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murzikaleks [220]
1 year ago
14

S Review. Two identical particles, each having charge +q , are fixed in space and separated by a distance.d. A third particle wi

th charge -Q is free to move and lies initially at rest. on the perpendicular bisector of the two fixed charges a distance x from the midpoint between those charges (Fig. P23.14). (b) Determine the period of that motion.
Physics
1 answer:
hammer [34]1 year ago
5 0

The period of motion is given as  ​π/2(√md³/KeqQ)

What is Coulombs law?

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

The period of motion from the instance given  ​π/2(√md³/KeqQ).

w²= (2π/T)²

T=2π/w

where w is gotten as 16KeqQ/md³

T=  ​π/2(√md³/KeqQ).

m is the mass of the body with charge-Q

In conclusion, the Coulomb's law equation provides an accurate description of the force between two objects whenever the objects act as point charges.

Learn more about coulombs law at :brainly.com/question/26892767

#SPJ1

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A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the
Finger [1]

After the collision, the momentum didn't change, so the total momentum in x and y are the same as the initial.

The x component was calculated by subtracting the initial momentum (total) minus the momentum of the first ball after the collision

In the y component, as at the beginning, the total momentum was 0 in this axis, the sum of both the first and struck ball has to be the same in opposite directions. In other words, both have the same magnitude but in opposite directions

\begin{gathered} Py=0.71kg\cdot2.17\cdot sin(30) \\ Py=0.77kg\cdot m/s \end{gathered}

This is for both balls after the collision, but one goes in a positive and the other in a negative direction.

7 0
1 year ago
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the
sashaice [31]

Answer:

52.47706 mph

54.5 mph

Explanation:

The average speed is given by

V_{av}=\dfrac{Distance}{Time}

V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph

Julie's average speed on the way to Grandmother's house is 52.47706 mph

V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph

Average speed on the return trip is 54.5 mph

4 0
4 years ago
Predict the speed of a car after 10.0 s when the car’s speed follows
andreev551 [17]

Answer:

48 m/s

Explanation:

The time is 10 seconds

The speed of the car after 10 seconds when the relationship is 4.8 m/s^2 × time can be calculated as follows

= 4.8 × 10

= 48

Hence the speed is 48 m/s

7 0
4 years ago
A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement to
Virty [35]
17.3 would be the correct answer.

4 0
4 years ago
Read 2 more answers
An ion's position vector is initially r with arrow = 7.0i hat â 7.0j + 1.0k, and 5.0 s later it is r with arrow = 7.0i hat + 7.0
harkovskaia [24]
<h2>Answer:</h2>

Shown in the explanation

<h2>Explanation:</h2>

Position vector of a  particle at a given instant is given by:

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

On the other hand, the average velocity is the change in the particle’s position vector divided by time interval:

\vec{v}=\frac{\Delta \vec{r}}{\Delta t}=\frac{\vec{r_{2}}-\vec{r_{1}} }{t_{2}-t_{1}} \\ \\ \\ Where: \\ \\ \\ \vec{r_{1}}: Initial \ position \\ \\ \vec{r_{2}}: Final \ position \\ \\ t_{1}:Initial \ time \\ \\ t_{2}:Final \ time

So we have:

\vec{r_{1}}=7\hat{i}+7\hat{j}+1\hat{k} \\ \\ \vec{r_{2}}=7\hat{i}+7\hat{j}+8\hat{k} \\ \\ \\ Then: \\ \\ \Delta \vec{r} = \vec{r_{2}}-\vec{r_{1}}=7\hat{i}+7\hat{j}+8\hat{k}-(7\hat{i}+7\hat{j}+1\hat{k}) \\ \\ \Delta \vec{r}=7\hat{k} \\ \\ \\ We \ also \ know: \\ \\ \Delta t=5s

Finally, the average velocity is:

\vec{v}=\frac{\Delta \vec{r}}{\Delta t} \\ \\ \\ \vec{v}=\frac{7\hat{k}}{5} \\ \\ \boxed{\vec{v}=\frac{7}{5}\hat{k} \ units/s}

3 0
4 years ago
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