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3 years ago

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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

of an equilateral triangle)?

1 answer:

Natasha_Volkova [10]3 years ago

7 0

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

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A uniformly charged conducting plate with area a has a total charge Q which is positive. The figure below shows a cross-sectiona

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Answer:

a) E = σ / 2 ε₀ = Q / 2A ε₀, b) E = 2Q/A ε₀

Explanation:

For this exercise we can use Gauss's Law

Ф = E. dA = $q_{int}$ / ε₀

Let us define a Gaussian surface as a cylinder with the base parallel to the plane. In this case, the walls of the cylinder and the charged plate have 90 degrees whereby the scalar product is zero, the normal vector at the base of the cylinder and the plate has zero degrees whereby the product is reduced to the algebraic product

Φ = E dA = q_{int} / ε₀ (1)

As they indicate that the plate has an area A, we can use the concept of surface charge density

σ = Q / A

Q = σ A

The flow is to both sides of loaded plate

Φ = 2 E A

Let's replace in equation 1

2E A = σA / ε₀

E = σ / 2 ε₀ = Q / 2A ε₀

This is in the field at point P.

b) Now we have two plates each with a load Q and 3Q respectively and they ask for the field between them

The electric field is a vector quantity

E = E₁ + E₂

In the gap between the plates the two fields point in the same direction whereby they add

σ₁ = Q / A

E₁ = σ₁ / 2 ε₀

For the plate 2

σ₂ = -3Q / A = -3 σ₁

E₂ = σ₂ / 2 ε₀

E₂ = -3 σ₁ / 2 ε₀

The total field is

E = σ₁ / 2 ε₀ + 3 σ₁ / 2 ε₀

E = σ₁ / 2 ε₀ (1+ 3)

E = 2 σ₁ / ε₀

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3 years ago

Suppose a lion running at a velocity of 10 m/s east slows

iris [78.8K]

Answer:

$a=0.34\ m/s^2$

Explanation:

Given that,

Initial velocity, u = 10 m/s

Final velocity, v = 0 (stops)

Time, t = 30 s

We need to find the acceleration of the lion.

We know that,

Acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-0}{30}\\\\a=0.34\ m/s^2$

So, the acceleration of the lion is $0.34\ m/s^2$ .

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3 years ago

At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another b

maks197457 [2]

Answer:

(a) The two balls collide $2\; \rm s$ after launch.

(b) The height of the collision is $4\; \rm m$ .

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of $g$ should be negative. The question states that the magnitude of $g\!$ is $10\; \rm m \cdot s^{-2}$ . Hence, the signed value of $\! g$ should be $\left(-10\; \rm m \cdot s^{-2}\right)$ .

Similarly, the initial velocity of the ball thrown downwards should also be negative: $\left(-8.0\; \rm m \cdot s^{-1}\right)$ .

On the other hand, the initial velocity of the ball thrown upwards should be positive: $\left(12\; \rm m \cdot s^{-1}\right)$ .

Let $v_0$ and $h_0$ denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time $t$ :

$\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0$ .

For both balls, $g = \left(-10\; \rm m \cdot s^{-2}\right)$ .

For the ball thrown downwards:

Initial velocity: $v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 40\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40$ (where $h$ is in meters and $t$ is in seconds.)

Similarly, for the ball thrown upwards:

Initial velocity: $v_0 = \left(12\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 0\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + 12\, t$ (where $h$ is in meters and $t$ is in seconds.)

Equate the two expressions and solve for $t$ :

$-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t$ .

$t = 2$ .

Therefore, the collision takes place $2\, \rm s$ after launch.

Substitute $t = 2$ into either of the two original expressions to find the height of collision:

$h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m$ .

In other words, the two balls collide when their height was $4\; \rm m$ .

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Answer:

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Explanation:

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