A static charge that is produced by a nearby objet is an?

Question 1 of 10

Neko [114]

Answer:

D: A mathematical model

Hope its right if not so sorry :)

3 0

3 years ago

In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist

marin [14]

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

$P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times 0.108^2}\pi\times 0.0262^2\\\Rightarrow F_1=847.45\ N$

Hence, force exerted to lift a 14400 N car is 847.45 N

8 0

4 years ago

While dragging a crate a workman exerts a force of 628 N. Later, the mass of the crate is increased by a factor of 3.8. If the w

Delvig [45]

Force applied = F = 628 N
Acceleration = a m/s² 
Newton's 2nd law of motion : F = Ma 
 a = F/M -------- (1) 
New mass of the crate = M1 = 3.8M kg 
New acceleration = a1 = F/M1 = F/(3.8 M) ----- (2) 
a1/a = {F/(3.8M)}/(F/M) = 1/3.8 = 10/38 = 5/19 ------- Answer

6 0

3 years ago

Read 2 more answers

HOLA, NECESITO AYUDA!

givi [52]

The electrostatic force between the two ions is $2.9\cdot 10^{-10} N$

Explanation:

The electrostatic force between two charged particle is given by Coulomb's law:

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

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, the ion of sodium has a charge of

$q_1 = +e = +1.6\cdot 10^{-19} C$

while the ion of chlorine has a charge of

$q_2 = -e = -1.6\cdot 10^{-19}C$

And the distance between the two ions is

$r=282 pm = 282\cdot 10^{-12} m$

Substituting, we find the electrostatic force between the two ions:

$F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(282\cdot 10^{-12})^2}=-2.9\cdot 10^{-10} N$

where the negative sign simply means that the force is attractive, since the two ions have opposite charge.

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric

sp2606 [1]

Answer:

a) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

b)E=0

c) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a) r < a

At a radius r ,from gauss theorem

$E.ds=\dfrac{q_i}{\varepsilon _o}$