Two point sources produce waves of the same wavelength that are in phase. At a point midway between the sources, what kind of interference would be observed?

Answer:

A) 2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Correct frequency is f2 = 724 rpm

C) Reason is stated in explanation

Explanation:

A) We are given;

Frequency of first propeller; f1 = 574 rpm

Beat frequency; f_beat = 2.5 Hz = 2.5 × 60 rpm = 150 rpm

Now, formula for the beat frequency is;

f_beat = |f1 - f2|

Now, |f1 - f2| means it is inside an absolute value.

Thus, it means,

f1 - f2 = 150 or f2 - f1 = 150

Thus;

574 - f2 = 150

Or f2 - 574 = 150

So,f2 = 574 - 150 = 424 rpm or f2 = 574 + 150 = 724 rpm

2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Now, if we increase the speed of the second propellar slightly, it means that f2 will increase as well.

Now, from the 2 possible values of f2 gotten, we can see that for f1 - f2 = 150, when we increase f2, the beat frequency will reduce while for f2 - f1 = 150, when we increase f2, the beat frequency will increase.

Thus, it's the frequency of f2 gotten in f2 - f1 = 150 that is the correct answer.

Thus, when we increase the speed of the second propellar slightly, it means that;

f2 = 724 rpm

C) Answer in part B is correct because as we increase the speed of the second propellar, the frequency will also increase and the value of f2 that corresponds with an increase in speed is 724 rpm.

Complete question:

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 6.0 m above the ground, what magnetic field does the line produce at ground level?

Answer:

The magnetic field the line produces at ground level is 5 x 10⁻⁶ T.

Explanation:

Given;

current in the dc transmission line, I = 150 A

height above ground level, R = 6 m

The transmission line will be treated as current in a long straight wire.

Thus, the magnetic field produced at the ground level is given as magnetic field in a long straight conductor.

[tex]B = \frac{\mu_o I}{2\pi R}[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

[tex]B = \frac{4\pi*10^{-7} * 150}{2\pi * 6}\\\\B = 5*10^{-6} \ T[/tex]

Therefore, the magnetic field the line produces at ground level is 5 x 10⁻⁶ T.

Answer:

Corkscrew, bicycle, scissors

Explanation:

they are made from many simple machine's

Answer:

compound meachiun include bicycles, cars, scissors, and fising rods with reels.

Answer:

Irms =226A

Explanation:

The current is high because the total impedance is relatively low. Actually, plugging such a circuit into a 120-V outlet would most likely burn out the circuit elements

Answer:

The final speed of small metal sphere is 12.6 m/s.

Explanation:

Given that,

Distance = 0.4 m

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the final speed of small metal sphere

Using conservation of energy

[tex]\dfrac{1}{2}mv_{1}^2+\dfrac{kq_{1}q_{2}}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{kq_{1}q_{2}}{r_{2}}[/tex]

[tex]\dfrac{1}{2}m(v_{1}^2-v_{2}^2)=kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{2}^2)=9\times10^{9}\times(-2)\times10^{-6}\times(-8)\times10^{-6}(\dfrac{1}{0.4}-\dfrac{1}{0.8})[/tex]

[tex]400-v_{2}^2=240[/tex]

[tex]-v_{2}^2=240-400[/tex]

[tex]v_{2}=\sqrt{160}\ m/s[/tex]

[tex]v_{2}=12.6\ m/s[/tex]

Hence, The final speed of small metal sphere is 12.6 m/s.

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

Answer:

force applied by red = 50 N

force applied by blue = - 63 N (since it is in the opposite direction)

net force = force by red + force by blue

net force = 50 + (- 63)

net force = - 13 N  

The answer is B

Explanation:

Every action has an equal reaction. If you punch a wall, the wall is going to exert the wall with exert the same amount of force back towards your hand, causing your hand to hurt.

Answer:

a.  According to Newton's Third Law, the force you put into the wall with your punch causes the wall to put the same force back onto your hand.

Explanation:

I got it correct.

Answer:

1)   α = 2.2 rad / s² , 2)   t = 7.068 s , 3) in this interval   s = 23.096 m

total distance    s = 57.4 m

Explanation:

For this exercise we use the angular kinematic relations, before starting the problem we reduce all the magnitude to the SI system

        v₀ = 88 km / h (1000 m / 1km) 1h / 3600s) = 24.44 m / s

        v = 56.0 km / h = 15.55 m / s

        θ = 90 rev (2pi rad / 1 rev) = 565,487 rad

        d = 0.84 m

          r = d / 2

         r = 0.84 / 2

         r = 0.42 m

1) ask for angular acceleration

        w² = w₀² - 2 α Δθ

        α = (w₀² -w²) / 2 Δθ

To find the angular velocities we use the acceleration between the linear and angular velocity

         v = w r

         w = v / r

         w₀ = 24.44 / 0.42

         w₀ = 58.20 rad / s

         w = 15.55 / 0.42

         w = 30.037 rad / s

we calculate

        α = (58.20² - 30.037²) / (2   565.487)

        α = 2.2 rad / s²

2) how much longer does it take to stop

         w₂ = 15.55 rad / s

         w = 0

         w = w₂ - α t

         t = (w₂ -0) / α

         t = 15.55 / 2.2

         t = 7.068 s

3) the distance that the car travels from the beginning of the movement, we can find it by looking for the number of revolutions until it stops and then using the relationship between the angular and linear variable

         w² = w₀² - 2 α θ

at the end of the movement speed is zero

         0 = w₀² - 2 α θ

         θ = w₀² / 2 α

         θ = 24.44² / (2 2.2)

          θ = 135.80 rad

If the angles are measured in radians, we can apply the relation

         θ = s / R

         s = R ttea

         s = 0.42 135

         s = 57.4 m

this is the distance from when the movement starts

the distance for the final part of the movement is

          w = 15.55 rad / s

          θ = w² / 2 α

          θ = 15.55 2 / (2 2.2)

          θ = 54.99 rad

the distance in this interval is

          s = 0.42 54.99

          s = 23.096 m

Answer:

height of cliff (h) = 112.38m

Explanation:

The time 5.12 s is the total time it takes for the rock to fall, and for the soundwave to travel back to the top of the cliff before it is heard.

[tex]5.12 = t_f\ +\ t_s - - - - -(1)\\where:\\t_f = time\ of\ fall\ of\ the\ piece\ of\ rock\\t_s = time\ travelled\ by\ the\ return\ sound[/tex]

Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:

[tex]t_f=\sqrt{\frac{2h}{g} } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2[/tex]

[tex]\therefore t_f = \sqrt{\frac{2h}{9.8}} \\squaring\ both\ sides\\(t_f)^2 = \frac{2h}{9.8}\\ 2h = 9.8 \timess\ (t_f)^2\\h = \frac{9.8 \timess\ (t_f)^2}{2} \\h= 4.9(t_f)^2 - - - - - (2)[/tex]

Next, let us calculate the time taken fot the sound to return

[tex]t_s = \frac{h}{v} \\where:\\t_s = time\ for\ sound\ to\ travel\ up\ the\ cliff\\h= distance\ tavelled\ = height\ of\ cliff\\v= speed\ = 340m \slash s\ (speed\ of\ sound)\\\therefore t_s = \frac{h}{340} - - - - - (3)\\[/tex]

now putting the values of  h from equation 2 into equation (3)

[tex]t_s = \frac{4.9(t_f)^2}{340}[/tex]

Putting the value of [tex]t_s[/tex] into equation (1)

[tex]5.12 = t_f +\frac{4.9(t_f)^2}{340} \\[/tex]

multiplying through by 340

[tex]1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0[/tex]

now let us solve the quadratic equationsss;

[tex]Let\ (t_f) = x[/tex]

[tex]4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-340 \pm\sqrt{(340)^2 -\ 4 \times4.9 \times(-1740.8)} }{2\times4.9}\\x = \frac{-340\ \pm\ 386.936}{9.8} \\x =\frac{386.938 - 340}{9.8} \\x = \frac{46.936}{9.8}\\ x = 4.789\\x = t_f\\t_f=4.789s[/tex]

note, time cannot be negative, so we ignored the negative answer

putting the value of  [tex]t_f[/tex]  into equation (2) to find height of cliff (h)

[tex]h= 4.9(t_f)^2\\h = 4.9 \times(4.789)^2\\h = 112.38m[/tex]

Therefore, height of cliff (h) = 112.38m

Answer:

2000Hz and 1500Hz

Explanation:

Using

a) f = f0((c+vr)/(c+vs))

=>>> f0((c)/(c-0.5c))

=>>>1000/0.5 = 2000Hz

b) f = f0((c+vr)/(c+vs))

=>>>f0((c+0.5c)/(c))

=>>>>1000 x 1.5 = 1500Hz

Here we have a problem referring to the Doppler effect, the solutions are:

The Doppler effect is an effect that explains how the perception of waves changes as the source moves or as the listener moves.

The formula, for sound, is:

[tex]f = \frac{v + v_0}{v - v_1}*f_0[/tex]

where:

a) First we have that the source moves towards a stationary listener, then we have:

[tex]f = \frac{v }{v - v/2}*1.0 kHz\\\\f = \frac{v }{v/2}*1.0 kHz = 2.0 kHz[/tex]

b) in this case, the listener moves towards the source, so we have:

[tex]f = \frac{v + v/2 }{v }*1.0 kHz\\\\f = \frac{ (3/2)*v }{v}*1.0 kHz = 1.5 kHz[/tex]

So in this case the perceived frequency is smaller than in the point a.

This is because the waves will move at a fixed rate in air, in one case, the successive waves are emitted from different points in space (each time closer to the listener) while in the other case the waves are emitted from a fixed point, and the listener moves towards them, thus feels that the waves move faster.

If you want to learn more about the Doppler effect, you can read:

https://brainly.com/question/3826119

Answer:

(a) 2.75 fm

(b) 2.89 fm

(c) 4.70 fm

(d) 7.12 fm

Explanation:

For a given element, the radius r of its nuclei is given by;

r = r₀[tex]A^{(1/3)}[/tex]

Where;

A = Atomic mass of the element

r₀ = 1.2 x 10⁻¹⁵m = 1.2fm

Now let's solve for the given elements

(a) ¹²₆C

Carbon element => This has an atomic mass number of 12

Therefore its radius is given by;

r = 1.2  x [tex]12^{1/3}[/tex]

r = 1.2 x 2.29

r = 2.75 fm

(b) ¹⁴₇N

Nitrogen element => This has an atomic mass number of 14

Therefore its radius is given by;

r = 1.2  x [tex]14^{1/3}[/tex]

r = 1.2 x 2.41

r = 2.89 fm

(c) ⁶⁰₂₇Co

Cobalt element => This has an atomic mass number of 60

Therefore its radius is given by;

r = 1.2  x [tex]60^{1/3}[/tex]

r = 1.2 x 3.92

r = 4.70 fm

(d) ²⁰⁸₈₂Pb

Lead element => This has an atomic mass number of 208

Therefore its radius is given by;

r = 1.2  x [tex]208^{1/3}[/tex]

r = 1.2 x 5.93

r = 7.12 fm

Answer:

The rebound speed of the mass 2m is v/2

Explanation:

I will designate the two masses as body A and body B.

mass of body A = m

mass of body B = 2m

velocity of body A = v

velocity of body B = -v    since they both move in opposite direction

final speed of mass A = 2v

final speed of body B = ?

The equation of conservation of momentum for this system is

mv - 2mv = -2mv + x

where x is the final momentum of the mass B

x = mv - 2mv + 2mv

x = mv

to get the speed, we divide the momentum by the mass of mass B

x/2m = v = mv/2m

speed of mass B = v/2

Answer:

The magnetic field at the center of the loop is 2.51 × 10⁻³ T

Explanation:

The magnetic field at the center of a circular loop is given by

B = μ₀I/2r

Where B is the magnetic field strength in Teslas (T)

μ₀ is the permeability of free space

μ₀ = 4π ×10⁻⁷ N/A²

I is the current in Amperes (A)

and r is the radius of the loop in meters (m)

From the question,

r = 1.0 cm

Convert this to meter (m)

1.0 cm = 1.0 × 10⁻² m = 0.01 m

∴ r = 0.01 m

I = 40 A

Hence, the magnetic field at the center of the loop is

B = μ₀I/2r

B = (4π ×10⁻⁷ × 40) / (2 × 0.01)

B = 5.0265 × 10⁻⁵ / 0.02

B =  2.51 × 10⁻³ T

Hence, the magnetic field at the center of the loop is 2.51 × 10⁻³ T

Answer:

The number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Explanation:

Number of photons of light produced is given by

[tex]Number of photons = \frac{Power}{Energy}[/tex]

From the question,

Power = 5 mW (milliwatts) = 5 × 10⁻³ W

Since 1 Watt = 1 Js⁻¹

Then, 5 × 10⁻³ W = 5 × 10⁻³ Js⁻¹

For the Energy,

As given from the question

Energy=(Power)x(time)

Time = 1 hour = (1 × 60 × 60) s = 3600 s

∴ Energy = 5 × 10⁻³ Js⁻¹ x 1 x 3600s

Energy =18.0 J

Now for the Number of photons produced,

[tex]Number of photons = \frac{Power}{Energy}[/tex]

Power = 5 × 10⁻³ Js⁻¹ = 0.005 Js⁻¹

[tex]Number of photons = \frac{0.005}{18}[/tex]

Number of photons = 2.78 × 10⁻⁴ Photons / sec

This is the number of photons produced in 1 second.

For the number of photons produced in 1 hour, we will multiply the result by 3600

(NOTE: 1 sec = [tex]\frac{1}{3600}[/tex] hour)

Number of photons = 2.78 × 10⁻⁴ Photons / sec

= 2.78 × 10⁻⁴ × 3600 Photons / hour

= 1.0008 Photons / hour

≅ 1 Photon / hour

Hence, the number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Answer:

TRICARE for Life (TFL), a program for Medicare-eligible military retirees and their dependents, acts as a supplement to Medicare.

Explanation:

Answer:

Average speed is 5 m/s

Explanation:

The initial speed = 0 m/s (since the runner starts from rest)

Final speed = 10 m/s

time interval = 2 sec

average speed = ?

The speed is assumed to change at a constant rate

Average speed = (final speed + initial speed)/2

==> (10 + 0)/2 = 10/2 = 5 m/s

Answer:

they are describing their displacement since displacement is nothing but distance along with the direction of motion

Answer:

1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Explanation:

1km = 10^3 m = 10^5 cm

1h = 60x60 = 3600s

--> 1km/h^2 = 10^5 cm / (3600^2) s^2

<=> 1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Answer:

8×10⁻¹⁷ N

Explanation:

from the question, Electric force is given as

F = QV/r.............. Equation 1

Where F = Electric Force,  Q = Charge, V = Electric potential, r = distance.

Given: Q = 8×10⁻¹⁹ C, V = 5 kV = 5000 V, r = 2 cm = 0.02 m.

Substitute into equation 1

F = 8×10⁻¹⁹(5000)(0.02)

F = 8×10⁻¹⁷ N

Hence the electric force on the dust particle is  8×10⁻¹⁷ N

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = [tex]\frac{u}{v} * 100[/tex]%

p = [tex]\frac{2.5}{92} * 100[/tex]%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

Answer:

The geocentric model of Ptolemy in which the sun and other planets were believed to move round the Earth.

Explanation:

The Copernicus heliocentric model of the solar system was put forward by  Nicolaus Copernicus and was published in 1543. The model proposed by Copernicus puts the Sun near the center of the Universe, motionless, with Earth and the other planets orbiting around the sun in circular paths, modified by epicycles, and at uniform speed. This is contrary to Ptolemy's geocentric solar system that puts the Earth at the center of the solar system, in which the sun and other planets were believed to move round the Earth.

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

Answer:

They should be connected in parallel

Explanation:

Because In Parallel circuit, voltage across all lamps are same and current will flow independent each other such that a malfunction of one will nor affect the rest unlike in series connection

Answer:3) variable affinities (stickiness) for something it is running past. Physical ... -measurement number (significant digits) unit (such as inches) -Significant ... Mass 1 oz. 28.25 g. Relations Between English and Metric Units Mass 1 dram. 1.772 g ... -graduated cylinder has an error of about 1% (± 0.1 mL in 10 mL). -Volumetric

Explanation:

Answer:

1. 28.25 mL, three significant digits.

2. 54.074 mL, three significant digits

3. 600.006 km, four significant digits

4. 1356 kg + 4.2 kg + 19.891 kg

Explanation:

Answer:

Lift, weight, thrust and drag.

Answer:

According to Newton's first law of motion (inertia), an object at rest will remain at rest, or an object in motion will continue in motion at the same speed and in the same direction, until an outside force acts on it. For an aircraft to taxi or fly, a force must be applied to it. It would remain at rest without an outside force. Once the aircraft is moving, another force must act on it to bring it to a stop. It would continue in motion without an outside force. This willingness of an object to remain at rest or to continue in motion is referred to as inertia.

Newton's Second Law of Motion

The second law of motion (force) states that if a object moving with uniform speed is acted upon by an external force, the change of motion (acceleration) will be directly proportional to the amount of force and inversely proportional to the mass of the object being moved. The motion will take place in the direction in which the force acts. Simply stated, this means that an object being pushed by 10 pounds of force will travel faster than it would if it were pushed by 5 pounds of force. A heavier object will accelerate more slowly than a lighter object when an equal force is applied.

Newton's Third Law of Motion

The third law of motion (action and reaction) states that for every action (force) there is an equal and opposite reaction (force). This law can be demonstrated with a balloon. If you inflate a balloon with air and release it without securing the neck, as the air is expelled the balloon moves in the opposite direction of the air rushing out of it.

Complete Question

Two spherical shells have a common center. A -1.6 x 10-6 C charge is spread uniformly over the inner shell, which has a radius of 0.030 m. A +5.1 x 10-6 C charge is spread uniformly over the outer shell, which has a radius of 0.15 m.What are the magnitude and direction of the electric field at a distance 43.5 mm from the center of the shells.

Answer:

The magnitude is  [tex]E = 7.6021*10^{6} \N/C[/tex]

The direction is radially inward toward the center

Explanation:

From the question we are told that

   The charge on the inner shell is  [tex]q_i = -1.6*10^{-6} \ C[/tex]

    The  radius of the inner shell is  [tex]c_1 = 0.030 \ m[/tex]

   The  charge on the outer shell is  [tex]q_o = 5.1*10^{-6}\ C[/tex]

     The  radius of the outer shell is  [tex]c_2 = 0.15\ m[/tex]

   The distance considered is  [tex]r = 43.5 \ mm = 0.0435 \ m[/tex]

Generally the electric field at the position considered is mathematically represented as

          [tex]E = \frac{Q_c }{4\pi r^2 \epsilon_o }[/tex]

Here  [tex]Q_c[/tex] is the charge  which is enclosed by the distance considered which in this case is the charge on the inner shell

  So  [tex]Q_c =q_i = -1.6*10^{-6} \ C[/tex]

Hence

          [tex]E = \frac{-1.6 *10^{-6} }{4* 3.142 *0.0435^2* 8.85*10^{-12} }[/tex]

 =>      [tex]E = -7.6021*10^{6} \N/C[/tex]

The negative sign is show that the direction of the field is radially inward toward the center