what wouldn't be the resonant frequency for series Rlc circuit with the following components a 1kv resistor ,a 900 mh inductor, and a 0.014mf capacitor?​

Answer:

hello your question has some missing information attached below is the missing information

answer : q3 = 0.300 nC

Explanation:

Given that Eay is in the Y axis ( upward ) , E3 will be downward and this will make the sign on the the charge ( q3 ) to be positive

E3 = Eay.   ( for an electric field to be neutral/zero the electric field in the opposite direction will have same magnitude )

To calculate the value of q3 we will apply the relation below

 K*q3/( 15 - 12 )^2  = 0.300

           = (9 * 10^9) * (q3 / 9 ) = 0.300

∴ q3 = 0.300 / 10^9

        = 0.300 nC

Note : The value of Eay is been calculated outside this solution as it is not part of the question asked  

Answer:

C. both A and B

Explanation:

Sana nakatulong

Answer:

5

Explanation:

From the question given above, the following data were obtained:

Proton number = 2

Neutron number = 3

Mass number =.?

The mass number of a given atom is simply the sum of the protons and neutrons in the nucleus of the atom. Mathematically, it is expressed as:

Mass number = proton + neutron

With the above formula, we can obtain the mass number of the atom as follow:

Proton number = 2

Neutron number = 3

Mass number =.?

Mass number = proton + neutron

Mass number = 2 + 3

Mass number = 5

Thus, the mass number the atom is 5.

Answer:

image distance = -8.47 cm

diameter of the image = 0.769 cm

Answer:

6. 9.4 m

7. 1050 m/s.

Explanation:

6. Determination of the wavelength

Velocity (v) = 30 m/s

Frequency (f) = 3.2 Hz

Wavelength (λ) =?

v = λf

30 = λ × 3.2

Divide both side by 3.2

λ = 30 / 3.2

λ = 9.4 m

Thus, the wavelength of the wave is 9.4 m.

7. Determination of the speed of the wave.

Wavelength (λ) = 350 m

Frequency (f) = 3 Hz

Velocity (v) =.?

v = λf

v = 350 × 3

v = 1050 m/s

Thus, the speed of the wave is 1050 m/s

Answer:its 2kg

Explanation:

Answer:

45.864N

Explanation:

Using the formula

F = nR

n is the coefficient of friction

R is the normal reaction

R = mg

F = nmg

F = 0.78 * 6 * 9.8

F = 45.864N

Hence the amount of force with which the hand pushes the box is 45.864N

Answer:

I hate to say it is a bit of the holiday but

Answer:

m = 24146.34 kg

Explanation:

The heat of fusion is defined as the amount of heat required to melt a unit mass of a substance completely. Hence, it can be calculated by the following formula:

[tex]H = \frac{Q}{m}\\\\m = \frac{Q}{H}[/tex]

where,

m = mass of copper = ?

Q = Heat absorbed by rod = 4950000000 J

H = Heat of fusion of copper = 205000 J/kg

Therefore,

[tex]m = \frac{4950000000\ J}{205000\ J/kg}[/tex]

m = 24146.34 kg

Answer:

Acoustics

Explanation:

Answer:

An illustration is a decoration, interpretation or visual explanation of a text, concept ... such as posters, flyers, magazines, books, teaching materials, animations, video games and films.

Explanation:

Answer:

The answer is D (It reduced his average

force.)

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

[tex]\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big)[/tex])

[tex]\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)[/tex]

Similarly;

[tex]\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)[/tex]

However; the relative difference is: [tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}[/tex]

[tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}[/tex]

[tex]= 0.2828 \\ \\ \mathbf{\simeq 29\%}[/tex]

Answer:

The speed of the knife after passing through the target is 9.33 m/s.

Explanation:

We can find the speed of the knife after the impact by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{k}v_{i_{k}} + m_{t}v_{i_{t}} = m_{k}v_{f_{k}} + m_{t}v_{f_{t}} [/tex]

Where:

[tex] m_{k}[/tex]: is the mass of the knife = 22.5 g = 0.0225 kg

[tex] m_{t}[/tex]: is the mass of the target = 300 g = 0.300 kg

[tex] v_{i_{k}}[/tex]: is the initial speed of the knife = 40.0 m/s

[tex] v_{i_{t}} [/tex]: is the initial speed of the target = 2.30 m/s

[tex]v_{f_{k}}[/tex]: is the final speed of the knife =?

[tex] v_{f_{t}} [/tex]: is the final speed of the target = 0 (it is stopped)

Taking as a positive direction the direction of the knife movement, we have:

[tex] m_{k}v_{i_{k}} - m_{t}v_{i_{t}} = m_{k}v_{f_{k}} [/tex]  

[tex] v_{f_{k}} = \frac{m_{k}v_{i_{k}} - m_{t}v_{i_{t}}}{m_{k}} = \frac{0.0225 kg*40.0 m/s - 0.300 kg*2.30 m/s}{0.0225 kg} = 9.33 m/s [/tex]

Therefore, the speed of the knife after passing through the target is 9.33 m/s.

I hope it helps you!              

Answer:

[tex]0.000835\ \text{m}[/tex]

Explanation:

d = Distance between plates = 2 mm

V = Potential difference = 500 V

v = Velocity of proton = [tex]2\times 10^5\ \text{m/s}[/tex]

a = Acceleration

m = Mass of proton = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]

Electric field is given by

[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{500}{2\times 10^{-3}}\\\Rightarrow E=250000\ \text{V/m}[/tex]

Force balance is given by

[tex]ma=qE\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 250000}{1.67\times 10^{-27}}\\\Rightarrow a=2.395\times 10^{13}\ \text{m/s}^2[/tex]

We have the relation

[tex]v^2=u^2+2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(2\times 10^5)^2-0}{2\times 2.395\times 10^{13}}\\\Rightarrow s=0.000835\ \text{m}[/tex]

The farthest distance from the negative plate that the proton reaches is [tex]0.000835\ \text{m}[/tex].

The farthest distance from the negative plate that the proton reaches will be s=0.000835 m

The electric field is defined as the force across the charged particles which attract or repel the other charged particles.

Now it is given in the question that

d = Distance between plates = 2 mm

V = Potential difference = 500 V

v = Velocity of proton =   [tex]2\times 10^5\ \dfrac{m}{s}[/tex]

a = Acceleration

m = Mass of proton = [tex]1.67\times10^{-27} kg[/tex]

The Electric field will be calculated as

[tex]E=\dfrac{V}{d}[/tex]

[tex]E=\dfrac{500}{2\times10^{-3}} =250000\ \frac{V}{m}[/tex]

Force balance is given by

[tex]ma=qe[/tex]

[tex]a=\dfrac{qE}{m}[/tex]

[tex]a=\dfrac{1.6\times10^{-19}\times250000}{1.67\times10^{-27}}[/tex]

[tex]a=2.395\times 10^{13}\frac{m}{s^2}[/tex]

Now from equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{(2\times10^5)^2-0}{2\times 2.395\times 10^{13}}[/tex]

[tex]s=0.000875 m[/tex]

Thus the farthest distance from the negative plate that the proton reaches will be s=0.000835 m

To know more about Electric field follow

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On lowering the temperature of the given reaction, H₂ gas and N₂ gas would react to produce more NH₃. Therefore, option (A) is correct.

When the chemical reaction is exothermic then increasing the temperature will cause the backward reaction to occur, decreasing the amounts of the products while increasing the amounts of reactants. Lowering the temperature will produce more reactants and cause the reaction to occur in the forward direction.

The formation of the ammonia is an exothermic reaction that can be represented as:

N₂  (g) +  3 H₂ (g) ⇄  2NH₃ (g)

If the temperature will be decreased, the chemical reaction will proceed forward to produce more heat. The effect of temperature on equilibrium will change the value of the equilibrium constant of the chemical reaction. The production of more ammonia on lowering the temperature.

Learn more about equilibrium, here:

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Answer:

A 2-kg ball is thrown at a speed of 5 m/s, exhibits 25 J of kinetic energy.

Explanation:

Given that,

The mass of a ball, m = 2 kg

Kinetic energy of the ball, K = 25 J

We need to find the speed of the ball. The formula for the kinetic energy is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 25}{2}} \\\\v=5\ m/s[/tex]

So,

A 2-kg ball is thrown at a speed of 5 m/s, exhibits 25 J of kinetic energy.

Answer:

6.003×10¯⁶ N

Explanation:

We'll begin by converting 1 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

1 cm = 1 cm × 1 m / 100 cm

1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 3 × / 0.01²

F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

Answer:

one or two three or four

Explanation:

animals of CO2 gas find the final temperature of gas

The final temperature of the 2 moles of gas has been 0.0014 K.

The gas has been assumed to be the ideal gas. The moles of the has been n. According to the ideal gas equation:

(a) PV= nRT.

For the reversible expansion of 1 mole of gas:

P = [tex]\rm \dfrac{RT}{V - b}\;\times\;\dfrac{a}{V^2}[/tex]

For n moles of gas, the reversible expansion will be:

P = [tex]\rm \dfrac{nRT}{V - b}\;\times\;\dfrac{a}{V^2}[/tex]

The final temperature of the reversible expansion has been:

[tex]\rm T_f\;=\;T_i\;+\;\dfrac{an^2\;[\frac{1}{Vf\;-\;\frac{1}{Vi} } ]}{Cv}[/tex]

(B) The ideal gas equation can be given as:

[tex]\rm \dfrac{P1V1}{nT1}\;=\; \dfrac{P2V2}{nT2}[/tex]

Substituting the values:

[tex]\rm \dfrac{40}{2\;\times\;350}\;=\;\dfrac{80}{2\;\times\;T2}[/tex]

T2 = 0.0014 K.

The final temperature of the 2 moles of gas has been 0.0014 K.

(c) The work done can be given by:

W = nRT In[tex]\rm \dfrac{V2}{V1}[/tex]

The internal energy of the system has been given by:

[tex]\Delta[/tex]U = [tex]\rm C_v[/tex][[tex]\rm T_f\;-\;T_i[/tex]] - a[tex]\rm n^2[/tex] [tex]\rm [\dfrac{1}{Vf}\;-\;\dfrac{1}{Vi} ][/tex]

For more information about the reversible expansion of gas, refer to the link:

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Answer:

Explanation:

In a standing wave function[tex]\psi (x,t) = A sin(kx)[/tex] characterized for x between (0.a). on the off chance that the amplitude of the wave interchange from positive to negative at the interval. there probably been a node at [tex]x_0[/tex], among 0 and a to such an extent that [tex]0<x_0 <a[/tex]. The reasoning is right that the likelihood of discovering the particle at the node [tex]x_0[/tex] is 0 in light of the fact that by definition, the nodes of the wave are the place where the wave function falls and is equivalent to 0. Since the likelihood of discovering a particle at a position [tex]x_0[/tex] at time [tex]t_0[/tex], is provided by [tex]P=|\psi(x_0,t_0)|^2 dx[/tex], this implies that at the nodes of a standing wave,

[tex]P = | \psi (x_0,t_0)|^2 \ dx \\ \\ P = |0|^2 dx \\ \\ P = 0[/tex]

 So the reasoning that the likelihood of the particle being at [tex]x_0[/tex] is 0 is right.  

However, to examine whether the particle can travel from a position  [tex]x <x_0[/tex] to a position of [tex]x_0>x[/tex]. All together words, can the molecule be found on one or the other side of the node?

The appropriate response is yes.  

Recall that in quantum mechanics. wave functions at most present with the likelihood of discovering a particle at a specific time inside a time frame. The wave function doesn't present with an old classical actual trajectory that a particle should follow to go in space: all things being equal, it simply yields chances of whether a particle can be found in a specific spot at a specific time. So the reasoning that a particle can't get from a position [tex]x <x_0[/tex] to a position of [tex]x>x_0[/tex], is incorrect.

Answer:

The answer is "[tex]1155\ \frac{kg}{m^3}[/tex]"

Explanation:

Please find the complete question in the attached file.

[tex]p = p_0 + ?gh[/tex]

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

[tex]h_1 = 11 \ cm\\\\h_2= 3 \ cm[/tex]

The pressures would be proportional to the quantity [tex]11-3 = 8[/tex] cm from below the surface at the interface between both the oil and the liquid.

[tex]\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\[/tex]

       [tex]= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}[/tex]

Answer:

a) the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]

b) the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]

c) the magnitude of the net momentum is 54102.5 kgm/s

Explanation:

Given the data in the question;

a) x-component of the net momentum of this system.

the second vehicle ( sedan ) doesn't have momentum along x-axis, the momentum along x-axis is strictly contributed by the pick up

so;

Px = 2600 kg × 17.0 m/s [tex](-x)[/tex]

Px = 44200 kgm/s [tex](-x)[/tex]

Therefore, the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]

b) y-component of the net momentum of this system

Also, momentum along y-axis is entirely provided by the sedan

Py = 1300 kg × 24.0 m/s [tex](y)[/tex]

Py = 31200 kgm/s [tex](y)[/tex]

Therefore, the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]

c) magnitude of the net momentum?

magnitude of the net momentum P = √( Px² + Py² )

so we substitute  

P = √( (44200)² + (31200)² )

P = √( 2927080000 )

P = 54102.5 kgm/s

Therefore, the magnitude of the net momentum is 54102.5 kgm/s

Answer:

Acceleration = 0.0282 m/s^2

Distance = 13.98 * 10^12 m

Explanation:

we will apply the energy theorem

work done = ΔK.E ( change in Kinetic energy )  ---- ( 1 )

where :

work done = p * t

                  = 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2   given ; m = 1200 kg  and initial V = 0

back to equation 1

473040000 * 10^6  = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

∴ Vf = 887918.92 m/s

i) Determine how fast the rocket is ( acceleration of the rocket )

a = Vf / t

  = 887918.92 / ( 1 year )

  = 0.0282 m/s^2

ii) determine distance travelled by rocket

Vf^2 - Vi^2 = 2as

Vi = 0

hence ; Vf^2 = 2as

s ( distance ) =  Vf^2 / ( 2a )

                     = ( 887918.92 )^2 / ( 2 * 0.0282 )

                    = 13.98 * 10^12 m

Answer:

two least important criteria are A and B.

Explanation:

In an industrial production process there are criteria on the location of the production plants. Among the most important criteria we have that a good quantity is produced from the producer and that the process is cost-beneficial - cats to increase the profit of the company.

There are other desirable but minor criteria

A) The size of the city is of minor importance

b) Finding raw material locally is interesting, but its importance is less, the raw material can be sent from other locations.

Consequently the two least important criteria are A and B.

Answer:

A. The process should be located in a major city.

B. The process should use local, organic reactants.

Explanation:

ap.ex

Answer:

final velocity = 0

because the train stoped

so,

Answer:

What is the question on the testtt

It's the one that has a concave line running upward.

Answer:

its the middle one

Explanation:

i did it on edge 2021

Answer:

conductor, insulator, resists, semiconductor, 3 and 5 group

Explanation: