To develop this problem use the concept of the sum given pressure in the tank. At the bottom of the tank the pressure of this will be given by atmospheric pressure, the pressure given by the oil and the pressure by the water, that is to say that mathematically the pressure would be
![P = P_{atm}+\rho_{oil}gt+\rho_{water}gh](https://tex.z-dn.net/?f=P%20%3D%20P_%7Batm%7D%2B%5Crho_%7Boil%7Dgt%2B%5Crho_%7Bwater%7Dgh)
<em>Note: Here the pressures are expressed in terms of density (
), gravity (g) and thickness (t) or height (h). If we rearrange this equation to find the oil thickness we will have to,</em>
<em>
</em>
![t = \frac{P-P_{atm}-\rho_{water}gh}{\rho_{oil}g}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BP-P_%7Batm%7D-%5Crho_%7Bwater%7Dgh%7D%7B%5Crho_%7Boil%7Dg%7D)
Our values are given as,
![h = 0.84m](https://tex.z-dn.net/?f=h%20%3D%200.84m)
![P= 113kPa](https://tex.z-dn.net/?f=P%3D%20113kPa)
![P_{atm} = 1.013*10^5Pa](https://tex.z-dn.net/?f=P_%7Batm%7D%20%3D%201.013%2A10%5E5Pa)
![g = 9.8m/s^2](https://tex.z-dn.net/?f=g%20%3D%209.8m%2Fs%5E2)
![\rho_{water} = 1000kg/m^3](https://tex.z-dn.net/?f=%5Crho_%7Bwater%7D%20%3D%201000kg%2Fm%5E3)
![\rho_{oil} = 920kg/m^3](https://tex.z-dn.net/?f=%5Crho_%7Boil%7D%20%3D%20920kg%2Fm%5E3)
Replacing we have that the thickness of the oil is:
![t = \frac{P-P_{atm}-\rho_{water}gh}{\rho_{oil}g}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BP-P_%7Batm%7D-%5Crho_%7Bwater%7Dgh%7D%7B%5Crho_%7Boil%7Dg%7D)
![t = \frac{113*10^3-1.1013*10^5-(1000)(9.8)(0.84)}{(920)(9.8)}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B113%2A10%5E3-1.1013%2A10%5E5-%281000%29%289.8%29%280.84%29%7D%7B%28920%29%289.8%29%7D)
![t = 0.5947m](https://tex.z-dn.net/?f=t%20%3D%200.5947m)
Therefore the thick of the oil is 0.5947m
-- With two resistors in parallel, the total effective resistance is
the reciprocal of (1/R₁ + 1/R₂).
1/R₁ + 1/R₂ = 1/15 + 1/40
= 8/120 + 3/120
= 11/120
So the total effective resistance is 120/11 = 10.9 ohms .
Current = (voltage) / (resistance)
= 12 / (120/11)
= (12 · 11) / 120
= 132/120 = 1.1 Amperes
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm
Answer: Multicellular organism
Explanation:
Multicellular organisms refer to living things that have more than a single cell as opposed to unicellular organisms such as bacteria. Humans as well as all animals and land plants fall under this classification.
Multicellular organisms can live longer because new cells can be produced when others die. They are also larger due to the presence of many different cells which then specialize in different roles to ensure the survival of the organism.
Answer:
20 m/s
Explanation:
The force experienced by a charged particle in an electric field is given by
![F=qE](https://tex.z-dn.net/?f=F%3DqE)
where, in this problem:
is the charge of the particle
E is the electric field
The electric field here has components:
![E_x=-2.5 N/C\\E_y=0\\E_z=0](https://tex.z-dn.net/?f=E_x%3D-2.5%20N%2FC%5C%5CE_y%3D0%5C%5CE_z%3D0)
So the components of the force experienced by the particle are:
![F_x=qE_x=(0.080)(-2.5)=-0.2 N\\F_y=0\\F_z=0](https://tex.z-dn.net/?f=F_x%3DqE_x%3D%280.080%29%28-2.5%29%3D-0.2%20N%5C%5CF_y%3D0%5C%5CF_z%3D0)
Now we can find the components of the acceleration experienced by the particle, using Newton's second law of motion:
![a=\frac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D)
where
m = 4.0 g = 0.004 kg is the mass of the particle
The 3 components of the acceleration are:
![a_x=\frac{F_x}{m}=\frac{-0.2}{0.004}=-50 m/s^2\\a_y=0\\a_z=0](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7BF_x%7D%7Bm%7D%3D%5Cfrac%7B-0.2%7D%7B0.004%7D%3D-50%20m%2Fs%5E2%5C%5Ca_y%3D0%5C%5Ca_z%3D0)
Now we can find the components of the velocity of the particle at time t using the suvat equation:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where:
are the initial components of the velocity
Therefore, at t = 2.0 s, we have:
![v_x=u_x+a_xt=80+(-50)(2.0)=-20 m/s\\v_y=u_y+a_yt=0+0=0\\v_z=u_z+a_zt=0+0=0](https://tex.z-dn.net/?f=v_x%3Du_x%2Ba_xt%3D80%2B%28-50%29%282.0%29%3D-20%20m%2Fs%5C%5Cv_y%3Du_y%2Ba_yt%3D0%2B0%3D0%5C%5Cv_z%3Du_z%2Ba_zt%3D0%2B0%3D0)
And so, the speed of the particle is the magnitude of the final velocity:
![v=\sqrt{v_x^2+v_y^2+v_z^2}=\sqrt{(-20)^2+0+0}=20 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%2Bv_z%5E2%7D%3D%5Csqrt%7B%28-20%29%5E2%2B0%2B0%7D%3D20%20m%2Fs)