I need help plzzz!!!!!

When you double the distance between a pair of charged particles what happens to the force between them?

Volgvan

Answer:

Force between the two charges becomes one fourth of the initial force.

Explanation:

The electrostatic force acting between any two charges is given as,

$F = k\frac{q_{1}q_{2}}{r^{2}}$

Here,

F = force

k = Coulomb's constant

$q_{1}$ = magnitude of charge of the first particle

$q_{2}$ = magnitude of charge of the second particle

$r$ = separation between the two charges

From the above relation,

$F \propto \frac{1}{r^{2}}$

Thus,

$\frac{F_{1}}{F_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2}$

$\frac{F_{1}}{F_{2}} = \left ( \frac{2r}{r} \right )^{2}$

$\Rightarrow \ F_{2} = \frac{1}{4}F_{4}.$

8 0

4 years ago

A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass

Nana76 [90]

V = 25,364.4 cm^3 Is volumer = 2.7g/cm^3 Is density
To calculate mass you use formula:m= V*rTo avoid remembering this formula you can see the type of unit on each given variable. We can see that we have g/cm^3 and cm^3. If we multiply them, we negate cm^3 and cm^3 and we are left with g which is unit for mass.
the answer is :
m = 68,486,6 g

6 0

3 years ago

Light travels in a transparent material at 2.5 x 10 m/s. Find the index of refraction of the

Andreyy89

I would say 5.6 to the 4 and I’ll I did was add it to both side

3 0

3 years ago

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

4 years ago

Which process is represented by the PV diagram?

Reika [66]

Hi!

The answer would be A. Isobaric Process

<h3>Explanation:</h3>

Isobaric process is a process where the pressure inside a system remains unchanged. In the Pressure Volume graph given, you can see that the pressure (y axis) remains constant with an increasing volume ( x axis). An example of this would be heating a container with a movable piston. Now, the degree of pressure is dependent on the frequency of collisions of particles inside a system on the walls. If this frequency changes, the pressure changes (proportionally). In our example, heating a container with a movable piston results in the particles inside the container to gain kinetic energy and move faster, meaning an increased frequency of collisions (higher pressure), but at the system time the increase in pressure results in the piston being pushed outwards, causing the volume of the container to increase. This results in decreased frequency of collision of the particles with the walls of the container (lesser pressure). This results in the a zero net effect on the pressure.

Hope this helps!

3 0

4 years ago