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Triss [41]
2 years ago
12

Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Whe

re must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Where must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? 50 cm 45.8 cm 9.12 cm 91.2 cm 4.58 cm
Physics
1 answer:
joja [24]2 years ago
8 0

Answer:

45.8 cm

Explanation:

To solve this, we will use the formula

5 / x² = 7/(1 - x)²

5 / x² = 7 / (1 - 2x + x²)

5 / 7 = x² / (1 - 2x + x²)

x = 0.5 * (√(35) - 5) meters

x = 0.5 * (5.916 - 5)

x = 0.5 * (0.916)

x = 0.458 or x = 45.8

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2 years ago
A man carries a load of mass 2.6kg from one end of a uniform pole 100cm long which has a mass 0.4kg. The pole rest on his should
miskamm [114]

Answer:

F = 39.2 N   (hand force) and    N = 68.6 N (shoulder force)

Explanation:

In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole

Let's set the reference system at the fit point of the shoulder

     ∑ τ = 0

We will assume that the counterclockwise turns are positive

    w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0

all distances are measured from the support of the man (x₀ = 0.60 m)

    F = (w₁ 0.60 + W 0.1) / 0.4

    F = (m₁ 0.6 + m 0.1) g / 0.4

let's calculate

    F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4

    F = 39.2 N

this is the force that the hand must exert to keep the system in balance

We apply the translational equilibrium condition

    -w₁ -W + N - F = 0

     N = w₁ + W + F

     N = (m₁ + m) g + F

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6 0
1 year ago
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
1 year ago
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