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Savatey [412]
3 years ago
8

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

charge on B one-fifth the charge on A. How far apart would the two spheres then have had to be for A to have had the same deflection that it had before
Physics
1 answer:
sdas [7]3 years ago
6 0

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

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Answer

Mass m = 78 kg

Vertical height in each stage h = 11 m

(a).

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Final speed v = 1.1 m / s

v^2=u^2 + 2 as

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a = 0.055 m/s²

Work done

W_a= m g h + \dfrac{1}{2}mv^2

W_a= 78\times 9.8 \times 11 + \dfrac{1}{2} 78 \times 1.1^2

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W_a = 8455.59 J

(b).Work done

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W_b = 78× 9.8× 11

W_b= 8408.4 J

c)

Work done

W= m g h + \dfrac{1}{2}m(v_f-v_i)^2

Where V = final speed

               = 0

            v = 1.1 m / s

for deceleration a = -0.055 m/s²

now,

F_L = 56 (-0.055+9.8) = 545.72 N

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W_c = 6003.25 J

   

7 0
3 years ago
the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel
Dmitriy789 [7]

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K

Therefore, the kinetic energy has quadrupled:

K' = 4(25)=100 J

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K

Therefore, the kinetic energy has increased by a factor of 9:

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3 years ago
Greg is in a bike race. At mile marker four (out of ten), his speed was measured at 13.5 mph. Which best describes the measured
Ipatiy [6.2K]

his speed/ velocity is 13.5 miles per hour


5 0
3 years ago
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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f
dsp73

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

I=\frac{\mu_0 I}{2 \pi r}

where \mu_0 is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

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We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

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Answer:

#see solution for details

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