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Rasek
2 months ago
10

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

, respectively. (Take V = 0 at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?
Physics
1 answer:
inna [3.1K]2 months ago
3 0

Answer:

A. 30.7 cm

B. 1.7*10^{-10}C

C. The electric field points outward from the charge

Explanation:

A.

because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm

B.

Using Gauss's law

EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C

C. The electric field emanates away from the point of charge if the charge is positive.

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In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
kicyunya [3294]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak mg-T=\frac{mv^2}{r}

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as mg-0=\frac{mv^2}{r}

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at the bottom E_b=\frac{mv_0^2}{2}

By comparing the two states using conservation of energy, we find v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}

6 0
2 months ago
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
2 months ago
To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to
kicyunya [3294]

Answer:

a)  τ = i ^ (y F_{z} - z F_{y}) + j ^ (z Fₓ - x F_{z}) + k ^ (x F_{y} - y Fₓ)

b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

Explanation:

a) The torque can be expressed as

        τ = r x F

To tackle this equation, using the determinant approach is the most straightforward method

        \tau =\left[\begin{array}{ccc}i&j&k\\x&y&z\\F_{x}&F_{y} &F_{z}\end{array}\right]  

The resulting expression is

      τ = i ^ (y F_{z} - z F_{y}) + j ^ (z Fₓ - x F_{z}) + k ^ (x F_{y} - y Fₓ)

b) Now let's compute

     τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)

     τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)

     τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) To find angular acceleration, we use

       τ = I α

       α = τ / I

The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.

           

     α = (-0.189i^  -5.6 j^  + 33.978k^) / 241

     α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

8 0
2 months ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
Keith_Richards [3271]

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%

3 0
2 months ago
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