What forces are acting on a heavy box sitting stationary on the floor?

A charge q1 of -5.00X10^-9 C and a charge q2 of -2.00X10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium posit

Alex73 [517]

Answer:

The equilibrium position for the third charge is 69.28 cm

Explanation:

Given;

q₁ = -5.00 x 10⁻⁹ C

q₂ = -2.00 x 10⁻⁹ C

q₃ = 15.00 x 10⁻⁹ C

distance between q₁ and q₂ = 40.0 cm = 0.4 m

(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)

At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃

According to Coulomb's law, repulsive or attractive force between charges is calculated as;

$F = \frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}$

where;

F is repulsive or attractive force between charges

K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²

r₁ is the distance between q₁ and q₂

q₁, q₂ and q₃ are the charge

distance between q₂ and q₃, r₂ is calculated as;

$\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m$

Therefore, the equilibrium position for the third charge is 69.28 cm

3 0

3 years ago

Consider a cylinder initially filled with 9.33 10-4 m3 of ideal gas at atmospheric pressure. An external force is applied to slo

Juli2301 [7.4K]

Answer:

Work done will be 78.76 J

Explanation:

We have given initial volume of the gas $V_1=9.33\times 10^{-4}m^3$

Pressure is given by $P=1.013\times 10^5Pa$

Final volume $V_2=\frac{V_1}{6}=\frac{9.33\times 10^{-4}}{6}=1.555\times 10^{-4}m^3$

Change in volume $\Delta V=V_2-V_1=1.555\times 10^{-4}-9.33\times 10^{-4}=-7.775\times 10^{-4}m^3$

We know that work done is given by

$W=-Pdv=-1.013\times 10^5\times 7.775\times 10^{-4}=78.760J$

Work done will be 78.76 J

8 0

3 years ago

Which of the following is strenuous?

Arada [10]

Answer:

I think it c

Explanation:

7 0

3 years ago

Read 2 more answers

Mass of 55 kg riding in a car at20m/s slams on brakes rests in.5 sec how much force from seatbelt

Evgen [1.6K]

2,200 Newtons.

F = ma

v f = v i + at
0mls = 20mls + a(.5s)

-20mls/.5s = -40mls squared

-40mls squared (55kg) = 2,200 Newtons.

Hope this helps!

5 0

3 years ago

An object of mass m is dropped from a height h above the surface of a planet of mass M and radius R. Find the speed of the objec

Shtirlitz [24]

Answer:

$v=\sqrt{\frac{2GMh}{R^{2}}}$

Explanation:

mass of object = m

Mass of planet = M

Radius of planet = R

Height = h

Let the speed of the object as it hits the earth's surface is v.

the value of acceleration due to gravity

g = G M / R^2

where, g is the universal gravitational constant.

Use third equation of motion

$v^{2}=u^{2}+2gh$

where, u is the initial velocity which is equal to zero.

So, $v^{2}=0 + 2 \times \frac{GM}{R^{2}}\times h$

$v=\sqrt{\frac{2GMh}{R^{2}}}$

8 0

4 years ago