11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

The formula of the pressure of the static fluid

P1 = P2+ρgh

Where,

P1 -  pressure at depth 'h'

P2 -  atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] =  1670900 Pa

h - Depth = 160m  

ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]

g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]

The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]  

[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]  

 For an ideal gas,  

PV =nRT  

[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]  

[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]

So,

[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]  

Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].

To know more air bubble volume,

https://brainly.com/question/10509397

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

By working with the vector forces, we will get:

Remember that we can directly add vector forces, so if our two forces are:

F₁ = <13.5 N, -7.5 N>

F₂ = < -12 N, -4.70 N>

Then the resultant force is:

F = F₁ + F₂ = <13.5 N + (-12 N), -7.5 N + ( -4.70 N) >

F = < 1.5 N, -12.2 N>

so we have:

a) The x-component is 1.5 N

b) The y-component is -12.2 N

c) The magnitude will be:

|F| = √( (1.5 N)^2 + (-12.2 N)^2) = 12.29 N

d) The direction of a vector <x, y> measured counterclockwise from the positive x-axis is given by:

θ = Atan(y/x)

Where Atan is the inverse tangent function, then here we have:

θ = Atan(-12.2 N/1.5 N) = 277.01°

If you want to learn more about vectors, you can read:

https://brainly.com/question/7802337

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]

Then the addition of these two velocities equals:

[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]

Answer:

50k/h is the answer to iy

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

Answer:

D

Explanation:

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 [tex]\theta _d =0.19 ^o[/tex]

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  [tex]n_F = 1.497[/tex]

     The  wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]

    The  refractive index of the transparent acrylic plastic for red light is  [tex]n_C = 1.488[/tex]

       The  wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]

    The incidence angle is  [tex]i = 45^o[/tex]

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       [tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]

      [tex]r_F = 28.18^o[/tex]

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       [tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]

      [tex]r_F = 28.37^o[/tex]

The angle between the blue beam and the red beam in the acrylic block

     [tex]\theta _d = r_C - r_F[/tex]

substituting values

       [tex]\theta _d = 28.37 - 28.18[/tex]

       [tex]\theta _d =0.19 ^o[/tex]

Answer:

a)   d sin θ = m λ₀ / n

b)   d sin θ = (m + ½) λ₀ / n

c)    d = 2,439 10⁻⁷ m

Explanation:

For the interference these rays of light we must take as for some aspects,

* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2

* when the ray penetrates the lens the donut length changes by the refractive index

            λ = λ₀ / n

now let's write the destructive interference equation for these lightning bolts

           d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n

           d sin θ = m λ₀ / n

b) now nc> np

in this case there is no phase change in the reflected ray and the equation for destructive interference remains

             d sin θ = (m + ½) λ₀ / n

c) select the value of nc = 2.05 of the ZnO

we calculate the thickness of the film (d)

            d = m λ / (n sin 90)

in this type of interference the observation is normal, that is, the angle is 90º)

           d = 1 500 10-9 / (2.05 1)

           d = 2,439 10⁻⁷ m

d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference

Answer:

the planet Earth is a good example

Answer:

The approximate displacement of the object is  23  m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]

= 23 m

Answer:

The voltage across the capacitor will remain constant

The capacitance of the capacitor will increase

The electric field between the plates will remain constant

The charge on the plates will increase

The energy stored in the capacitor will increase

Explanation:

First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.

The charge on the capacitor is equal to

Q = CV

Since the voltage is constant, and the charge increases, the capacitance will also increase.

The energy in a capacitor is given as

E = [tex]\frac{1}{2}CV^{2}[/tex]

since the capacitance has increased, the energy stored will also increase.

Answer:

N=3176.5rulling

Explanation:

We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with

wavelengths whose mean is 540 nm

Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)

Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then

Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)

the formular below can be used to can be used to calculate our minimum number of lines

N= λ /(m Δλ)

Where N is number of fillings i.e number of lines

λ= wavelength

Δλ= difference in wavelength

m=1

Then if we substitute the values we have

,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]

N =3176.5rulling

Therefore, minimum number of lines = =3176.5rulling

Answer:

The  angle is  [tex]\theta = 15.48^o[/tex]

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  [tex]d = 3.66 \ m[/tex]

     The time taken is  [tex]t = 0.455 \ s[/tex]

The  horizontal component of the speed of the dart is mathematically represented as

      [tex]u_x = ucos \theta[/tex]

where u is the the velocity at dart is lunched

  so

      [tex]distance = velocity \ in \ the\ x-direction * time[/tex]

substituting values

      [tex]3.66 = ucos \theta * (0.455)[/tex]

 =>   [tex]ucos \theta = 8.04 \ m/s[/tex]

From projectile kinematics the time taken by the dart can be mathematically represented as

         [tex]t = \frac{2usin \theta }{g}[/tex]

=>    [tex]usin \theta = \frac{g * t}{2 }[/tex]

       [tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]

      [tex]usin \theta = 2.23[/tex]

=>   [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]

       [tex]\theta = tan^{-1} [0.277][/tex]

      [tex]\theta = 15.48^o[/tex]

Answer:

Explanation:

Let the weight of the truck be W . reaction force R = W

Maximum frictional force = μ R

= .8 x W

So for movement of truck

Pulling force = frictional force

10560 = .8W

W = 13200 N

weight of heaviest truck required = 13200 N .

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

use a hammer to hit it

Explanation:

if u hit it u will be able to hear the shattered noise

Answer:

It can coast uphill 6.2m

Explanation:

See attached file pls

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Explanation:

From the question we are told that

   The  width of the slit is  [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]

     The angle is  [tex]\theta = 11^o[/tex]

The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as

        [tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]

Where [tex]\beta[/tex] is mathematically represented as

        [tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]

substituting values

      [tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]

     [tex]\beta = 708.1 \ rad[/tex]

So

  [tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Answer:

The  potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]

Explanation:

From the question we are told that

    The inner radius is  [tex]r_i = 5 \ mm = 0.005 \ m[/tex]

      The outer radius is  [tex]r_o = 11 \ mm = 0.011 \ m[/tex]

     The  common length is  [tex]l = 160 \ m[/tex]

      The  potential  difference is   [tex]V = 6 \ V[/tex]

Generally the capacitance of the cylindrical capacitor is mathematically represented as

       [tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]

Where  [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

and  k  is the dielectric constant  of the dielectric material here the  dielectric material is free space so  k  =   1

     Substituting values

             [tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]

             [tex]C = 1.129 *10^{-8} \ F[/tex]

The potential energy stored is mathematically represented as

       [tex]PE = \frac{1}{2} * C * V ^2[/tex]

substituting values

      [tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]

      [tex]PE = 2.031 *10^{-7} \ J[/tex]

Answer:

0.46N

Explanation:

See attached file

Answer:

4.31 N

Explanation:

Given:

Δy = -55 m

v₀ = 0 m/s

v = -29 m/s

Find: a

v² = v₀² + 2aΔy

(-29 m/s)² = (0 m/s)² + 2a (-55 m)

a = -7.65 m/s²

Sum of forces in the y direction:

∑F = ma

R − mg = ma

R = m (g + a)

R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)

R = 4.31 N

Answer:

The wave takes 0.132 seconds to travel from one end of the string to the other.

Explanation:

The velocity of a transversal wave ([tex]v[/tex]) travelling through a string pulled on both ends is determined by this formula:

[tex]v = \sqrt{\frac{T\cdot L}{m} }[/tex]

Where:

[tex]T[/tex] - Tension, measured in newtons.

[tex]L[/tex] - Length of the string, measured in meters.

[tex]m[/tex] - Mass of the string, measured in meters.

Given that [tex]T = 15\,N[/tex], [tex]L = 7.6\,m[/tex] and [tex]m = 0.034\,kg[/tex], the velocity of the tranversal wave is:

[tex]v = \sqrt{\frac{(15\,N)\cdot (7.6\,m)}{0.034\,kg} }[/tex]

[tex]v\approx 57.522\,\frac{m}{s}[/tex]

Since speed of transversal waves through material are constant, the time required ([tex]\Delta t[/tex]) to travel from one end of the string to the other is described by the following kinematic equation:

[tex]\Delta t = \frac{L}{v}[/tex]

If [tex]L = 7.6\,m[/tex] and [tex]v\approx 57.522\,\frac{m}{s}[/tex], then:

[tex]\Delta t = \frac{7.6\,m}{57.522\,\frac{m}{s} }[/tex]

[tex]\Delta t = 0.132\,s[/tex]

The wave takes 0.132 seconds to travel from one end of the string to the other.