure of the gravitational force on an object. I hope it helps :)</span>
Answer:
<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>
Explanation:
<u>Operations With Vectors</u>
Given two vectors in rectangular components:
![\vec a=(ax,ay)\ ,\ \vec b=(bx,by)](https://tex.z-dn.net/?f=%5Cvec%20a%3D%28ax%2Cay%29%5C%20%2C%5C%20%20%5Cvec%20b%3D%28bx%2Cby%29)
The sum of the vectors is:
![\vec a+\vec b=(ax+bx,ay+by)](https://tex.z-dn.net/?f=%5Cvec%20a%2B%5Cvec%20b%3D%28ax%2Bbx%2Cay%2Bby%29)
The difference between the vectors is:
![\vec a-\vec b=(ax-bx,ay-by)](https://tex.z-dn.net/?f=%5Cvec%20a-%5Cvec%20b%3D%28ax-bx%2Cay-by%29)
The magnitude of
is:
![|\vec a|=\sqrt{ax^2+ay^2}](https://tex.z-dn.net/?f=%7C%5Cvec%20a%7C%3D%5Csqrt%7Bax%5E2%2Bay%5E2%7D)
The angle
makes with the horizontal positive direction is:
![\displaystyle \tan\theta=\frac{ay}{ax}\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctan%5Ctheta%3D%5Cfrac%7Bay%7D%7Bax%7D%5C%5C)
The question provides the vectors:
![\vec a=(2,6)](https://tex.z-dn.net/?f=%5Cvec%20a%3D%282%2C6%29)
![\vec b=(2,22)](https://tex.z-dn.net/?f=%5Cvec%20b%3D%282%2C22%29)
![\vec d=\vec a-\vec b](https://tex.z-dn.net/?f=%5Cvec%20d%3D%5Cvec%20a-%5Cvec%20b)
Calculate:
![\vec d=(2,6)-(2,22)=(0,-16)](https://tex.z-dn.net/?f=%5Cvec%20d%3D%282%2C6%29-%282%2C22%29%3D%280%2C-16%29)
The magnitude of
is:
![|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16](https://tex.z-dn.net/?f=%7C%5Cvec%20d%7C%3D%5Csqrt%7B0%5E2%2B%28-16%29%5E2%7D%3D%5Csqrt%7B0%2B256%7D%3D16)
The angle is calculated by:
![\displaystyle \tan\theta=\frac{-16}{0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctan%5Ctheta%3D%5Cfrac%7B-16%7D%7B0%7D)
The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.
The magnitude of vector d is 16 and the angle with the x-axis is 270°