9. What happens to the particles of a medium when the temperature is high?

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the

Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is $W= 9*10^{-7}J$

D

The speed of the charge is $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of $k = 9*10^9 N m^2 /C^2$

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as $d =5m$

Now we are given that $q_1 = q$ and $q_2 = 3q$ and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at $q_2$

Now potential at $q_1$ is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

$V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

$v=600m/s$

8 0

4 years ago

What is generally TRUE about diagnosing psychological disorders?

Allisa [31]

The statement that says "Psychological disorders can be very difficult to diagnose" is true about diagnosing psychological disorders.

<h2>What are psychological disorders?</h2>

Psychological disorders are those mental, behavioral, emotional and thinking conditions that interfere with the normal performance of the individual in society.

Mental disorders are psychiatric conditions that are expressed in a syndrome, verifiable from different diagnostic criteria.

The steps to obtain a diagnosis include a medical history, physical examination, and possibly laboratory tests and a psychological evaluation.

Therefore, we can conclude that a psychological disorder is an alteration in the mental balance of a person that requires specialized attention adapted to the characteristics of the dysfunction.

Learn more about psychological disorders here: brainly.com/question/6367767

3 0

3 years ago

What does “use f=ma, where a=v-u divided by t”mean

m_a_m_a [10]

$f = m( \frac{v - u}{t} )$

4 0

3 years ago

A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam

iogann1982 [59]

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

$F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]$

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

$W=F*d$

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

$W=4713.25*66\\W=311074.5[J]$

5 0

3 years ago

2. What is the mass of an object if it accelerates at 3 m/s2 and has a force of 1 N?

Dvinal [7]

Answer:

hey what is this 2 in between the question

please tell

3 0

2 years ago