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Answer:

The intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

[tex]I(dB) = 10Log(\frac{I}{I_o} )[/tex]

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

[tex]120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2[/tex]

The intensity of sound of a whisper

[tex]20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2[/tex]

Thus, the intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

Answer:

0.1 kg or 100 g

Explanation:

The length of the wire = 0.5 m

the field magnitude = 0.98 T

the current through the wire = 2.0 A

magnetic force due to a wire carrying current is

F = [tex]IlB[/tex]

where

F is the force

[tex]I[/tex] is the current = 2 A

[tex]l[/tex] is the length of the wire

B is the magnetic field strength

Substituting, we have

F = 2 x 0.5 x 0.98 = 0.98 N

This force balances the weight of the mass

weight = mg

where m is the mass of the wire

g is acceleration due to gravity = 9.81 m/s^2

therefore, weight = m x 9.81 = 9.81m

equating this weight with the force, we have

0.98 = 9.81m

m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g

Answer:

100 g

Explanation:

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m

Answer:

v = speed

c = speed of light

using the equation

L = Lc Sqrt( 1 - (v/c)2)

9 = 14 Sqrt( 1 - (v/c)2)

v/c = 0.765

v = 0.765 c

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

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Answer:

0.273m/s

Explanation:

first find out the meaning of 0.90×10−4m3/s

literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s

1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?

cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²

so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²

they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s

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:

D. evacuated tube collector due to reduced heat loss

Explanation:

Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.

During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

Check attached photo

Check attached photo

Answer:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

Answer:

far away

Explanation:

There are different types of eye defect ranging from short sightedness, longsighted, astigmatism, presbyopia etc.

If someone is only able to see close ranged object clearly but not far distant object, then such person is suffering from short sightedness or myopia. This occurs when the light rays entering the eye does not converge on the retina. Instead of converging on the retina, the light ray is formed on a point in front of the retina. This causes the distance from the eye's lens to the retina shorter compared to that of a normal eye. This eye defect is usually corrected using concave lens in order to diverge the rays thereby allowing it to focus on the retina.

Hence, if the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are far away (at a far distant).

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

Answer:

1. Berenice  = 0.67 D

2. Avishka  = 1.50 D

3. Francesca  = 2.00 D

4. Edouard = 2.75 D

Explanation:

The farsighted people are those with near point greater than 25 cm.

They include Avishka, Berenice, Edouard and Francisca.

A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative

From the lens formula, 1/f = 1/v  1/u; but v is negative

Therefore, 1/f = 1/u - 1/v

But, power of a lens = 1/f in meters.

Therefore,  P =  1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

P = 1/ 0.25 - 1/0.4 = 1.5 D

For Berenice, u = 0.25 m, v = 0.3 m

P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v = 0.80 m

P = 1/0.25 - 1/0.80 =2.75 D

For Francesca, u = 0.25 m, v = 0.50 m

P = 1/0.25 - 1/0.5 = 2.0 D

When The farsighted people are those with a near point greater than 25cm.

Berenice is = 0.67 D

Avishka is = 1.50 D

Francesca is = 2.00 D

Edouard is = 2.75 D

When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.

Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.

Also, From the lens formula, 1/f = 1/v 1/u; but v is negative

Thus, 1/f = 1/u - 1/v

But, when the power of a lens = 1/f in meters.

Thus, P = 1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

After that, P = 1/ 0.25 - 1/0.4 = 1.5 D

Then, For Berenice, u = 0.25 m, v = 0.3 m

Now, P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v is = 0.80 m

Then, P = 1/0.25 - 1/0.80 is =2.75 D

Now, For Francesca, u = 0.25 m, v is = 0.50 m

Therefore, P = 1/0.25 - 1/0.5 = 2.0 D

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Answer:

whats the formula

Explanation:

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]

[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]

[tex]\alpha=\cos^{-1}(0.3846)[/tex]

[tex]\alpha=67.38^{\circ}[/tex]

We need to calculate the angle β

Using cosine law

[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]

[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]

[tex]\beta=\cos^{-1}(0.923)[/tex]

[tex]\beta=22.63^{\circ}[/tex]

We need to calculate the force on 130 cm side

Using formula of force

[tex]F_{130}=ILB\sin\theta[/tex]

[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]

[tex]F_{130}=0[/tex]

We need to calculate the force on 120 cm side

Using formula of force

[tex]F_{120}=ILB\sin\beta[/tex]

[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]

[tex]F_{120}=0.1385\ N[/tex]

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

[tex]F_{50}=ILB\sin\alpha[/tex]

[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]

[tex]F_{50}=0.1385\ N[/tex]

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.

b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.

c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.

Given the following data:

Let us assume the current is flowing in a counterclockwise direction in the right-angle triangle.

First of all, we would determine the angles by using cosine rule:

[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]

[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]

a. To the determine the magnitude of the magnetic force on the 130 cm side:

Mathematically, the force acting on a current in a magnetic field is given by the formula:

[tex]F = BILsin\theta[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]

b. To the determine the magnitude of the magnetic force on the 120 cm side:

[tex]F_{120}=BILsin\beta[/tex]

[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]

c. To the determine the magnitude of the magnetic force on the 50 cm side:

[tex]F_{50}=BILsin\alpha[/tex]

[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]

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Answer:

The time taken is  [tex]t = 2.739 *10^{8} \ s[/tex]

Explanation:

From the question we are told that

    The speed of the spacecraft is [tex]v = 0.99c[/tex]

    where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]

    =>   [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]

    The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]

      [tex]t = 2.739 *10^{8} \ s[/tex]

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

Answer:

364days

Explanation:

Pls see attached file

Explanation:

The moon will take 112.7 days to make one revolution around the planet.

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

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Answer:

1) f = 214 Hz , 2)  answer is c , 3) f = 428 Hz , 4)   f₂ = 428 Hz ,   f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

                L = λ / 2

               λ  = 2L

               λ  = 2 0.8

               λ  = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

                v =λ  f

                f = v / λ  

                f = 343 / 1.6

                f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

           λ' = 2 L ’

          as L ’<L₀

          λ' <λ₀

          f = v / λ'

          f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

          λ = 2 0.4 = 0.8 m

          f = 343 / 0.8

          f = 428 Hz

4) the different harmonics are described by the expression

         λ = 2L / n           n = 1, 2, 3

         λ₂ = L

         f₂ = 343 / 0.8

         f₂ = 428 Hz

         λ₃ = 2 0.8 / 3

         λ₃ = 0.533 m

         f₃ = 343 / 0.533

         f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

       the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

        L = λ / 4

        L = 1.6 / 4

        L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

           λ = 4L / n         n = 1, 3, 5 (2n + 1)

therefore only odd values ​​can produce standing waves

           λ₃ = 4L / 3

           λ₃ = 4 0.4 / 3

           λ₃ = 0.533

           f₃ = 343 / 0.533

           f₃ = 643 Hz

The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

The given parameters include:

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]

To learn more about projectile calculations please visit: https://brainly.com/question/14083704

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]