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Nostrana [21]
2 years ago
9

An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the gr

ound. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of q on the plane and a point charge of -q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 53.5 J. Find the magnitude of the charges.
Physics
1 answer:
amm18122 years ago
5 0

Answer:

q=3.5*10^-4

Explanation:

<u>concept:</u>

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

<u>The tension force:</u>

<u><em>when the plane is uncharged </em></u>

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

<u><em>when the plane is charged </em></u>

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

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2 years ago
A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners
guapka [62]

Answer:

a)   x₀ = - 2 m  , b)     y = 4.47 m

Explanation:

A wave travels in the middle with constant speed, let's use the equation of uniform motion

    v = d / t

    t = d / v

The distance to the first listeners, see attached

    d₁ = x₀-x

     t = (x₀ +7) / v

The distance to the second listener

    d₂ = x - x₀

     t = (+ 3- x₀) / v

As the wave arrives at the same time, we can equal the two equations

     (x₀ +7) / v = (3 -x₀) / v

      x₀ + 7 = 3 - x₀

      2 x₀ = 3 - 7

      x₀ = -4/2

      x₀ = - 2 m

b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s

          t = 5/340

          t = 0.0147 s

Let's look for the distance the wave travels for the listener axis and

       v = d₃ / t

       d₃ = v.t

       d₃ = 340 * 0.0147

       d₃ = 5 m

For the distance component we use the Pythagorean triangle

      d₃² = x₀² + y²

      y² = d₃² - x₀²

     y = √ (d₃² -4)

      y = √ (5² -4)

     y = 4.47 m

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What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed
tekilochka [14]

Answer:

charge will be equal to 2.03\times 10^{-5}C  

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity g=9.8m/sec^2

Electric field E = 700 N/C

Electric force will be equal to F=qE, here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C

So charge will be equal to 2.03\times 10^{-5}C

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