plzzz help its for my civics class what answer is it????????????

Plzzz Help Its For My Civics Class What Answer Is It????????????

Answers

Answer 1

Answer:

the correct answer is The English Bill of Rights

Explanation:

This is the answer because i have taken the test.

Answer 2
Bill of rights djdjdkdodpdd

Related Questions

I NEED HELP PLEASEE ITS AN ECONOMICS QUESTION ABOVE

Answers

Answer:

I believe the answer is Property taxes

Explanation:

Answer: I'm pretty sure property taxes

Explanation:

A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational field strength of Planet X?


A. The gravitational field strength of Planet X is mg.

B. The gravitational field strength of Planet X is Wx/m.

C. The gravitational field strength of Planet X is 9.8 N/kg.

D. The gravitational field strength of Planet X is mWx.

Answers

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

Answers

Answer:

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

[tex]x=x_{2}-x_{1}[/tex]

Put the value into the formula

[tex]x=4-(0.75+0.75)[/tex]

[tex]x=2.5\ m[/tex]

We need to calculate the displacement of the canoe

Using conservation of momentum

[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]

[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]

[tex]86.5\times2.5=(61+86.5)\times x'[/tex]

[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]

[tex]x'=1.46\ m[/tex]

Hence, The displacement of the canoe is 1.46 m

in a controlled experiment do none of the variables change?

Answers

Answer:

Yes

Explanation:

The variables change in and experiment.

Answer:

If you are carefully enough to control everything, then everything that could change the result of your experiment won't happen.

Explanation:

Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.

Answers

Answer:

Explanation

According to Newton's second law of motion,

F = ma

m is the mass

a is the acceleration

If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g

The formula becomes;

F = m(3.2g)

F = 3.2mg

m= 75kg

g = 9.81m/s²

F = 3.2(75)(9.81)

F = 2,354.4N

Hence the force exerted by the jumper is 2,354.4N

An LED lamp powered by a USB-based portable battery has an effective resistance of 500 Ohm. If the battery is rated for 10,000 mAh, then how long can the lamp be powered and what is the total power consumed by the lamp? The operating voltage of a USB powered device is 5V. Assume that the battery is also rated at 5V.

Answers

Answer:

Time = 1000 h

Power = 0.05 W = 50 mW

Explanation:

First we will find the current consumed by the lamp. For this purpose we can use the Ohm's Law. The equation of Ohm's Law is written as follows:

V = IR

I = V/R

where,

I = Current = ?

V = Voltage = 5 V

R = Resistance = 500 Ω

Therefore,

I = 5 V/500 Ω

I = 0.01 A = 10 mA

Now the time duration of the operation f lamp can be found by:

Time = Rating/Current

Time = 10000 mAh/ 10 mA

Time = 1000 h

The power consumption of lamp is given as follows:

Power = IV

Power = (0.01 A)(5 V)

Power = 0.05 W = 50 mW

The scientific method is the only way of learning about Nature used by scientist today *

A. true
B. false

Answers

Answer:

false

Explanation:

which statement is correct about the strength of forces?
-Electrostatic forces are exactly 10 times stronger than gravitational forces.
-Electrostatic forces are exactly 10 times weaker than gravitational forces.
-Electrostatic forces are trillions of times stronger than gravitational forces.
-Electrostatic forces are trillions of times weaker than gravitational forces.

Answers

Answer:

Thanks!!!!! adding this so it doesn’t get deleted.

Explanation:

1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.

The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.

What is an Electrostatic Force?

One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.

When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.

Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.

To know more about Electrostatic Force:

https://brainly.com/question/9774180

#SPJ2

Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?

Answers

Answer:

Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?

Which property of a substance can be determined using a pH indicator.​

Answers

Answer:

Acidity

Explanation:

A pH indicator measures how acidic or basic a substance is.

Hope this helped :)

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:

29 m/s2

31 m/s2

33 m/s2

30 m/s2

32 m/s2

Answers

Explanation:

The question is incomplete. Here is the complete question.

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of  30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...

Given

initial velocity of rocket u = 0m/s

final velocity of rocket = 30m/s

Height reached by the rocket = 98m

Required

upward acceleration of the rocket

Using the equation of motion below to get the acceleration a:

[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]

Hence  upward acceleration of the rocket during the burn phase is closest to 5m/s²

Note that the velocity used in calculation was assumed.

Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist her and parks their car 14 meters from the front of her car. Marcella grabs the car bumper and pulls very hard, with 200 N of force. The work she does transfers energy out, it reduces the K of the car, as it gradually approaches the other car. a) What is the initial kinetic energy before she tries to stop the car? b) What is the final kinetic energy, when her car hits the other car? What is the speed? c) Suppose the other person also slowed her car, pushing it from the front. How much force would be needed to stop her car 1 meter from the other car? [1 m allows the person not to be crushed!]

Answers

Answer:

Explanation:

a) KE = (1/2) * m * ([tex]v^{2}[/tex]) = F * d = 14m * 200N = 2800 m/N or 2.8 * [tex]10^{3}[/tex] m/N

b) 0J and 0m/s (if Marcella stopped after going 14 meters)

c)  Known from part (a) that KE = 2800 J = F1 * d1,

    2800J = F1 * (14m - 1m)  => F1 = 2800J/13m = 215.384 N

   

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device?

Answers

Complete Question

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:

a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.

b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position

Answer:

a

[tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

b

[tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  50  

   The radius is  r =  1.55 cm  =  0.0155 m

   The  magnetic field is  B  =  0.35 T

   The induced current is  [tex]I =  45 \mu A =  45 *10^{-6} \  A[/tex]

   

Generally the area of  loop is mathematically represented as

      [tex]A =  \pi r^2[/tex]

=>   [tex]A =3.142 *  0.0155^2[/tex]

=>   [tex]A =0.000755\ m^2[/tex]

Generally the maximum torque is mathematically represented as

     [tex]\tau_{m} =  N  *  B  * I  *  A[/tex]

=>  [tex]\tau_{m} =  50  *  0.35  * 45 *10^{-6} *  0.000755[/tex]

=>  [tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

Generally the  torque 60° from the spring's equilibrium position is mathematically represented as

      [tex]\tau  =  N  *  B  *  I  *  A  *  sin (60)[/tex]

=>    [tex]\tau  =  50  *  0.35  *  45 *10^{-6} *  0.000755  *  sin (60)[/tex]

=>   [tex]\tau  = 2.973 *10^{-7} \  N  \cdot m [/tex]

Generally the toque constant of the spring is mathematically represented as

     [tex]\beta =  \frac{\tau}{60}[/tex]

=>   [tex]\beta =  \frac{\tau}{\frac{\pi}{3}}[/tex]

=>   [tex]\beta =  \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]

=>    [tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .

Answers

Answer:

The magnitude is [tex]B = \frac{450}{7.9* 10^5}[/tex]

The direction is  the positive x axis

Explanation:

From the question we are told that

   The  electric field is  E = 450 V/m in the negative y ‑direction

   The speed of the proton is  [tex]v= 7.9* 10^5\  m/s[/tex] in the positive z direction

Generally the overall force acting on  the proton is mathematical represented as

              [tex]F_E =  q(\vec E + \vec v  * \vec B)[/tex]

Now for the beam of protons to continue un-deflected along its straight-line trajectory then  [tex]F_E =0[/tex]

So

          [tex] 0  =  q( E (-y) + v(z)  * \vec B)[/tex]

=>      [tex]E\^y = v \^ z  * \vec B[/tex]

Generally from unit vector cross product vector multiplication

         [tex]\^ z  \ *  \  \^ x  =  \^  y[/tex]

So the direction of  B (magnetic field must be in the positive x -axis )

So

       [tex]E\^y = v \^ z  *  B\^ x [/tex]

=>     [tex]E\^y = vB ( \^ z  *  \^ x) [/tex]

=>     [tex]E\^y = vB ( \^y) [/tex]        

=>      [tex]E = vB [/tex]  

=>      [tex]B = \frac{450}{7.9* 10^5}[/tex]

=>      [tex]B =  0.0005696 \ T [/tex]  

   

   

At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?

Answers

Answer:

At a distance of 100 m from the source the intensity will be 40 dB.

Explanation:

Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.

The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.

The conversion between intensity and decibels corresponds to:

[tex]L=10*log\frac{I}{I0}[/tex]

where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.

In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:

[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]

[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]

Being L1= 70 dB and L2= 40 dB

[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]

[tex]30=10*[log(\frac{I1}{I2})][/tex]

[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]

[tex]3=log(\frac{I1}{I2})[/tex]

[tex]10^{3} =\frac{I1}{I2}[/tex]

[tex]1,000=\frac{I1}{I2}[/tex]

The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:

[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]

Then:

[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]

Being d1= 10 m

[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]

[tex]1,000=\frac{d2^{2} }{100}[/tex]

1,000*100= d2²

10,000= d2²

√10,000= d2

100 m= d2

At a distance of 100 m from the source the intensity will be 40 dB.

PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV by a person's 65th birthday? Solve using Dimensional Analysis.

Answers

Answer:

5694000 min

Explanation:

Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.

(4 h/day) × (60 min/1 h) = 240 min/day

They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.

240 min/day × (365 day/year) = 87600 min/year

They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.

87600 min/year × 65 year = 5694000 min

A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside the solenoid near its center

Answers

Answer:

The value  is  [tex]B =  0.0452 \  T [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  1000

    The length is  L =  50 cm =  0.50 m  

    The radius is  r =  2.0 cm  =  0.02 m

     The current is I  =  18.0 A

   

Generally the magnetic field is mathematically represented as

         [tex]B = \mu_o  * \frac{N }{L}  *  I[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value  

     [tex]\mu_o  =  4\pi * 10^{-7} N/A^2[/tex]

So

     [tex]B =  4\pi * 10^{-7}  *   \frac{1000}{0.50} *  18.0[/tex]

=>   [tex]B =  0.0452 \  T [/tex]

On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)

Answers

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁  - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀  - h₂) = ¹/₂ (v₂² - v₁²)

h₀  - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]

Therefore, the height of the cliff is 121.276 m

What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?

Answers

Answer:

Answer and steps in the pic

a. In one short sentence, explain why we call the force of gravity an attractive force.
b. Does a force of gravity exist between any two objects

Answers

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G[tex]\frac{m1m2}{r}[/tex]

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground

Answers

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value

Answers

Complete Question

The complete  question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The original voltage is  [tex]V_o[/tex]

     The new voltage is [tex]V  =\frac{V_o}{2}[/tex]

     The capacitance is  [tex]C = 150\ nF = 150 *10^{-9} \  F[/tex]

     The first resistance is  [tex]R_i =  26 \Omega[/tex]

      The second resistance is [tex]R_E =  200 \Omega[/tex]

Generally the equivalent resistance is  

        [tex]R_e =  R_1 + R_E[/tex]

=>     [tex]R_e =  26 +200 [/tex]

=>     [tex]R_e = 226 \ \Omega [/tex]

Generally the time constant is mathematically represented as

     [tex]\tau  =  RC[/tex]

=>  [tex]\tau  =  226 * 150 *10^{-9}[/tex]

=>  [tex]\tau  =  3.39 *10^{-5} \  s [/tex]

Generally the voltage is mathematically represented as

    [tex]V =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]\frac{V_o}{2} =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]0.5 =    e^{-\frac{t}{\tau} }[/tex]

=>   [tex]ln(0.5) =    {-\frac{t}{ 3.39 *10^{-5} } }[/tex]

=>  [tex]ln(0.5)   * 3.39 *10^{-5}  =   -t [/tex]

=>  [tex]t = 2.35*10^{-5} \  s  [/tex]

cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?

Answers

Answer:

Displacement is 3 km North

Explanation:

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *

A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s

Answers

Answer:

D. 10 m/s

Explanation:

What are the standard international (si) units of distance

Answers

Answer:

meter

Explanation:

Answer: The International System of Units is a system of measurement based on 7 base units

Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .

Answers

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  [tex]  v  =  350 \  m/s [/tex]  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by [tex]k  =  60 cm  =  0.6 \  m[/tex]

    The frequency of both speakers is [tex]f =  700 \  Hz[/tex]

Generally the distance of the listener to the first speaker is mathematically represented as

       [tex]L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]

       [tex]L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]

        [tex]L_1  =   5.39 \  m [/tex]

Generally the distance of the listener to second speaker at its new position is  

          [tex]L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]

       [tex]L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]

        [tex]L_2  =   5.64  \  m [/tex]  

Generally the path difference between the speakers is mathematically represented as

        [tex]pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}[/tex]

Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as

         [tex]\lambda =  \frac{v}{f}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}[/tex]

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

=>    [tex] L_2 - L_1  =  \frac{n  *  v}{2f}[/tex]  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    [tex]  5.64 - 5.39   =  \frac{1  *  v}{2*700}[/tex]      

=>    [tex]  v  =  350 \  m/s [/tex]  

The speed of sound in air is 350 m/s

Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,

L = √[(5 m)² + (4/2 m)²]

= √[25 m² + (2 m)²]

= √[25 m² + 4 m²]

= √29 m² = 5.39 m.

Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now

L' = √[(5 m)² + (4/2 + 0.6 m)²]

= √[25 m² + (2 m + 0.6 m)²]

= √[25 m² + (2.6 m)²]

= √[25 m² + 6.76 m²]

= √31.76 m²

= 5.64 m.

Now, the path difference when we first have destructive interference is

ΔL = L' - L

= 5.64 - 5.39

= 0.25

Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.

Now, since we have destructive interference for the first time, n = 0.

So,  ΔL = (n + 1/2)λ

ΔL = (0 + 1/2)v/f

ΔL = v/2f

Making v subject of the formula, we have

v = 2fΔL

Substituting the values of the variables into the equation, we have

v = 2fΔL

v = 2 × 700 Hz × 0.25 m

v = 350 m/s

So, the speed of sound in air is 350 m/s

Learn more about interference of sound here:

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Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years

Answers

Answer:

T = 1.4696 10⁴ years

Explanation:

For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law

               F = ma

               G m M / r² = m a_c = m v² / r

                G M / r = v²

the speed of the circular orbit is

               v = 2π r / T

           

we substitute

               G M / r = 4π² r² / T²

               T² = (4π² / G M)  r³

Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse

               T² = (4π² /GM)  a³

the constant is worth

                (4π² / GM) = 2.97 10⁻¹⁹    s² / m³

let's reduce the distance to SI units

AU is the distance from the Earth to the Sun

               a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)

               a = 8.976 10¹³ m

               T² = 2.97 10⁻¹⁹ (8.976 10¹³)³

               T² = 21.4786 10²²

               T = 4.63 10¹¹ s

Let's reduce to years

               T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)

               T = 1.4696 10⁴ years



When driving at slower speeds you need to use what type of steering

wheel movements compared to when driving at faster speeds? *

Answers

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

1 (4 points) A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the magnitude of the net force on the ball

Answers

Answer:

100 N

Explanation:

The following data were obtained from the question:

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 50 cm

Force (F) =.?

Next, we shall convert 50 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

50 cm = 50 cm × 1 m / 100 cm

50 cm = 0.5 m

Therefore, 50 cm is equivalent to 0.5 m.

Finally, we shall determine the magnitude of the net force on the ball by using the following formula:

F = mv²/r

Mass (m) = 2 Kg

Velocity (v) = 5 m/s

Radius (r) = 0.5 m

Force (F) =.?

F = mv²/r

F = 2 × 5²/ 0.5

F = 2 × 25/ 0.5

F = 50 / 0.5

F = 100 N

Therefore, the magnitude of the net force on the ball is 100 N.

The magnitude of the net force on the ball will be "100 N".

Force and speed

According to the question,

Mass, m = 2 kg

Velocity, v = 5 m/s

Radius, r = 50 cm or,

               = 50 × [tex]\frac{1}{100}[/tex]

               = 0.5 m

We know the relation,

Force, F = [tex]\frac{mv^2}{r}[/tex]

By substituting the values, we get

              = [tex]\frac{2\times (25)^2}{0.5}[/tex]

              = [tex]\frac{2\times 25}{0.5}[/tex]

              = [tex]\frac{50}{0.5}[/tex]

              = 100 N

Thus the above response is appropriate.

Find out more information about force here:

https://brainly.com/question/6504879

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