g four small masses 0.2 kg each are connected by light rods 0.4m long to form a square.what is the moment of interia axis

Explanation:

bro I have no idea fam......

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]

The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

The frequency is 4,6E14 Hz.

What is the frequency?

Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.

Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.

Learn more about frequency here:-https://brainly.com/question/254161

#SPJ2

Answer:

Hey,. its a simple question. hope you learn from the solution. check attached picture

Explanation:

Answer:

the density of the salt water is 1030 kg/m³

Explanation:

Given;

radius of the cylindrical pool, r = 2 m

depth of the pool, h = 1.3 m

specific gravity of the salt water, γ = 1.03

The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa

Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³

The density of the salt water is calculated as;

[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]

Therefore, the density of the salt water is 1030 kg/m³

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d =  9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards

The electric field due to the charge q' at midpoint is

[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.

Explanation:

Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.

Answer:

C. A force must be acting on the object.

Explanation:

This is due to the action of its momentum direction.

[tex].[/tex]

Answer:

B. -40 kgm/s is the answer

Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.

The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.

Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.

Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.

Learn more at https://brainly.com/question/24372632

Answer:

I think 9.5

Explanation:

............

Answer:

Various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

The vector sum is the algebraic sum if the two vectors have the same direction.

The sum vector is zero if the two vectors have the same magnitude and opposite direction

Vector addition is a process that can be performed graphically using the parallelogram method, see  attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.

There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel

case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector

case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.

In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector

a) Magnitudes: [tex]\| \vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| \ge 0[/tex]; Directions: [tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex], [tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex], [tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

b) Magnitudes: [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| = 0[/tex]; Directions: [tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex], [tex]\theta_{C}[/tex] is undefined.

a) Let suppose that [tex]\vec A \ne \vec O[/tex], [tex]\vec B \ne \vec O[/tex] and [tex]\vec C \ne \vec O[/tex], where [tex]\vec O[/tex] is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex]:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}[/tex]

[tex]\| \vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}[/tex]

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}[/tex]

[tex]\|\vec C\| \ge 0[/tex]

Direction ([tex]\vec A[/tex])

[tex]\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}[/tex]

[tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex].

Direction ([tex]\vec B[/tex])

[tex]\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}[/tex]

[tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex].

Direction ([tex]\vec C[/tex])

[tex]\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}[/tex]

[tex]\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}[/tex]

[tex]\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}[/tex]

[tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

Please notice that [tex]\vec u[/tex] is the Vector Unit.

b) Let suppose that [tex]\vec A \ne \vec O[/tex] and [tex]\vec B \ne \vec O[/tex] and [tex]\vec C = \vec O[/tex]. Hence, [tex]\vec A = -\vec B[/tex]. In other words, we find that both vectors are antiparallel to each other, that is, that angle between [tex]\vec A[/tex] and [tex]\vec B[/tex] is 180°. From a) we understand that [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], but [tex]\|\vec C\| = 0[/tex].

Then, we have the following conclusions:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = 0[/tex]

Directions ([tex]\vec A[/tex], [tex]\vec B[/tex]):

[tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex]

Direction ([tex]\vec C[/tex]):

Undefined

Answer:

21.55 m

Explanation:

Answer:

1. True

2. False

3. True

4. True

5. True

Answer:

that is pure falsereeeeeeeee

Explanation:

Answer:

The SI unit of power is watt

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]

Where:

[tex]v_{i}[/tex]: is the initial velocity = 20 m/s

[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²  

[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³

[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]

[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]

[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]

Therefore, the exit area of the nozzle is 23.6 cm².

You can find another example of mass flow rate here: https://brainly.com/question/13346498?referrer=searchResults

I hope it helps you!                                                                   

a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

[tex]\dot m_{in} = \dot m_{out}[/tex] (1)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.

[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)

Where:

[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.

[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.

[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.

a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:

[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]

[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]

[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]

[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]

[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]

[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]

[tex]A_{out} = 23.202\,cm^{2}[/tex]

The exit area of the nozzle is 23.202 square centimeters.

Answer:

The correct solution is:

(a) 1.66

(b) 1.05

Explanation:

Given:

Bending stress,

[tex]\sigma_b = 25 \ kpsi[/tex]

Torsional stress,

[tex]\tau= 15 \ kpsi[/tex]

Yield stress of steel bar,

[tex]\delta_y = 60 \ kpsi[/tex]

As we know,

⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]

        [tex]= \sqrt{(25)^2+3(15)^2}[/tex]

        [tex]=36.055 \ kpsi[/tex]

(a)

The factor of safety against static failure will be:

⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]

By putting the values, we get

        [tex]=\frac{60}{36.055}[/tex]

        [tex]=1.66[/tex]

(b)

According to the Goodman line failure,

[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]

[tex]S_e = 40 \ kpsi[/tex]

[tex]\sigma_m = \sqrt{3} \tau[/tex]

     [tex]=\sqrt{3}\times 15[/tex]

     [tex]=26 \ kpsi[/tex]

[tex]Sut = 80 \ kpsi[/tex]

⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]

      [tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]

              [tex]\eta_y = 1.05[/tex]

Answer:

Lateral inversion will occur in a plane mirror.

Explanation:

When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.

A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.

Answer:

Here , mass of bucket ,m = 3.2 Kg

Now , let the tension in upper rope is T1

the tension in the middle rope is T2

a)

For lower bucket, balancing forces in vertical direction

T2 - mg = 0

T2 = mg

T2 = 3.2 *9.8

T2 = 31.36 N

tension in the middle rope is 31.36 N

For the upper bucket , balancing forces in vertical direction

T1 - T2 - mg = 0

T1 = T2 + 3.2 *9.8

T1 = 62.72 N

the tension in the upper rope is 62.72 N

B)

for a = 1.25 m/s^2

Using second law of motion ,for both the buckets

Fnet = ma

T1 - 2mg = 2m*a

T1 = 2*3.2*(9.8 +1.25)

T1 = 70.72 N

the tension in the upper rope is 70.7 N

Now , the lower bucket

Using second law of motion,

T2 - mg = ma

T2 = 3.2 * (9.8 + 1.25)

T2 = 35.36 N

the tension in the lower rope is 35.36 N

Answer:

Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)

Explanation:

Time=Distance/Speed

=14.5/6.63

=2.19s(to 3s.f.)

Acceleration=Final Velocity(v)-Initial Velocity(u)/Time

9.81=v-6.63/2.19

v-6.63=21.5

v=28.1m/s

Answer:

Explanation:

That is an amazing fact.

The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.

The answer is D

Answer:

I hope this is correct!

Explanation:

a) work = (force)(displacement)

We know that d = 5m, now we just need to find the force

We need to calculate the acceleration first using a = v^2 - u^2 / 2d

final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m

a = 2.1^2 - 0^2 / 2(5)

  = 0.441 m/s^2

Now we can find force using f = ma

f = (14kg)(0.441m/s^2)

 = 6.174 newtons

Finally, you can calculate work now that you have force!

work = (6.174n)(5m)

       = 30.87 J (you may need to round sig figs!)

b) Work done by friction = (coefficient of kinetic friction)(mg)(v)

m = 14kg, g = 9.8, coefficient of friction = 0.2

force due to friction = (0.2)(14kg x 9.8)(2.1)

                                 = 57.624 J (again you may need to round to sig figs)

c) Work you perform = total work in direction of motion + work done by friction

total work = 30.87 J, work done by friction = 57.624 J

Work you perform = 30.84 + 57.624

                               = 88.464 J

d) Force you applied = work you performed / distance

work you performed = 88.464 / 5 m

                                   = 17.6929 N (again you may need to round to sig figs)

Hope this helps ya! Best of luck <3

Answer:

Force,

[tex]F = \frac{kQ_{1} Q_{2} }{ {r}^{2} } \\ F = \frac{(9 \times {10}^{9}) \times (25 \times {10}^{ - 6}) \times (10 \times {10}^{ - 6} ) }{ {(0.85)}^{2} } \\ \\ F = 3.114 \: newtons[/tex]

The magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away would be 311.4 N.

Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.

The math mathematical expression for the coulomb's law given as

F= k Q₁Q₂/r²

where F is the force between two charges

k is the electrostatic constant which is also known as the coulomb constant,it has a value of 9×10⁹

Q₁ and Q₂ are the electric charges

r is the distance between the charges

As given in the problem two charges a 25μC charge exerts on a -10μC charge 8.5cm away

By substituting the respective values in the above formula of Coulomb law

F =9×10⁹×(25×10⁻⁶)×(-10×10⁻⁶)/(8.5×10⁻²)²

F= -311.4 N

A negative sign represents that the force is attractive in nature

Thus, the magnitude of the force is 311.4 N.

Learn more about Coulomb's law from here

https://brainly.com/question/506926

#SPJ2

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz

Answer:

[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]

Answer:

No ejection of photo electron takes place.

Explanation:

When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.

The minimum energy required to just  eject an electron is called work function.

The photo electric equation is

E = W + KE

where, E is the incident energy, W is the work function and KE is the kinetic energy.

W = h f

where. h is the Plank's constant and f is the threshold frequency.

Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]

Therefore,

[tex]$AD-BD = \frac{\lambda}{2}[/tex]

[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2 \times 0.4985$[/tex]

[tex]$\lambda = 0.99714 \ m$[/tex]

Thus the frequency is :

[tex]$f=\frac{v}{\lambda}$[/tex]

[tex]$f=\frac{340}{0.99714}$[/tex]

[tex]f=340.9744[/tex] Hz

Answer:

THIS IS YOUR ANSWER:

☺✍️HOPE IT HELPS YOU ✍️☺