Ask question

Ask question

All categories

3 years ago

5

Charge A has 10 electric field lines coming out, Charge B has 20 lines coming out, and Charge C has 30 lines coming in. Which pa

ir of these charges will have the largest force between them if placed one cm apart?

1 answer:

Whitepunk [10]3 years ago

3 0

Answer:

c will have the largest charge

Explanation:

because more lines are coming more line means more foce is required

You might be interested in

Which scientific law describes the relationship between action and reaction force pairs? Newton's third law of motion Kepler's l

Hitman42 [59]

Newton's third law of motion

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>

It is important to note that this law is always valid, even when it seems it is not.

Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.

It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.

Learn more about Newton's third law of motion:

brainly.com/question/11411375

#LearnwithBrainly

7 0

4 years ago

Find the wavelength in meters for a transverse mechanical wave with an amplitude of 10 cm and a radian frequency of 20π rad/s if

nydimaria [60]

Answer:

The wavelength of the wave is 20 m.

Explanation:

Given that,

Amplitude = 10 cm

Radial frequency $\omega = 20\pi\ rad/s$

Bulk modulus = 40 MPa

Density = 1000 kg/m³

We need to calculate the velocity of the wave in the medium

Using formula of velocity

$v=\sqrt{\dfrac{k}{\rho}}$

Put the value into the formula

$v=\sqrt{\dfrac{40\times10^{6}}{10^3}}$

$v=200\ m/s$

We need to calculate the wavelength

Using formula of wavelength

$\lambda =\dfrac{v}{f}$

$\lambda=\dfrac{v\times2\pi}{\omega}$

Put the value into the formula

$\lambda=\dfrac{200\times2\pi}{20\pi}$

$\lambda=20\ m$

Hence, The wavelength of the wave is 20 m.

5 0

3 years ago

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?

zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

$h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m$

3 0

3 years ago

How do u get rid of a hacker? plz help my friend needs this

Alja [10]

IP address then call the cops

3 0

3 years ago

Read 2 more answers

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago