11. Trait theory claims that

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 hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

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

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s

Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity $v$ and at a distance of $r$ from the axis of rotation is given as:

$a=\frac{v^2}{r}$

Here, $v=8\ m/s,r=0.70\ m$

∴ $a=\frac{8}{0.70}=11.43\ m/s^2$

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0

3 years ago

What order shows increasing frequency for gamma rays, microwaves, visible light, and X-rays?

sergejj [24]

Answer:

Person above me is right, it's B: microwaves, visible light, X-rays, gamma rays

Explanation:

6 0

3 years ago

. The Soviet Yuri Gagarin was the first human to orbit Earth true or false

Dmitrij [34]

Answer:

true , I searched and got u the answer

3 0

2 years ago