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Karolina [17]
3 years ago
13

A point charge q is located at the origin. A charge q0 can be placed at a point P1, which is a distance r from the origin (top d

rawing). Or, a charge 2q0 can be placed at P2, which is a distance 2r from the origin (bottom drawing). All charges are positive. Which statement is true about the electric potentials due charge q at P1 and P2?
a.The electric potential at P1 is greater than that at P2, because r is smaller than 2r.
b. The electric potential at P1 is the same as that at P2.
c. The electric potential at P1 is less than that at P2, because q0 is smaller than 2q0.
d. The electric potential at P1 is less than that at P2, because r is smaller than 2r.
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

Option A is correct

The electric potential at P1 is greater than that at P2, because r is smaller than 2r.

Explanation:

Electric potential at a point due to a particular charge (q) at r distance from that point is given as

V = kQ/r

where k = Coulomb's constant

For point P₁, the electric potential due to charge q, r distance away is given as

V₁ = kq/r

For point P₂, the electric potential due to charge q, (2r) distance away is given as

V₂ = kq/(2r)

This shows that the electric potential due to charge q at P1 is twice that experienced at P2 because of the same charge.

The electric potential at a point due to a particular only depends on the charge in question and the distance of that charge from that point.

If the charge and other parameters are constant, the electric potential at some distance away is inversely proportional to that distance. So, smaller r, indicates bigger electric potential.

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'C' is the only true statement on the list.

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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3 years ago
Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very
timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ v=\frac{502.1}{\sqrt{3} }

      =289.9 \ m/s

The time will be:

⇒ t=\frac{d}{v}

      =\frac{2\times 6}{289.9}

      =\frac{12}{289.9}

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hence,

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4 0
3 years ago
You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas
stepan [7]
Refer to the diagram shown below.

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The mass near the right end is 1.8 kg.
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