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lutik1710 [3]
3 years ago
5

Find the acute angles between the curves at their points of intersection. (The angle between two curves is the angle between the

ir tangent lines at the point of intersection. Give your answers in degrees, rounding to one decimal place. Enter your answers as a comma-separated list.) y = 8x2, y = 8x3
Physics
1 answer:
ladessa [460]3 years ago
3 0

Answer:

1.190390345 degrees.

Explanation:

As stated in the problem that the angle between the two curves is also the angle between slopes of the two cures at the point of intersection, which is 1 (when we set two equation equal and solve for x).

We know, if you don't  you could verify it for yourself and it will be a nice mathematical exercise for you, that when there are two lines with slopes m_1 and m_2 then the angle between them and their two slopes has following relationship.

tan(\alpha ) = |\frac{m_2-m_1}{1+m_1m_2}|   where \alpha is angle between the two lines.

As it is clear that we can easily get the angle between the two curves if we know slopes at that point, which you will see in second is very straight forward to calculate.

Slopes are simply derivatives evaluated at the point of intersection and the two derivatives are D_1 16x and D_2 =24x^2, substituting x =1 we get, m_1 = 16 and m_2 = 24.

Now put m_1 and m_2 in this relationship, rearrange and solve for angle, it will come out to be

\alpha = 1.190390345 degrees.

that is our angle that we want to know between the two cures at the point of their intersection.

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Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr
Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

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Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

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Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

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6 0
4 years ago
What is the difference between VELOCITY and SPEED?​
valentina_108 [34]

Answer:

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Explanation:

5 0
4 years ago
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Explanation:

Just did the test myself. Hope this helps!

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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
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To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

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dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

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But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

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Arte-miy333 [17]

Answer:

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Explanation:

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