True or false gravity is a force

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass = $volume\times density$

Using the formula

Mass of solution= $1.12\times 473=529.76 g$

Mass % of KBr= $\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100$

Mass % of KBr= $\frac{49.3}{529.76}\times 100$

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

7 0

3 years ago

A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.

Nastasia [14]

<h3>Solution for the above question : -</h3>

Ohm's law states that :

$v = ir$

the terms used are :

$r = resistance$
$v = potential \: \: difference$
$i = current$

let's solve for electric current :

$2 = i \times 40$

$i = \dfrac{2}{40}$

$i = 0.05 \: A$

$i = 50 \: mA$

$\mathfrak{good\: \: luck \: \: for \: \: your \: \: assignment}$

8 0

2 years ago

What is the law of variation of the period T of a simple pendulum

DIA [1.3K]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum and $g=9.81 m/s^2$ is the gravitational acceleration. As we can see, the period of a simple pendulum depends only on its length.

3 0

3 years ago

Suppose the wavelength of the light is 460 nm. how much farther is it from the dot on the screen in the center of fringe e to th

gavmur [86]

The difference is 920nm.

<h3>What is called Wavelength?</h3>

The wavelength can be defined as the separation between wave crests. Additionally, a wide variety of objects move in waves, including light, the earth or ground, water, strings, air, and sound waves. Furthermore, we use the Greek letter lambda to denote the wave's wavelength. In addition, the wave's wavelength is determined by dividing its velocity by its frequency. Additionally, we employ meters (m), which is the unit used to indicate wavelength in meters.

Positive interference is produced by a bright fringe, and the wavelength is always multiple.

The difference at the center fringe is (460 nm)(0) = 0.

The difference for the first maximum is (460 nm)(1) = 460 nm.

The difference for the second maximum is (460 nm)(2) = 920 nm.

As a result, the difference for maxima n is (460 nm) (n)

Since C is on the second maximum in the image that was submitted, the difference is stated as (460 nm)(2) = 920 nm.

Consequently, the screen's central dot on the fringe E to the left slit is 920 nm.

To learn more about Wavelength, visit:

brainly.com/question/13533093

#SPJ4

3 0

1 year ago