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Whitepunk [10]
3 years ago
12

Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c

ertain point and the force on charge C is zero. How far from charge A is charge C?
Physics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

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What is a real-life example of destructive interference of light?
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Destructive Interference. An example of destructive interference is noise canceling headphones. Noise canceling headphones work by using a microphone to pick up the frequencies of incoming waves. The headphone then sends out a wave that is the exact opposite, canceling out the sound.
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Who discovered the basic laws of planetary orbits?.
Phoenix [80]

Answer:

kepler

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6 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Nutka1998 [239]

Answer:

6.99535\times 10^{-6}\ V/m

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity of Electric radiation is given by

I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2

Intensity of Electric radiation is given by

I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m

The amplitude of the electric field vector is 6.99535\times 10^{-6}\ V/m

6 0
4 years ago
An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8
Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of both momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

m_2v_2-m_2v'=m_1v'-m_1v_1

Factoring m2 and m1:

m_2(v_2-v')=m_1(v'-v_1)

Solving:

\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}

Substituting:

\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}

\displaystyle m_2=\frac{-2.8896}{-16.8}

\displaystyle m_2=0.172\ kg

The mass of the apple is 0.172 kg (172 g)

3 0
3 years ago
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