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tekilochka [14]
3 years ago
8

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro

static force between the two points, and is this force a repulsive force or an attractive force (k = 9.0 E9 Nm2/C2)
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

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

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

q_1 =q_2 = +4.5\cdot 10^{-6}C are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

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5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
The centripetal force acting on the space shuttle
KIM [24]
The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttles momentum
4 0
3 years ago
A current of 0.4 A flows through a wire. How many electrons flow through a cross section of
Free_Kalibri [48]

9 × 10²¹ electrons flow through a cross section of the wire in one hour.

<h3>What is the relation between current and charge?</h3>
  • Mathematically, current = charge / time
  • In S.I. unit, Charge is written in Coulomb and time in second.

<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
  • Charge= current × time
  • Current= 0.4 A, time = 1 hour= 3600 s
  • Charge= 0.4× 3600

= 1440 C

<h3>How many numbers of electrons present in 1440C of charge?</h3>
  • One electron= 1.6 × 10^(-19) C
  • So, 1440 C = 1440/1.6 × 10^(-19)

= 9 × 10²¹ electrons

Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.

Learn more about current here:

brainly.com/question/25922783

#SPJ1

4 0
2 years ago
If a 100 N net force act on a 50-kg car, what will the acceleration be
inna [77]
Acceleration = force / mass.

A = 100/50 = 2 m/s^2 .
7 0
2 years ago
g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

3 0
3 years ago
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