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denpristay [2]
3 years ago
11

Three 100 nCcharged objects are equally spaced on a straight line. The separation of each object from its neighbor is 0.3 m. Fin

d the force exerted on the center object if the rightmost charge is negative and the other two are positive.
Physics
1 answer:
Rainbow [258]3 years ago
6 0

Answer:

0N

Explanation:

Parameters given:

Q1 = Q2 = Q3 = 100 * 10^(-9) C

r = 0.3 m

The net force on the middle charge is:

F = F(1,2) + F(2,3)

Where F(1,2) = force due to Q1 on Q2

F(2,3) = force due to Q3 on Q2

F(1,2) = k*Q(1)*Q(2)/r²

F(1,2) = [9 * 10^9 * 100 * 10^(-9) * 100 * 10^(-9)]/(0.3²)

F(1,2) = 0.001N

F(2,3) = k*Q(2)*Q(3)/r²

F(2,3) = [9 * 10^9 * 100 * 10^(-9) * -100 * 10^(-9)]/(0.3²)

F(2,3) = -0.001N

Therefore,

F = 0.001 - 0.001

F = 0N

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A small apple weighs 0.8N. How much would an apple that is 4 times as massive weigh?
-BARSIC- [3]

Answer:

W=31.99 N

Explanation:

W= 8 N (N= kg*m/ s^2)

W= m* g

8 N= m * 9.8 m/s^2

m= 0.816 kg

m*4= (0.816)4

= 3.264 kg

W= mg

= (3.264 kg)(9.8 m/s^2)

W=31.99 N

4 0
3 years ago
Which of these experiments tests a chemical property of an object??
Ilia_Sergeevich [38]
<span>B. shining a bright light on the objects and testing for decomposition </span> <span>      

In explanation, chemical property is a characteristic of a certain substance came from an outcome due to chemical change or reaction. In the situation above, more specifically toxicity is involved in the chemical property/change. Hence, when the object is tested for decomposition. Like for an example of decomposition simply in metals, rusting. Rusting a process of degeneration of metals. Here it works the same. Toxicity is how much damage did a certain entity do to the object. </span>



8 0
3 years ago
Read 2 more answers
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart
marta [7]

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          x_{cm} = \frac{1}{M} \sum x_i m_i

in this case we have two bodies

          x_{cm} = \frac{1}{M} (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = \frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )

          x_{cm} = \frac{1}{M} ( m_1 v_1 + m_2 v_2 )

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = \frac{m_1}{m_1+m_2}  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         v_{cm o} = \frac{1}{m_1 +m_2} (m₁ vo)

final

         v_{cm f} = \frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o ) ( v) = v

         v_{cm f} =  \frac{m_1}{M^2} v_o

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = \frac{1}{M} = \frac{1}{m_1 +m_2}

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = \frac{m_1 -m_2}{m1 +m2 } v_o

             v_{2f} = \frac{2 m_1}{m-1+m_2} vo

now let's find the velocity of the center of mass

initial

          v_{cm o} = \frac{1}{M} m₁ v₀

final

          v_{cm f} = \frac{1}{M}  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = \frac{1}{M} [  m_1  \frac{m_2}{M} + m_2  \frac{2 m_1}{M} ] v₀

          v_{cm f} = \frac{1}{M^2} ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = \frac{1}{M^2} (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = \frac{1}{M} M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

4 0
3 years ago
a ball is dopped and falls with an accelerationof 9.8 m/s downward it hits the ground with a velocity of 49 m/s downward how lon
juin [17]

In order to know how long it has been falling for you take the final velocity "49m/s" and divide it by the acceleration "9.8m/s" and get 5, since you have been using seconds in the calculations the answer is 5 seconds. (Fun fact, it is actually 9.82m/s per second since it accelerates and they rounded it down.)

3 0
3 years ago
a bat experts a 120 N force on a 0.134 kg baseball for 0.04 seconds. what impolse is excerted on the bat
musickatia [10]
The impulse exerted by a force F on an object in a time \Delta t is given by
I = F \Delta t
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I=F \Delta t = (120 N)(0.04 s)=4.8 Ns
3 0
3 years ago
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