0.119cm/s is the radius of the balloon increasing when the diameter is 20 cm.
<h3>How big is a circle's radius?</h3>
The radius of a circle is the distance a circle's center from any point along its circumference. Usually, "R" or "r" is used to indicate it.
A circle's diameter cuts through the center and extends from edge to edge, in contrast to a circle's radius, which extends from center to edge. Essentially, a circle is divided in half by its diameter.
dv/dt = 150cm³/s
d = 2r = 20cm
r = 10cm
find dr/dt
Given that the volume of a sphere is calculated using
v = 4/3πr³
Consider both sides of a derivative
d/dt(v) = d/dt( 4/3πr³)
dv/dt = 4/3π(3r²)dr/dt = 4πr²dr/dt
Hence,
dr/dt = 1/4πr².dv/dt
dr/dt = 1/4π×(10)²×150
dr/dt = 1/4π×100×150
dr/dt = 0.119cm/s.
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Answer:
4.763 × 10⁶ N/C
Explanation:
Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.
Resolving E₂ into horizontal and vertical components, we have
E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.
Summing the horizontal components we have
E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²
= -k/r²(q₁ + q₂cos60)
= -k/r²(4 μC + (-6.0 μC)(1/2))
= -k/r²(4 μC - 3.0 μC)
= -k/r²(1 μC)
= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²
= -9 × 10⁵ N/C
Summing the vertical components, we have
E₄ = 0 + (-E₂sin60)
= -E₂sin60
= -kq₂sin60/r²
= -k(-6.0 μC)(0.8660)/(0.10 m)²
= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²
= 46.77 × 10⁵ N/C
The magnitude of the resultant electric field, E is thus
E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴
= 476.28 × 10⁴ N/C
= 4.7628 × 10⁶ N/C
≅ 4.763 × 10⁶ N/C
Answer:
Earths shadow covering the moon would create a lunar eclipse.
Explanation:
because i just know
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)
