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Vlada [557]
3 years ago
8

A capacitor stores 7.6 × 10–11 C when the electric potential difference between the plates is 5.8 V. What is the electric potent

ial energy stored in the capacitor?
Physics
1 answer:
AlekseyPX3 years ago
8 0

Energy stored in the capacitor= 2.2 x 10⁻¹⁰ J

Explanation:

The energy stored in the capacitor is given by

E= 1/2 q V

E= energy stored

q= charge= 7.6 x 10⁻¹¹ C

V= potential difference=5.8 V

so energy stored= E= 1/2 (7.6 x 10⁻¹¹) (5.8)

Energy stored= 2.2 x 10⁻¹⁰ J

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3 years ago
A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite
DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx  is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x  4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x  4 π

≥  . 5254 x ⁻²⁵

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3 years ago
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

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