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Marrrta [24]
3 years ago
14

Why are citizens obligated to respond to such documents?

Physics
1 answer:
Zina [86]3 years ago
5 0
To protect the constitutional right to confront ones accused
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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s
Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, R=50\ K\Omega=5\times 10^4\ \Omega

Capacitance, C=2.1\ \mu F=2.1\times 10^{-6}\ F

We need to find the expected time constant for this RC circuit. It can be calculated as :

\tau=R\times C

\tau=5\times 10^4\times 2.1\times 10^{-6}

\tau=0.105\ s

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0
3 years ago
A primary succession does not include any invasive species <br><br> True or false
yan [13]
To be referenced, it would be true
7 0
3 years ago
You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
3 years ago
Use Percentages Given that the molecular mass of magnesium
ira [324]

Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

5 0
3 years ago
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