A primary coil has 360 turns, while secondary has 120 turns in a transformer. The output voltage is 220V. What is the input voltage, and what type of transformer is it

Answer: F = 498.04 lbs

Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.

Force of friction is given by the formula

[tex]F_{f}=\mu.F_{N}[/tex]

where

μ is coefficient of friction

[tex]F_{N}[/tex] is normal force

Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.

In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So

[tex]F_{N}=m.g[/tex]

Coefficient of friction is how much friction exists between two surfaces.

Rearraging friction force is

[tex]F_{f}=\mu.m.g[/tex]

Mass for this system is the sum of Julie and the sled, therefore

m = 109 + 12

m = 121 lb

Calculating Friction Force:

[tex]F_{f}=0.42.121.9.8[/tex]

[tex]F_{f}=[/tex] 498.04 lbs

LBS is a unit of measurement referred as pound by weight.

In conclusion, force Mary needs to start moving the sled is 498.04 lbs

Answer:

40 N

Explanation:

According to the scenario, computation of given data are as follows:

Mass (m) = 20 kg

Initially moving (v) = 3

Actual distance (d) = 2.25 m

So, we can calculate friction (f) by using following formula,

f × d = [tex]\frac{1}{2} mv^{2}[/tex]

By putting the value, we get

f × 2.25 = [tex]\frac{1}{2}[/tex] × 20 × [tex]3^{2}[/tex]

f × 2.25 = 10 × 9

f = 90 ÷ 2.25

= 40 N.

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm

Answer:

1 kg

Explanation:

Assuming that,

Δx(2) = v(2)t, where Δx(2) = d and v(2) = 2m1 / (m1 + m2) v1i

On the other hand again, if we assume that

Δx(1) = v(1)t, where Δx(1) = -2d, and v(1)t = m1 - m2 / m1 + m2 v1i

From the above, we proceed to dividing Δx(2) by Δx(1), so that we have

d/-2d = [2m1 / (m1 + m2) v1i] / [m1 - m2 / m1 + m2 v1i], this is further simplified to

1/-2 = [2m1 / (m1 + m2)] / [m1 - m2 / m1 + m2]

1/-2 = 2m1 / (m1 + m2) * m1 + m2 / m1 - m2

1/-2 = 2m1 / m1 - m2, if we cross multiply, we have

m1 - m2 = -2 * 2m1

m1 - m2 = -4m1

m2 = 5m1

From the question, we're told that m1 = 0.2 kg, if we substitute for that, we have

m2 = 5 * 0.2

m2 = 1 kg

I believe the answer would be protentional because they have the potential energy in them to lift the hammer.

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

The value of  20m/s, east is a vector quantity is Hence, option (C) is correct.

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

Given values  200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).

But The value of  20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a  vector quantity is Hence, option (C) is correct.

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Answer:

b. Designing an uncontrolled experiment.

Explanation:

They always have it controlled.

Answer:

B. Designing an uncontrolled experiment.

Explanation:

Correct Answer!!!!!!

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Answer:

cognitive, humanistic, behavioral, sociocultural

Explanation:

Behavioral, sociocultural, cognitive, and humanistic are approaches to psychological science.

Psychology is a term to refer to the discipline that focuses on the study of various topics related to human thought such as:

Due to the above, several subdisciplines have emerged that focus on the study of each of the topics. For example:

Behavioral psychology: focused on the study of human behavior.

Sociocultural psychology: focused on the study of human behavior and thought in different social situations.

Cognitive psychology: focused on mental processes related to learning.

Humanistic psychology: focused on the study of human thought from a comprehensive approach.

According to the above, options A, E, F, and G are correct because they mention different sub-disciplines of psychology while the other options mention terms that are not related to sub-disciplines or psychological sciences.

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Answer:

[tex]m=1.63\times 10^{-27}\ kg[/tex]

Explanation:

The velocity of a particle is 90% of the speed of light.

The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]

We need to find the mass of the particle.

The formula for the wavelength of a particle is given by :

[tex]\lambda=\dfrac{h}{mv}[/tex]

h is Planck's constant

v is 90% of speed of light

m is mass of the particle

[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]

So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].

Answer:

3kg

Explanation:

Given parameters:

Force  = 9N

Acceleration  = 3m/s²

Unknown:

Mass = ?

Solution:

From Newton's second law of motion:

        Force  = mass x acceleration

So;

             9  = mass x 3

             mass  = 3kg

Answer:

2.59m/s

Explanation:

Using the equation of motion

v = u+at

v is the final velocity = 35ms

u is the initially velocity = 9m/s

t is the time = 13.5s

a is the acceleration

Substitute into the formula

35 = 0+13.5a

a = 35/13.5

a = 2.59m/s²

Hence the acceleration of the motorcycle is 2.59m/s

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]   (1)

Where:

x: is the displacement

[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)                    

[tex]v_{0}[/tex]: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

Answer:

hello your question has a missing information

The other is Δ-connected with an impedance of (60 - j45) ohm per phase.

answer : A) 5A ∠0° ,

               p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

 B) 193.64 v

C) current at load 1 = 2.236 A , current at load 2 = 4.472 A

 D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )

      Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )

Explanation:

First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution

A) determine the current, real power and reactive power delivered by the sending-end source

current power delivered (Is)  =  5A ∠0°

complex power delivered ( s ) = 3vs Is  

                                                  = 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )

also s = p + jQ  ------ ( 2 )

comparing equation 1 and 2

p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

B) determine Line-to-line voltage at the load

Vload = √3 * 111.8

           = 193.64 v

c) Determine current per phase in each load

[tex]I_{l1} = Vl1 / Zl1[/tex]

     = [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A   hence current at load 1 = 2.236 A

[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]  

     = [tex]\frac{111.8<-10.3}{25<-36.87}[/tex]  = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A

D) Determine the Total three-phase real and reactive powers absorbed by each load

For load 1

3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR

for load 2

3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex]  = 1200 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR

The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.

(a) The impedance per phase of the equivalent Y,

[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]

The phase voltage,

[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]

Total impedance from the input terminals,

[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]

The three-phase complex power supplied  [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]  

P =1800 W and Q = 0 VAR delivered by the sending-end source.

(b) Phase voltage at load terminals will be,  

[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]  

The line voltage magnitude at the load terminal,  

[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]    

(c) The current per phase in the Y-connected load,  

[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} ​}[/tex]

The phase current magnitude,  

[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]

(d) The three-phase complex power absorbed by each load,

[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]

The three-phase complex power absorbed by the line is  

[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]

Since, the sum of load powers and line losses,  

[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]

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Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

[tex]H_m=1.65m[/tex]

[tex]H_E=1.16307m[/tex]

Explanation:

From the question we are told that

Mass of ball [tex]M=2kg[/tex]

Length of string [tex]L= 2m[/tex]

Wind force [tex]F=13.2N[/tex]

Generally the equation for [tex]\angle \theta[/tex] is mathematically given as

[tex]tan\theta=\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]

[tex]\theta=73.14\textdegree[/tex]

Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]

Generally the equation for max Height [tex]H_m[/tex] is mathematically given as

[tex]H_m=L(1-cos146.28)[/tex]

[tex]H_m=0.9(1+0.8318)[/tex]

[tex]H_m=1.65m[/tex]

Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as

[tex]H_E=L(1-cos73.14)[/tex]

[tex]H_E=0.9(1+0.2923)[/tex]

[tex]H_E=1.16307m[/tex]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Answer:

electrons are located around the nucleus of an atom.

Explanation:

Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.

Answer:

The Carnot engine has zero power

Explanation:

Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.

This is because all its processes are reversible (that is isothermic and adiabatic).

So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.

Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.

This is not practical.

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

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Answer:

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a

Answer:

V= -3.6*10⁻¹¹ V

Explanation:

      [tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]

       [tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]

      [tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]

       [tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]

The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]

We can arrive at this answer as follows:

[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]

[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]

[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]

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Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2[tex]\pi[/tex]  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

Answer:

La aceleración del autobús es -5.80 m/s².

Explanation:

Podemos encontrar la aceleración del autobús usando la siguiente ecuación:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)

[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h

d: es la distancia recorrida = 60 m

Por lo tanto, la aceleración es:

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]

El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).

Entonces, la aceleración del autobús es -5.80 m/s².

Espero que te sea de utilidad!                      

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units