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ololo11 [35]
3 years ago
15

A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

traight line to a new position at a distance R away from its starting position. The final location of q is at a distance r2 from Q. 1)What is the change in the potential energy of charge q during this process
Physics
1 answer:
topjm [15]3 years ago
7 0

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

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Two people travel from Columbus to Cleveland, one by car at an average velocity of 90km/h, and one by plane at an average veloci
Flura [38]

Answer:

Explanation:

Yes , their displacement may be equal .

Suppose the displacement is AB where A is starting point and B is end point .

The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .

7 0
3 years ago
An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta
icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

v=\sqrt{\frac{T}{\mu} }

where T= 200 N is tension in the string , \mu=1.0 g/m is the linear mass density

v=\sqrt{\frac{200}{1\times 10^{-3} }

v=447.2 m/s

Wavelength of the wave in the string is

\lambda =2L=2\times 0.8=1.6 m

The frequency is

f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m

8 0
3 years ago
A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together a
slavikrds [6]

Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

5 0
3 years ago
The height of the mercury column in a barometer is 756 mm Hg on
notsponge [240]

Answer:

    p = 1.0076 10⁵ Pa

Explanation:

Atmospheric pressure is given by the relation

         P = rho g h

In this case they indicate that the height of the column of mercury is h = 756 mm Hg

let's reduce the height to the SI system

          h = 756 mm (1m / 1000 mm)

          h = 0.756 m

let's calculate

        P = 13600 9.8 0.756

        p = 1.0076 10⁵ Pa

7 0
2 years ago
How much work is required to turn an electric dipole 180° in a uniform electric field of magnitude E = 46.0 N/C if the dipole mo
chubhunter [2.5K]

Answer:

W=1.22*10^{-23}J

Explanation:

Torque and energy of an electric dipole in an electric field we find:

W=U(\alpha_{o}+\pi  )-U(\alpha_{o} )=-pE(cos(\alpha_{o}+\pi )-cos(\alpha_{o} ))\\W=2pECos\alpha_{o}\\ W=2(3.02*10^{-25}C.m )(46.oN/C)Cos64\\W=1.22*10^{-23}J

3 0
3 years ago
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