How many valence electrons do most stable atoms have?

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

sergejj [24]

The distance between the charges is 13.86 X 10⁴m

<u>Explanation:</u>

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2} \\\\$

Substituting the value in the formula we get:

$1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m$

Therefore, the distance between the charges is 13.86 X 10⁴m

5 0

3 years ago

A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

Sauron [17]

Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

8 0

3 years ago

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A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th

almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

$\frac{1}{2} *m*v_{i}^{2}-(F*d)= \frac{1}{2} *m*v_{f}^{2}$