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Troyanec [42]
2 years ago
5

The electric force between two charged vinyl balloons is 3.0 × 10-3 newtons. If the distance between them is 6.0 × 10-2 meters a

nd the charge of one vinyl balloon is 3.3 × 10-8 coulombs, what is the charge on the other vinyl balloon? (k = 9.0 × 109 newton·meter2/coulombs2) 3.3 × 10-8 coulombs 3.6 × 10-8 coulombs 4.8 × 10-8 coulombs 6.0 × 10-8 coulombs 3.3 × 10-3 coulombs
Physics
1 answer:
irakobra [83]2 years ago
8 0
The electric force between two objects can be understood by looking at Coulomb's Law. The law states that the electric force is directly proportional to the charges of the 2 objects, and inversely proportional as the square of the distance. This is shown below:

F = (k * Q1 * Q2) / d^2

Where: k = proportionality constant; Q1 = charge of object 1; Q2 = charge of object 2; d = distance between objects.

Given:

F = <span>3.0 × 10^-3 N
Q1 = </span>3.3 × 10^-8 coulombs<span>
Q2 = ?
d = </span><span>6.0 × 10^-2 m
</span><span>k = 9.0 × 10^9

Substituting:

3 x 10^-3 = (9 x 10^9 * 3.3 x 10^-8 * Q2) / (6 x 10^-2)^2
[(</span>3 x 10^-3) * (6 x 10^-2)^2] / [(<span>9 x 10^9) * (3.3 x 10^-8)</span>] = Q2
Q2 = 3.636 x 10^-8 coulombs
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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
2 years ago
charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³  N / C .

3 0
2 years ago
Part A
irina [24]

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

       m_{b} * v_{b} -M_{c}*V_{c}  = m_{b} * 18 m/s + (-M_{c}* 20 m/s)  (2)

  • The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
  • So, we can write again (3) as follows:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (4)

  • Replacing (2) and (4) in (1), we get:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (5)

  • Simplifying, and rearranging, we can solve for v'b, as follows:
  • v'_{b} = -18 m/s (6), which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had  inverted its speed after the collision.
3 0
1 year ago
What potential difference vwc between the wire and the cylinder produces an electric field of 2.00Ã104 volts per meter at a dist
vovikov84 [41]
<span>The potential difference Vwc between the wire and the cylinder produces an electric field of 2.00Ã104 volts per meter at a distance of 1.20 centimeters from the axis of the wire is 1160 V</span>

In case of a long wire, electric field E is inversely proportional to r

Therefore E= x/r or x = E*r

Now, 

5 0
2 years ago
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