Điện tích Q = 8. 10-6C đặt cố định trong

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

Which toy car had the greatest applied force?

andreyandreev [35.5K]

what are the choices

7 0

3 years ago

Read 2 more answers

The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. What is the magnitude and direction of the electric field?

Stells [14]

The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. The magnitude and direction of the electric field will be 400 V/m directed parallel to the +x-axis

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

The electric potential energy of any given charge or system of changes is termed as the total work done by an external agent in bringing the charge or the system of charges from infinity to the present configuration without undergoing any acceleration.

The relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.”

Electric field = - d V / dx

-(-2000-400) = $\int\ {E} \, dx$

2400 = E (8-2)

2400 V = E (6)

E = 400 V/m

To learn more about electric potential energy here

brainly.com/question/16890427

#SPJ4

3 0

2 years ago

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

HELP!!! What role might electrostatic force play in spider dispersal, according to a recent study?

Marat540 [252]

The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.

Answer: Option C

<u>Explanation:</u>

The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.

Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.

6 0

4 years ago