"Pipe A has length L and is open at one end and closed at the other. Pipe B is open at both ends and has length 2L. Which harmonic of pipe B matches in frequency the fundamental of pipe A?"

Answer:

The frequency increases by 4 because it is inversely proportional to the wavelength.

Answer:

   h = 4 sin (314.15 t)

Explanation:

This is a kinematics exercise, as the system is rotating at a constant speed.

          w = θ / t

          θ = w t

in angular motion all angles are measured in radians, which is defined

         θ = s / R

   we substitute

          s / R = w t

          s = w R t

let's reduce the magnitude to the SI system

    w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s

let's calculate

   s = 314.16 4 t

   s = 1,256.6 t

this is the value of the arc

Let's find the function of this system, let's use trigonometry to find the projection on the x axis

                  cos θ = x / R

                  x = R cos θ

                  x = R cos wt

projection onto the y-axis is

               sin θ = y / R

how is a uniform movement

               θ = w t

               y = R sin wt

In the case y = h

              h = R sin wt

              h = 4 sin (314.15 t)

Answer:

Maximum height, h = 3062.5m

Total time of flight, T = 49.49secs or 50secs approx.

Range, R = 12250m

Explanation:

Given data:

U = 350m/s

Angle = 45°

Assume g = 10m/s

At the greatest height, v = 0

Therefore,

V^2 = U^2 sin^2 × angle - 2×g×h

Substituting values:

0^2 = 350^2 sin^2 (45) - 2 × 10 × h

Let h = maximum height reached

Rearranging gives:

350^2 sin^2(45) = 2 x 10 x h

h = 350^2 sin^2(45)/2×10

h = 122500 x 0.5/20

h = 61250/20

h = 3062.5m

2)Total time of flight, T

T = 2U sin(angle)/g

= 2x350 sin(45)/10

= 494.9747/10

= 49.49secs or 50sec approx.

3) Range of projectile, R

R = U^2 sin2(angle)

= 350^2 sin2 (45)

= 122500 x 1/10

= 12250m

Answer:

[tex]I=2.67\times 10^{-6}\ W/m[/tex]

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]

So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Answer:

8.9*10^6 V/m

Explanation:

The expression for electric field strength E is given as

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

where V= voltage

           d= distance of separation

Given data

[tex]voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9[/tex]

substituting our given data into the electric field strength formula we have

[tex]E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6 V/m[/tex]

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:

neither will happen

Explanation:

cause the water is already defreezed

Answer:

A- to simplify the study of things

Explanation:

a visual reference

Answer:

Radiation is transferred through electromagnetic waves so D.

Explanation:

Answer:

D. Waves

Explanation:

a and b don't make much sense, conduction is transfer of energy through touch

Answer:

64 times

Explanation:

if increase of 1 gives you 32

then increase of 2 will give you its double

64

If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.

An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.

The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.

The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.

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Answer:

10 kgm/s

Explanation:

Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.

From the question,

ΔM = m(v-u)...................... Equation 1

Where ΔM = change in momentum, u = initial velocity, v = final velocity.

Note: Let upward direction be negative, and downward direction be positive.

Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s

Substitute into equation 1

ΔM = 0.2(-20-30)

ΔM = 0.2(-50)

ΔM = -10 kgm/s.

The negative sign shows that the change in momentum is Upward

The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

Given data:

The mass of rubber ball is, m = 0.2 kg.

The initial speed of ball is, u = 30 m/s.

The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)

The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,

[tex]p= m ( v-u)[/tex]

Solving as,

[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]

Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

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Answer:

E = 12.25 x 10³ N/C = 12.25 KN/C

Explanation:

In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,

Electrostatic Force = Weight

E q = mg

where,

E = Electric Field = ?

m = Mass of the Charge = 5 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C

Therefore,

E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)

E = 0.049 N/4 x 10⁻⁶ C

E = 12.25 x 10³ N/C = 12.25 KN/C

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 answer is c

Explanation:

the balloon is lighter than the air around displaced by it

Answer:

1) c. Helium

2) Iron

3) False.

Explanation:

1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.

2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.

3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.

Answer:

a. 8.465 m/s

b.22.3659

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a          --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r          [v = linear velocity of particle, r = radius of circular path]

Therefore, equation (i) becomes;

F = m v²/ r             --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ          -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

Combine equations (ii) and (iii) as follows;

m (v² / r) = qvBsinθ         [divide both side by v]

m v / r = qBsinθ              [make r subject of the formula]

r = (m v) / (qBsinθ)              ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

Answer:

W = 1014 J = 1.014 KJ

Explanation:

As, Sam has to stop the boats in the log ride. Therefore, the work Sam needs to do, in order to stop a boat must be equal to the kinetic energy of the boat:

Work Done by Sam = Kinetic Energy of the Boat

W = K.E

W = (1/2)mv²

where,

m = mass of boat and its rider = 1200 kg

v = speed of the boat = 1.3 m/s

Therefore,

W = (1/2)(1200 kg)(1.3 m/s)²

W = 1014 J = 1.014 KJ

Answer:

b)using

DT = (LAl - LBr) / (LBr aBr - LAl aAl)

DT = (10.02-10)/(10*19x10-6 –10.02*24x10-6)

DT = -396 C°

20°C + -396 C° < -273.15 °C;

So the temperature will be -396° this is unattainable because we can’t go below absolute zero

Answer:

20J

Explanation:

The computation of the change in kinetic energy is shown below:

As we know that

Work was done on system =  change in potential energy + change in kinetic energy + change in thermal energy

-430J = -510J + change in kinetic energy + 100J

-430J = -410J + change in kinetic energy

So, the change in kinetic energy is 20J

We simply applied the above formula to find out the change in kinetic energy

Answer:

When we have completely polarized light with intensity I0, and it passes through a polarizing filter with its axis at an angle θ with respect to the light's polarization direction, the new intensity of the light will be:

I = I0*cos^2(θ)

This is called the "Malus' law".

in this case, we have:

I0 =  125 W/m^2

θ = 89.5°

then:

I = (125 W/m^2)*cos^2(89.9°) = 0.00038 W/m^2

Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V

The root mean square current is given as;

[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]

[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]

Inductive reactance is given by;

[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]

Inductance is given by;

[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.

Given the following data:

To determine the value of the inductance (L):

First of all, we would find the root mean square (rms) current by using the formula:

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]

Next, we would calculate the inductive reactance of the oscillator by using the formula:

[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]

Now, we can solve for the value of the inductance (L):

[tex]L = \frac{X_L}{2\pi f}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]

L = [tex]9.55 \times 10^{-4}[/tex] Henry.

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Answer:

He was going at 54.5mph

Explanation:

Measuring distances in feet and time (starting with the brakes being applied) in seconds,

we have a(t) = −16, s(0) = 0, and v(T) = 0, s(T) = 200,

where T is the time of braking.

So Finding antiderivatives, v(t) = −16t + A and so s(t) = −8t²+ At +B. Since s(0) = 0,

we have B = 0. Since v(T) = 0,

we have T = A /16.

Putting this into s() gives

200 = −8(A/ 16)² +A· (A /16) = A²/32.

This gives A2 = 6,400, so A = 80 = v(0). That is, the initial speed was 80 ft/s ≈ 54.5 mph.

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

Answer:

15.01 m/s

Explanation:

Applying newtons equation of motion,

v² = u²+2gs.................. Equation 1

Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.

Given: u = 21 m/s, s = 11 m

Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).

Substitute these values into equation 1

v² = 21²-2(11)(9.8)

v² = 441-215.6

v² = 225.4

v = √225.4

v = 15.01 m/s

Answer:

Potential gradient = 1.6 v/m  

E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l

l=1.2\1.6

=3\4

3\4=0.75m

convert m into cm 0.75x10

0.75m=75 cm

ans=75 cm

Explanation:

Answer:

Velocity of bear (v) = 20 m/s

Explanation:

Given:

Mass of bear (m) = 500 kg

Mechanical kinetic energy (K.E) = 100,000 J

Find:

Velocity of bear (v) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2(m)(v)²

100,000 = 1/2(500)(v)²

200,000 = 500(v)²

400 = (v)²

Velocity of bear (v) = 20 m/s

Answer:

   ΔP = - 0.262 ρ v₁²

Explanation:

This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

suppose the pipe is horizontal, therefore

           y₁ = y₂

         P₁ -P₂ = ½ ρ (v₁² -v₂²)             1

Now let's use the continuity equation

         v₁ A₁ = v₂ A₂                              2

The area of ​​a circle is

         A = π r²

also the radius is half the diameter r = d /2

         A₁ = π d₁² / 4

let's substitute in equation 2

         v1 π d₁² / 4 = v2 π d₂² / 4

          v₁ d₁² = v₂ d₂²

          v₂ = v₁ d₁² / d₂²

let's substitute in equation 1

          P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )

Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1

           d₂ = D -0.1D = 0.9 D

we substitute in the previous equation

          P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)

          P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )

          P1 - P2 = ½ ρ v12 (- 0.524)

          ΔP = - 0.262 ρ v₁²

in this solution we assume that the data in zone I is known