A ball is thrown up in the air for 1 second and has a displacement of 5 m.

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago

Question 6 of 10

sergejj [24]

The answer is C.89,866. I multiplied 262x343.

3 0

2 years ago

If you push a 200 kg box with a horizontal force of 1,172 N and kinetic friction resists the motion with a force of 962 N, what

Lynna [10]

Answer:

a = 1.05m.s²

Explanation:

Fnet = m×a

Fapplied - friction = m×a

1172 - 962 = 200 × a

210 = 200a

a = 1.05

5 0

1 year ago

Altitude is the angle measured above ____. North Pole horizon equator zenith

juin [17]

Altitude is the angle measured above the horizon

5 0

3 years ago

Read 2 more answers

The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i

gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

8 0

3 years ago