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Vadim26 [7]
3 years ago
7

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

f 0.3m apart in a vacuum.​
Physics
1 answer:
adell [148]3 years ago
7 0

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

        F =k \frac{q_1q_2}{r^2}

In this case they indicate that the load is of equal magnitude

       q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

       q = \sqrt{\frac{F \ r^2}{k} }

we calculate

        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

        q = \sqrt{7 \ 10^{-12} }Ra 7 10-12

        q = 2.65 10⁻⁶ C

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TP hydrolysis distinct from any incorporation into the chain.

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If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m
charle [14.2K]

Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

I=envA   ....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

I=Ne      ....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

eN = envA

N=nvA

But area of wire, A=\pi \frac{d^{2} }{4}

Here d is diameter of wire.

So, N = nv\pi \frac{d^{2} }{4}

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}

N = 2.42 x 10¹⁹ s⁻¹  

8 0
3 years ago
2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters
Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0
2 years ago
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