All categories

2 years ago

How do gravitational and electic force compare

1 answer:

3 0

Electric Forces. ... Just like objects that have mass exert gravitational forces on each other, objects that are charged will also exert electric forces on each other. The electric force is directly proportional to the charge of the two objects and inversely proportional to the distance between them squared.

You might be interested in

Explain the reason for following in science. A small stone trapped in your shoe, under your foot, can be very painful.

igor_vitrenko [27]

Answer:

bvvfalbvenvea;vfahvfahvfna.fvn.adnvfad.nvfa;vnfavnfdavnfdanv.VHFvna.vnfad.vnfa;

Explanation:nvfad;bfvf vf fvdnva,e vdfvf dfvfeva

6 0

2 years ago

S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initiall

telo118 [61]

The sphere has a constant potential. It is the electric field.

$E = V_{0} = 0$

In the sphere, then

$E_{x} = 0, E_{y}=0, E_{z}=0$

Outside the sphere, then

$V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2} }$

The elements of the electric field include

$E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}$

Which becomes,

$=E_{0} (1-\frac{a^{3}}{x^{2} +y^{2}+z^{2})^{3/2}}+\frac{3a^{3}z^{2}}{(x^{2} +y^{2}+z^{2})^{5/2}})$

<h3>In a consistent electric field, is force constant?</h3>

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

To learn more about uniform electric field, visit

brainly.com/question/17426130

#SPJ4

5 0

1 year ago

Velocity (m/s)

alekssr [168]

Explanation:

⠀

(a) The segment A shows acceleration as velocity increases with the increase in time.

⠀

(b) The segment C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

⠀

(c) The velocity is segment B is 40m/s. And in the diagram there is no change in velocity.

⠀

(d) The acceleration of segment B is zero. As there in no change in curve and it is moving with uniform velocity.

⠀

$\:$

<h2>Thank you!</h2>

3 0

1 year ago

The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl

aleksandr82 [10.1K]

$acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}$

6 0

3 years ago

How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?

svet-max [94.6K]

We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!

8 0

3 years ago