Answer:
, 
, 
Explanation:
The cube root of the complex number can determined by the following De Moivre's Formula:
![z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D%20%3D%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D%5Ccdot%20%5Cleft%5B%5Ccos%5Cleft%28%5Cfrac%7Bx%20%2B%202%5Cpi%5Ccdot%20k%7D%7Bn%7D%20%5Cright%29%20%2B%20i%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7Bx%2B2%5Cpi%5Ccdot%20k%7D%7Bn%7D%20%5Cright%29%5Cright%5D)
Where angles are measured in radians and k represents an integer between 
and 
.
The magnitude of the complex number is 
and the equivalent angular value is 
. The set of cubic roots are, respectively:
k = 0
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B1.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B1.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)
k = 1
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B3.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B3.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)
k = 2
![z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]](https://tex.z-dn.net/?f=z%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%203%5Ccdot%20%5Cleft%5B%5Ccos%20%5Cleft%28%5Cfrac%7B5.817%5Cpi%7D%7B3%7D%20%5Cright%29%2Bi%5Ccdot%20%5Csin%5Cleft%28%5Cfrac%7B5.817%5Cpi%7D%7B3%7D%20%5Cright%29%5D)
Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine 
= 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine 
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
YES. 6√10
Explanation:
apply E=1/2mv^2 and find the velocity of carriage which is 6√10 ms-1
as our velocity is higher than that of cart. we can catch the carriage
Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
<u>v₂ = 306.12 m/s</u>
i’m not 100 percent sure but I need points to ask questions good luck th
