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AlladinOne [14]
3 years ago
12

Solve for X, please help!!

Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

x=12

Explanation:

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Components connected in series are connected along a single path, so the same current flows through all of the components. If the light bulbs are connected in parallel, the currents through the light bulbs combine to form the current in the battery, while the voltage drop is across each bulb and they all glow.
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4 years ago
Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i
kkurt [141]

The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

Charge, Q = 3 nC = 3 × 10⁻⁹ C

Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

Calculation:

We know that the electric potential is given as:

V = k Q/r   - (1)

where, V is the electric potential

            k is Coulomb's force constant

            Qis the charge on the  sphere

            r is the  separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]

   = 1350 V

The electric potential at point D due to charged sphere can be given as:

V₂ = k Q/r₂

   = (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]

   = 450 V

Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

     = 450 V - 1350 V

     = -900 V

The negative sign indicates that the work is done against the electric field in moving the charge from C to D.

Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.

Learn more about the electric potential here:

<u>brainly.com/question/12645463</u>

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8 0
2 years ago
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A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, th
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Answer:

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Explanation:  

After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.

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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.

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3 years ago
One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s
tiny-mole [99]

Answer:

B

Explanation:

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Which of these is a good example of Newton's First Law of inertia
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