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3 years ago

Find equation tangent to a circle at given point

1 answer:

7 0

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. 
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.

Let's find the equation of the line parallel to the radius.

The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product),

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent

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In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

4vir4ik [10]

Answer:

The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is

$k = 0.903$

Explanation:

From the question we are told that

The time constant $\tau = 3$

The potential across the capacitor can be mathematically represented as

$V = V_o (1 - e^{- \tau})$

Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

$V^2 = (0.950213 V_o )^2$

$V^2 = 0.903 V_o ^2$

Hence the fraction of the energy stored in an initially uncharged capacitor is

$k = 0.903$

4 0

3 years ago

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

Lana71 [14]

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$

7 0

3 years ago

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0

3 years ago

which of the following cannot be determined by looking at the phase diagram? A. melting point, B. boiling point, C. subilmation

bulgar [2K]

I am pretty sure that the following whihc cannot be determined by looking at the phase diagram is definitely D. system pressure. I consider this one to be correct because only this point is not included into phase diagram and can't be determined itself. Hope it will help! Regards!

7 0

3 years ago

What is the angular displacement of a minute hand of a clock after 3 minutes?

Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0

2 years ago