Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000 Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), there is no change in height. Use the Bernoulli equation to calculate the velocity of the water exiting the nozzle. The velocity of the water exiting the nozzle is_______m/s. (Hint: The density of water is 1000 kg/m3).

Answer:

The  self-inductance is  [tex]L = 0.0053 \ H[/tex]

Explanation:

From the question we are told that  

      The number of loops is  [tex]N = 900[/tex]

      The  length of the rod is  [tex]l =6 \ cm = 0.06 \ m[/tex]

      The radius of the rod is  [tex]r = 1 \ cm = 0.01 \ m[/tex]

The  self-inductance for the solenoid is mathematically represented as

        [tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]

Now the cross-sectional of the solenoid is mathematically evaluated as

        [tex]A = \pi r^2[/tex]

substituting values  

         [tex]A =3.142 * 0.01 ^2[/tex]

        [tex]A = 3.142 *10^{-4} \ m^2[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with a value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

    substituting values into above equation

          [tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]

          [tex]L = 0.0053 \ H[/tex]

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:                  

From above;  the distance between speaker  A and speaker B can be expressed as:

[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]

The path length difference  will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length  will be half the wavelength; which is

= [tex]\dfrac{1}{2}*4 \ m[/tex]

= 2 m

The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.

Given data:

The distance between the speakers is, d = 3 m.

The distance between the observer and speaker is, s = 4 m.

The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then  the distance between speaker  A and speaker B can be expressed as:

[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]

And the path length difference is,

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive

interference for that path length  will be half the wavelength; which is

 [tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]

= 2 m

Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.

Learn more about the interference here:

https://brainly.com/question/16098226

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

Answer:

Drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].

Explanation:

Formula for Drift Velocity is:

[tex]u = \dfrac{I}{nAq}[/tex]

Where I is the current

n is the number of electrons in 1 [tex]m^3[/tex] or the electron density

A is the area of cross section and

q is the charge of one electron.

We are given the following:

n = [tex]8.49 \times 10^{28}\ electrons/m^3[/tex]

I = 1 A

A = 0.40 [tex]cm^2[/tex] = 40 [tex]\times 10^{-4}[/tex] [tex]m^{2}[/tex]

We know that q = [tex]1.6\times10^{-19} C[/tex]

Putting all the values to find drift velocity:

[tex]u = \dfrac{1}{8.49 \times 10 ^{28} \times 40 \times 10^{-4}\times 1.6 \times 10^{-19}}\\u = \dfrac{1}{543.36 \times 10 ^{5} }\\u = 1.84 \times 10^{-8}\ m/s[/tex]

So, drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].

Answer:

Maximum speed of the block, [tex]v_{max} = 4.58 m/s[/tex]

Explanation:

Mass of the block, m = 2.0 kg

Spring constant, k = 600 N/m

Spring extension, x = 20 cm = 0.2 m

Speed of the block due to the extension, v = 3.0 m/s

First, Potential energy, PE stored in the spring:

PE = 0.5 kx²

PE = 0.5 * 600 * 0.2²

PE = 12 J

Calculate the kinetic energy of the block due to the extension:

[tex]KE_x = 0.5 mv^2\\KE_x = 0.5 * 2 * 3^2\\KE_x = 9 J[/tex]

The maximum Kinetic Energy of the block will be:

[tex]KE_{max} = 0.5 m v_{max}^2\\KE_{max} = 0.5 * 2 * v_{max}^2\\KE_{max} = v_{max}^2[/tex]

[tex]KE_{max} = KE_x + PE\\v_{max}^2 = 9 + 12\\ v_{max}^2 = 21\\ v_{max} = \sqrt{21} \\ v_{max} = 4.58 m/s[/tex]

Answer:

The workdone is [tex]W_v = - 20 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the object is [tex]m = 2.5 \ kg[/tex]

     The speed of fall is [tex]v = 2.5 \ m/s[/tex]

     The depth of fall is  [tex]d = 80\ cm = 0.8 \ m[/tex]

Generally according to the work energy theorem

      [tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]

Now here given the that the velocity is  constant  i.e  [tex]v_1 = v_2 = v[/tex] then

We have that

    [tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]  

So in terms of workdone by the potential energy of the object and that of the viscous liquid we have

       [tex]W = W_v - W_p[/tex]

Where  [tex]W_v[/tex] is workdone by viscous liquid

             [tex]W_p[/tex] is the workdone by the object which is mathematically represented as

            [tex]W_p = mgd[/tex]

So  

       [tex]0 = W_v + mgd[/tex]

=>    [tex]W_v = - m * g * d[/tex]

substituting values

       [tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]

      [tex]W_v = - 20 \ J[/tex]

Answer:

Electric dipole

Explanation:

the dipole axis makes an angle with the electric field. depending on direction (clockwise/aniclockwise) you get the torque

Hope this helps

Answer:

all of the above

Explanation:

because it is.

Answer:

Gas 1

Explanation:

Answer:

The maximum number of bright spot is [tex]n_{max} =5001[/tex]

Explanation:

From the question we are told that

     The  slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]

      The  wavelength is  [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]

Generally the condition for interference is  

        [tex]n * \lambda = d * sin \theta[/tex]

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  [tex]\theta = 90[/tex]

=>     [tex]sin( 90 )= 1[/tex]

So

     [tex]n = \frac{d }{\lambda }[/tex]

substituting values

     [tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]

     [tex]n = 2500[/tex]

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        [tex]n_{max} = 2 * n + 1[/tex]

The  1  here represented the central bright spot

So  

      [tex]n_{max} = 2 * 2500 + 1[/tex]

     [tex]n_{max} =5001[/tex]      

Answer:

The mass of the second ball is [tex]m_2 = 2.175m \ kg[/tex]

Explanation:

From the question we are told that  

     The mass of the first  ball is  m

Let assume

    The original speed of the first ball is  [tex]u_1[/tex]

     The rebound speed is  [tex]v_1 = 0.370u_1 \ m/s[/tex]

According to the law of momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = -m_1 v_1 + m_2 v_2[/tex]

The  second ball is at rest so  

      [tex]u_2 = 0[/tex]

And the negative sign shows that it is a rebound velocity

        [tex]mu_1 + = -m v_1 + m_2 v_2[/tex]

=>   [tex]mu_1 + = -0.370 mu_1 + m_2 v_2[/tex]

Now  we are told that the collision is  elastic, this means that the velocity of approach will be equal to the velocity of separation

  So  

          [tex]u_1 = v_1 + v_2[/tex]

=>       [tex]u_1 = 0.370u_1 + v_2[/tex]

=>       [tex]v_2 = 0.63 \ u_1[/tex]

substituting this into equation above

[tex]mu_1 + = -0.370 mu_1 + m_2 (0.63 u_1)[/tex]

[tex]1.370 mu_1 = m_2 (0.63 u_1)[/tex]

=>    [tex]m_2 = 2.175m \ kg[/tex]

The mass of the second ball in terms of the first ball, m, is 2.175 m.

The given parameters;

Apply the principle of conservation of linear momentum to determine the mass of the second ball as shown below;

[tex]m_1 u_1 + m_2 u_2 = m_1v_1 + m_2 v_2\\\\mu_1 + m_2(0) = m(-0.37u_1) + m_2v_2\\\\mu_1 = -0.37u_1m + m_2v_2\\\\mu_1 + 0.37mu_1 = m_2v_2\\\\1.37 mu_1 = m_2 v_2 \ --(1)[/tex]

Apply one-dimensional velocity equation for elastic collision;

[tex]u_1 + v_1 = u_2 + v_2\\\\u_1 + (-0.37u_1) = 0 + v_2\\\\u_1 -0.37u_1 = v_2\\\\v_2 = 0.63 u_1[/tex]

substitute the value of v₂ into the above equation;

[tex]1.37mu_1 = m_2(0.63u_1)\\\\1.37mu_1 = 0.63m_2u_1\\\\m_2 = \frac{1.37mu_1}{0.63u_1} \\\\m_2 = 2.175 \ m[/tex]

Thus, the mass of the second ball in terms of the first ball, m, is 2.175 m.

Learn more here:https://brainly.com/question/3227235

The container has the greatest total force acting on its base is the conatainer with the greatest total weight.

For each container, there is a pressure, P acting on its base given by

P = F/A where

Now, since the force on the base is the weight of water acting on the base, W, we have that

P = W/A where w = total weight of water acting on base

Now, for each container, the area of base is the same. So, we have that

P ∝ W

So, we see that the pressure is directly proportional to the weight on the container.

Since pressure is directly proportional to force, we see that the force on the base is directly proportional to the weight of water acting on the base.

So, for the greatest force acting on the base, we would have the greatest total weight of water acting on the base.

So, the container has the greatest total force acting on its base is the conatainer with the greatest total weight.

Learn more about force here:

https://brainly.com/question/25239010

#SPJ1

Answer:

dV/dt = 150.79 m^3/s

Explanation:

In order to calculate the rate of change of the volume, you calculate the derivative, respect to the radius of the sphere, of the volume of the sphere, as follow:

[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)[/tex]                     (1)

r: radius of the sphere

You calculate the derivative of the equation (1):

[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)=3\frac{4}{3}\pi r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex](2)

where dr/dt = 3m/s

You replace the values of dr/dt and r=2m in the equation (2):

[tex]\frac{dV}{dt}=4\pi (2m)^2(3\frac{m}{s})=150.79\frac{m^3}{s}[/tex]

The rate of change of the sphere, when it has a radius of 2m, is 150.79m^3/s

The question is not complete, the value of z is not given.

Assuming the value of z = 4.0m

Answer:

the magnitude of the electric field at any point having z(4.0 m)  =

E = 5.65 N/C

Explanation:

given

σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²

z = 4.0 m

Recall

E =F/q (coulumb's law)

E = kQ/r²

σ = Q/A

A = 4πr²

∴ The electric field at point z =

E = σ/zε₀

E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)

E = 5.65 N/C

sorry but I don't understand

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

[tex]KE=\dfrac{1}{2}mv^2[/tex]

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].

The kinetic energy of an object depends on the mass and velocity with which it moves.

While under free-fall, the mass of an object does not affect the velocity with which it falls.

So, the velocities of both the balls are the same.

Let the mass of ball A is 'm'

So, the mass of ball B is '4m'

The kinetic energy of ball A is given by;

[tex]KE_{A}=\frac{1}{2} mv^2[/tex]

The kinetic energy of ball B is given by;

[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]

Therefore, the ratio of kinetic energies of A and B is,

[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]

Learn more about kinetic energy here:

https://brainly.com/question/11580018

Answer:

Explanation:

The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :

D. Quadruple

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.

The stored energy shifts with the square of the electric charge put away within the capacitor.

In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.

Thus, the correct answer is D.

Learn more about "energy":

https://brainly.com/question/1932868?referrer=searchResults

Answer:

non of the above

Explanation:

Quantity of heat = mass× specific heat× change in temperature

m= 7kg c= 4.18 temp= 46-25=21°

.......H= 7×4.18×21= 614.46kJ

Answer:610 KJ

Explanation:A P E X answers

Answer:

B and D

Explanation:

Because

R= resistivity xlenght/ Area

Where R= resistance

Answer:

(a)    vo = 24.98m/s

(b)    t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex]      (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]        

[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]

The time that the ball takes to arrive to the ground is 5.09s

We have that for the Question, it can be said that the speed of  ball and How much longer would it take the ball to reach the ground is

From the question we are told

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.

(a) How fast was the ball originally moving when it was kicked.

(b) How much longer would it take the ball to reach the ground?

a)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]

b)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]

Therefore

[tex]X=5.095-5[/tex]

X=0.095sec

For more information on this visit

https://brainly.com/question/23379286

Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]

Where:

[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.

[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.

[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:

[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]

Roots of the polynomial are, respectively:

[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]

[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.

[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.

[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:

[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]

[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

[tex]v_{1} = v_{1,o} + g \cdot t[/tex]

Second stone

[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]

Where:

[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.

[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.

If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:

First stone

[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]

[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]

Second stone

[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]

[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

This question involves the concepts of the law of conservation of energy.

The temperature change of the bullet is "84.38°C".

According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.

[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]

where,

Therefore,

[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]

ΔT = 84.38°C

Learn more about the law of conservation of energy here:

https://brainly.com/question/20971995

#SPJ2

2Q

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = [tex]\frac{1}{2}qV[/tex]            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = [tex]\frac{1}{2} QV[/tex]              ----------------(ii)

Make V subject of the formula in equation (ii)

V = [tex]\frac{2E_{p}}{Q}[/tex]

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = [tex]\frac{1}{2}qV[/tex]

Substitute the value of V into the equation above;

2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])

Solve for q;

[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]

[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]

[tex]q = 2Q[/tex]

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q

Answer:

Planet A = planet B > planet D > planet A = planet E

Explanation:

Gravitational force obeys the inverse square law. That is,

Gravitation force F is inversely proportional to the square of the distance between the two masses.

The larger the distance, the weaker the gravitational force F.

From largest to smallest, the rank of gravitational force that each planet exerts on the star with mass M. Distances are as follows:

Planet B, planet C, planet D, planet planet A, planet E.

Planet B and C may experience different the same gravitational force depending on their masses. This is also applicable to planet A and E.

Therefore,

Planet A = planet B > planet D > planet A = planet E

Answer:

Explanation:

Using the relationship to calculate the time that elapse 2x = vt

x = distance between the source and the reflector

v is the velocity of sound in air

t = time elapsed

Given x = 30.0m and v = 343m/s

substituting the values into the formula above to get the total time elapsed t;

t = 2x/v

t = 2(30)/343

t= 60/343

t = 0.175sec

If you are seated in the middle row, the time that will elapse between the sound from the stage reaching your ear directly will be;

0.175/2 = 0.0875 secs

Answer:

The  gauge pressure is  [tex]P = 4.2*10^{5} \ N/m^2[/tex]

Explanation:

From the question we are told that

     The height of the 14th floor from the point where the water entered the building  is  h = 43 m

The  gauge pressure is  mathematically represented as

           [tex]P = mgh[/tex]

Where  the m is the mass of the water which is  mathematically represented as  

      [tex]m = \frac{\rho}{V}[/tex]

Where  [tex]\rho[/tex] is the density of the water which has a constant value of  [tex]\rho = 1000 \ kg/m^3[/tex] and this standard value of density the volume is  [tex]1 m^3[/tex] so

     [tex]m = \frac{1000}{1}[/tex]

     [tex]m = 1000 \ kg[/tex]

Thus  

     [tex]P = 1000 * 9.8 * 43[/tex]

     [tex]P = 4.2*10^{5} \ N/m^2[/tex]

Answer:

11.7 m

Explanation:

The radius of the circle is 13 m.

The central angle of the arc is 0.9 radians

The length of an arc is given as:

L = r θ

where θ = central angle in radians = 0.9

=> L = 0.9 * 13 = 11.7 m

Length of the arc will be 11.7 m ≈ 10 m

Arc length refers to the distance between two points along a curve’s section.

Arc length = radius * theta

where

Arc length  = ? to find

given :

radius = 13 m

theta ( central angle) = 0.9 radians

Arc length = 13 m * 0.9 radians

                = 11.7 m ≈ 10 m

length of the arc will be 11.7 m ≈ 10 m

Learn more about Arc length

https://brainly.com/question/16403495?referrer=searchResults

#SPJ2

Answer:

The  tension on the rope  is  T  =  900 N

Explanation:

From the question we are told that  

     The mass of the person on the left is  [tex]m_l = 100 \ kg[/tex]

      The force of the person on the left is  [tex]F_l = 1000 \ N[/tex]

       The mass of the person on the right  is  [tex]m_r = 70 \ kg[/tex]

       The force of the person on the right is  [tex]F_r = 830 \ N[/tex]

Generally the net force is  mathematically represented as

         [tex]F_{Net} = F_l - F_r[/tex]

substituting  values

        [tex]F_{Net} = 1000-830[/tex]

       [tex]F_{Net} = 170 \ N[/tex]

Now the acceleration net acceleration of the rope is mathematically evaluated as

        [tex]a = \frac{F_{net}}{m_I + m_r }[/tex]

substituting  values

     [tex]a = \frac{170}{100 + 70 }[/tex]

     [tex]a = 1 \ m/s ^2[/tex]

The  force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by  a  is  mathematically represented as

        [tex]m_l * a = F_r -T[/tex]

Where T  is  the tension on the rope  

      substituting values

        [tex]100 * 1 = 1000 - T[/tex]

=>      T  =  900 N

Answer:

a

 [tex]\theta = 0.0022 rad[/tex]

b

 [tex]I = 0.000304 I_o[/tex]

Explanation:

From the question we are told that  

   The  wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]

    The  distance of the slit separation is  [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]

Generally the condition for two slit interference  is  

     [tex]dsin \theta = m \lambda[/tex]

Where m is the order which is given from the question as  m = 2

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

 substituting values  

      [tex]\theta = 0.0022 rad[/tex]

Now on the second question  

   The distance of separation of the slit is  

       [tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]

The  intensity at the  the angular position in part "a" is mathematically evaluated as

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

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

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

  substituting values

     [tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]

    [tex]\beta = 0.06581[/tex]

So the intensity is  

    [tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]

   [tex]I = 0.000304 I_o[/tex]

Answer:

The  horizontal displacement is  [tex]Adj = 114.71 \ m[/tex]

Explanation:

From the question we are told that  

    The  angle at which the string is maintained is  [tex]\theta = 41 ^o[/tex]

      The length of string reeled in is  [tex]l = 152 \ m[/tex]

Using the SOHCAHTOA formula

   We have that the hypotenuse(Hyp) is  l  =  152

Hence the  horizontal displacement of the kite  which is the Adjacent(Adj)  can be evaluated as  

     [tex]cos \theta = \frac{Adj}{Hyp }[/tex]

substituting values

     [tex]cos(41) = \frac{Adj}{152}[/tex]

=>   [tex]Adj = 114.71 \ m[/tex]