What would scratch a mineral that registers 8 on the Moh’s hardness scale?

Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot

Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

q₂ = -6.0 nC = -6.0 10⁻⁹ C

x₂ = 4.0 cm = 4.0 10⁻² m

q₃ = 5.0 nC = 5.0 10⁻⁹ C

x3 = 0 m

Charges q1 and q3

r = x₁ -x₃

r = 2.0 10⁻² -0

r = 2.0 10⁻² m

F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

r = r₂ –r₃

r = 4.0 10⁻² - 0 = 4.0 10⁻² m

F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

F = F₂₃ - F₁₃

F = 16.86 10⁻⁵ - 33.7 10⁻⁵

F = - 16.84 10⁻⁵ N

F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0

4 years ago

If planet A is three times as far from planet C, then the period of its orbit will be __ times as long

liubo4ka [24]

I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.

6 0

3 years ago

Help please

Naddik [55]

Answer:

acid and bases

Explanation:

4 0

3 years ago

Ety ratio

horrorfan [7]

3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is $3\cdot 10^4 W$

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

When the force is applied perpendicular to the direction of motion,

$\theta=90^{\circ}$

Therefore, the work done is:

$W=Fd(cos 90^{\circ})=0$

4)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the girl in this problem, we have

m = 40 kg

v = 3 m/s

Therefore her kinetic energy is

$K=\frac{1}{2}(40)(3)^2=180 J$

5)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

$g=10 m/s^2$ is the acceleration of gravity

h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

$PE=(0.4)(10)(30)=120 J$

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height $\Delta h$ between the two points, not on the path taken during the fall
When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

$P=Fv$

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

v = 3000 m/s is its velocity

Substituting into the equation, we find the power expended:

$P=(10)(3000)=30,000 W = 3\cdot 10^4 W$

9.

The efficiency of a machine is given by

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{in}$ is the energy in input to the machine

$W_{out}$ is the useful work in output from the machine

For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so

$W_{out}=W_{in}$

And therefore the efficiency is

$\eta=1$

which means 100%.

10.

The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

$VR=\frac{d_{eff}}{d_{load}}$

In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is

$VR=\frac{5}{1}=5$