All categories

2 years ago

How will the electrostatic force between two electric charges change if one charge is doubled and the second charge is tripled?

Physics

1 answer:

rewona [7]2 years ago

7 0

Answer:

b. The force is 6 times the original

Explanation:

Coulomb's Law

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose one of the charges is doubled, i.e., q1=2q1 and the other charge is tripled (q2'=3q2). The new force F' would be:

$\displaystyle F'=k\frac{(2q_1)(3q_2)}{d^2}$

Operating:

$\displaystyle F'=k\frac{6q_1q_2}{d^2}$

$\displaystyle F'=6\cdot k\frac{q_1q_2}{d^2}$

Substituting the original value of the force:

$F'=6\cdot F$

b. The force is 6 times the original

You might be interested in

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick

rewona [7]

Answer:

The new time period is $T_2 = 3.8 \ s$

Explanation:

From the question we are told that

The period of oscillation is $T = 5 \ s$

The new length is $l_2 = 0.76 \ m$

Let assume the original length was $l_1 = 1 m$

Generally the time period is mathematically represented as

$T = 2 \pi \sqrt{ \frac{ I }{ mgh } }$

Now I is the moment of inertia of the stick which is mathematically represented as

$I = \frac{m * l^2 }{12 }$

$T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }$

Looking at the above equation we see that

$T \ \ \ \alpha \ \ \ l$

=> $\frac{ T_2 }{T_1} = \frac{l_2}{l_1}$

=> $\frac{ T_2}{5} = \frac{0.76}{1}$

=> $T_2 = 3.8 \ s$

3 0

3 years ago

An object traveling at constant speed v in a circle of radius R has an acceleration am 5 m/s2. If both R and v are doubled, what

MakcuM [25]

Answer:

im pretty sure its 10 m/s but its kinda hard sorry

Explanation:

7 0

3 years ago

On both the spring and fall equinoxes, the sun's rays are vertically overhead at the _______

Anni [7]

It is overhead at the equator, it is because the sun ray’s will be moving vertically as this will be directed at the equator. It is because if it moves vertically, it will hit or overhead the equator and this usually happens in spring and fall.

5 0

2 years ago

A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d

Ipatiy [6.2K]

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

6 0

3 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago