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posledela
3 years ago
5

Two point charges carrying charges q1 = +3 μC and q2 = −4 μC are fixed on the x − y plane at positions (x1 = 3.5 cm, y1 = 0.5 cm

) and (x2 = −2 cm, y2 = 1.5 cm) respectively.
a) Find the magnitude and direction of the electrostatic force on charge q2.
b) Where in the x − y plane can you place a third charge q3 = 4 μC, such that the net electrostatic force on charge q2 is zero?

Physics
1 answer:
Stells [14]3 years ago
3 0

Answer:

Explanation:

The concept of coulomb's law was applied to analyze the problem and other mathematical application as is shown in the attached file.

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Explanation:

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T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

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Projectile Motion: A hobbyist launches a projectile from ground level on a horizontal plain. It reaches a maximum height of 72.3
marin [14]

Answer:

The angle of launch from the horizontal direction is 20.99° .

Explanation:

Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.

The equations for projectile motion are :

H = ( u² sin²θ)/ 2g      ......(1)

Here H is maximum height of the projectile motion and g is acceleration due to gravity.

R = ( u² sin2θ)/g         .......(2)

Here R is the maximum horizontal displacement of the object.

Rearrange equation (1) in terms of u².

u² = (2gH)/sin²θ

Substitute this equation in equation (2).

R = (2gH sin2θ) / (sin²θ x g)

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So, above equation becomes,

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Substitute 111 m for R and 72.3 m for H in the above equation.

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3 years ago
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