A cyclist is riding along at a speed of 20.7 when she decides to apply the brakes which gave a deceleration applied was a rate of -3.4 m/s2 over the span of 7.8 s. What distance does she travel over that period of time.

Answers

Answer 1

Answer:

The distance is 58.03 m

Explanation:

Constant Acceleration Motion

It occurs when the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

[tex]v_f=v_o+at[/tex]

The distance traveled by the object is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]

The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.

The final speed is:

[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]

[tex]v_f=20.7-26.52[/tex]

[tex]v_f=-5.82\ m/s[/tex]

Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:

[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]

[tex]\displaystyle x=161.46-103.43[/tex]

x=58.03 m


Related Questions

A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is launched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?

Answers

We know, by conservation of energy :

[tex]\dfrac{kx^2}{2}=mgh[/tex]

Therefore,

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]

Putting given values, we get :

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

define these terms about speed and state their units
speed​
distance covered
time taken

Answers

Explanation:

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

  Speed  = [tex]\frac{distance}{time}[/tex]

The unit is m/s or km/hr or mile/hr

Distance covered is simply the length of the path traveled.

 The unit is m or km or miles

Time taken is the duration of an event.

  The unit is seconds or minutes or hour.

What ate the two safety precautions that should be taken before driving your car?

Answers

Answer:

When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.

Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.

During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.

Answers

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

How do lenses and mirrors compare in their interactions with light?
A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.​

Answers

Answer:

C. lenses refract light; mirrors do not

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

LENSES AND MIRRORS

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the  bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

Learn more about reflection and refraction here:

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Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2 A.If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?Express your answer in terms of V but do not type in the symbol "V"

Answers

Answer:

V' = V/2

Explanation:

The voltage across a parallel plate capacitor is given as follows:

V = Q/C

where,

V = Voltage across capacitor

Q = Charge on Capacitor

C = Capacitance of Capacitor = A∈₀/d

Therefore,

V = Qd/A∈₀

where,

A = Area of plate

d = distance between plates

∈₀ = permittivity of free space

FOR CAPACITOR 1:

Q = Q

d = d

A = A

V = V

Therefore,

V = Qd/A∈₀   --------------- equation (1)

FOR CAPACITOR 2:

V' = ?

Q' = Q

d' = d

A' = 2A

Therefore,

V' = Q'd'/A'∈₀

V' = Qd/2A∈₀

V' = (1/2)(Qd/A∈₀)

using equation (1):

V' = V/2

A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N

Answers

Answer:

(d) 2.93 x 10⁻³ N

Explanation:

Given;

current in the wire, I = 11.4 A

angle of inclination, θ = 11.4⁰

magnetic field on the wire, B = 11. 4  x 10⁻³

length of the wire, L = 11.4 cm = 0.114 m

The magnitude of magnetic force on the wire is given by;

F = BILsinθ

F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)

F = 0.00293 N

F = 2.93 x 10⁻³ N

Therefore, the correct option is "D"

Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s).
Let B = 0.86 T , I = 2300 A , m = 20 kg , and L = 55 cm . For simplicity, assume the net force on the object is equal to the magnetic force, as in parts A and B, even though gravity plays an important role in an actual launch into space.
Express your answer using two significant figures.

Answers

Answer:

The distance of the bar D = 1153 km

Explanation:

The electric force is the one that takes place between electric charges.

The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.

Recall that:

Electrical force(F) = I*B*L

where;

I = the current,

B = the magnetic field strength,

L = the length of the bar

However;

From the second equation of motion,

F = Ma

Since; (F) = I*B*L

Then,

Ma = IBL,

where;

M is the mass;

a is the acceleration

Making the acceleration (a) the subject of the formula, we have

a = IBL/M

Similarly;

From the third equation of motion;

v^2= u^2+2as,

where v and u are the final velocity and the initial velocity respectively

Here u = 0

Also; let distance s = D

Then

v^2 = 2aD

where;

a = IBL/M

Making the distance D  the subject of the formula, we get:

D = v^2/2a = v2*M/(2IBL)

D = 11200² × 20/(2×2300×0.86×0.55)

D = 1153047.155 m

D = 1153 km

Can someone help me with my physics

A pendulum is a body that is suspended from a fixed point so that it can swing back and forth through an exchange of kinetic energy and gravitational potential energy. Using 1–2 sentences, explain what happens to the kinetic energy and gravitational potential energy of the pendulum at the highest point and at the lowest point of its swing.

Answers

In a simple pendulum with no friction, mechanical energy is conserved. Total mechanical energy is a combination of kinetic energy and gravitational potential energy. As the pendulum swings back and forth, there is a constant exchange between kinetic energy and gravitational potential energy.

Answer:

Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.

Use the information below for the next five questions:



An open organ pipe emits B (494 Hz) when the temperature is 14°C. The speed of sound in air is v≈(331 + 0.60T)m/s, where T is the temperature in °C



Determine the length of the pipe.



What is the wavelength of the fundamental standing wave in the pipe?



What is frequency of the fundamental standing wave in the pipe?



What is the frequency in the traveling sound wave produced in the outside air?



What is the wavelength in the traveling sound wave produced in the outside air?



How far from the mouthpiece of the flute should the hole be that must be uncovered to play D above middle C at 294 Hz? The speed of sound in air is 343 m/s.

Answers

Answer:  Please see answer in explanation column.

Explanation:

Given that

v≈(331 + 0.60T)m/s

where Temperature, T =  14°C

v≈(331 + 0.60 x 14)m/s

v =331+ 8.4 = 339.4m/s

In our solvings, note that

f= frequency

 λ=wavelength

L = length

v= speed of sound

a) Length of the pipe is calculated using the fundamental frequency formulae that

f=v/2L

Length = v/ 2f

= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m

b) wavelength of the fundamental standing wave in the pipe

L = nλ/2,

λ = 2L/ n

λ( wavelength )= 2 x 0.3435/ 1

= 0.687m

c) frequency of the fundamental standing wave in the pipe

F = v/  λ

= 339.4m/s/0.687m=

494.03s^-1 = 494 Hz

d) the frequency in the traveling sound wave produced in the outside air.

This is the same as the frequency in the open organ pipe = 494Hz

e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m

f) To play D above middle c . the distance is given by

L =v/ 2 f

= 343/ 2 x 294

=0.583m

I have a pen
I have a apple
what do I have now?

Answers

Answer:

You have an apple pen. :)

Answer:

I have a pen

I have a apple

apple pen

Explanation:

3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall? ​

Answers

work done = force x distance
10kg is 98.1 newtons
98.1x200=19620
i’m not sure though sorry if this is wrong :)

What is the electric force between two point charges when 91 = -4e, 92 = +3 e,
and r= 0.05 m?
e = 1.6 x 10-1C, k = 9.00 x 10°Nom/C2)
Kouch
A. -1.1 * 10-24 N
B. 1.1 * 10-24 N
C. 5,5 x 10-25 N
D. -5.5 x 10-25 N

Answers

Answer:option B

Explanation:

7. What does the changing colour perceived by the person as the filter changes indicate to you
about white light?

Answers

Answer:

lo

Explanation:

A 37.0-kg child swings in a swing supported by two chains, each 3.06 m long. The tension in each chain at the lowest point is 410 N. (a) Find the child's speed at the lowest point.______m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)_______ N(upword)

Answers

Answer:

1. 6.15m/s

2. 820N

Explanation:

The total upward force

= 410x2

= 820

g = 9.81

a = v²/r

= 2xT - msg = m x v²/r

= 820-37*9.81 = 37v²/3.06

= 820-362.97 = 37v²/3.06

= 457.03 = 12.09v²

To get v²

V² = 457.03/12.09

V² = 37.8

V = √37.8

V = 6.15m/s

B. We already have the answer to this question

The force exerted is simply gotten by this calculation

2x410

= 820N

What is the result of increasing the speed at which a magnet moves in and
out of a wire coil?
A. The current in the wire increases.
B. The magnetic field around the magnet decreases.
C. The current in the wire decreases.
D. The magnetic field around the magnet increases.

Answers

Answer:

A. The current in the wire increases.

Explanation:

Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.

This phenomenon shows the inter-relationship between electricity and magnetic fields.

Magnetic fields are induced by passage of electric current. Also, electric current can be produce by magnetic fields. When the speed at which a magnet moves in and out of a wire coil increases, the current also increases.

A nonmechanical water meter could utilize the Hall effect by applying a magnetic field across a metal pipe and measuring the Hall voltage produced.What is the average fluid velocity in m/s for a 4.25 cm diameter pipe, if a 0.575 T field across it creates a 60.0 mV Hall voltage?

Answers

Answer:

The velocity is  [tex]v =2.455 \ m/s[/tex]

Explanation:

From the question we are told that  

   The diameter of the pipe is  [tex]d = 4.25 \ cm = 0.0425 \ m[/tex]

    The magnetic field is  [tex]B = 0.575 \ T[/tex]

     The hall voltage is  [tex]V_H = 60.0 mV = 60 *10^{-3} \ V[/tex]

Generally the average fluid velocity is mathematically represented as

          [tex]v = \frac{V}{ B * d }[/tex]

=>       [tex]v = \frac{60*10^{-3}}{ 0.575 * 0.0425 }[/tex]

=>       [tex]v =2.455 \ m/s[/tex]

During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young's modulus of 1.5×109 N/m2?

Answers

Answer:

[tex]3978873.58\ \text{Pa}[/tex]

[tex]0.00002122\ \text{m}[/tex]

Explanation:

F = Force = 5000 N

r = Radius of circular cross section = 2 cm

l = Length of disk = 0.8 cm

A = Area = [tex]\pi r^2[/tex]

Y = Young's modulus = [tex]1.5\times 10^9\ \text{N/m}^2[/tex]

Pressure is given by

[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}[/tex]

The pressure on the cross section is [tex]3978873.58\ \text{Pa}[/tex]

The change in length of the cross section is given by

[tex]\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}[/tex]

The deformation produced is [tex]0.00002122\ \text{m}[/tex]

HELP ASAP!

What happens to a circuit's resistance (R), voltage (V), and current (I) when you increase the length of the wire in the circuit?

Answers

Answer:

B R is constant V increases /increases

If one increase the length of the wire in the circuit, then, R increases, V decreases, I decreases. The correct option is D.

The resistance (R) of a wire increases as its length is increased in a circuit. This is because longer wires have more resistance because resistance increases with length.

Ohm's Law states that the current (I) in the circuit will drop if the resistance (R) rises while the voltage (V) stays constant. This is due to the inverse proportionality between current and resistance.

Thus, the correct option is D.

For more details regarding Ohm's law, visit:

https://brainly.com/question/1247379

#SPJ5

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The other end of the spring is fixed to a wall. The spring is compressed by 10.0 cm from its equilibrium position and released from rest.
1) What is the speed of the object when it is 8.00 cm from equilibrium? (Express your answer to three significant figures.)
2) What is the speed when the object is 5.00 cm from equilibrium? (Express your answer to three significant figures.)
3) What is the speed when the object is at the equilibrium position? (Express your answer to three significant figures.)

Answers

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]                              

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]      

 

3) When the object is at the equilibrium position, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]

I hope it helps you!                                                                                        

(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s

(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s

(3) the speed of the object when the spring is at equilibrium is 0.748 m/s

Compression of spring and conservation of energy:

Given that the mass of the object, m = 5 kg

spring constant, k = 280 N/m

compression of the spring , x = 10 cm = 0.1m

(i) the spring compression is at d = 8cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]

v = 0.449 m/s

(ii) the spring compression is at d = 5cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]

v = 0.648 m/s

(iii) the spring is at equilibrium so compression is at d = 0cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]

v = 0.748 m/s

Learn more about conservation of energy:

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Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface on the ground. The spring is compressed 0.5 m past its equilibrium. What is the initial energy of the system.

Answer: 75 joules

Answers

glad it’s figured out

Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled eastward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.

Answers

Answer:

lucky mauld mauldgomary was an british poet...

A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?

Answers

Answer:

The number is  [tex]N = 300[/tex]

Explanation:

From the question we are told that

   The  net charge is  [tex]Q = -4.8 *10^{-17 } \ C[/tex]

Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]

Generally the number of excess electrons is mathematically represented as

      [tex]N = \frac{Q}{e}[/tex]

=>  [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]

=>  [tex]N = 300[/tex]

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the horizontal. The ball rolls without slipping down the incline and at the bottom has a speed of 4.9 m/s. How many revolutions does the ball rotate through as it rolls down the incline

Answers

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!  

The ball rotates 6.78 revolutions.

     

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!        

At the bottom the ball has the following angular speed:

[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]

To find the revolutions we need the time, which can be found using the following equation:                

[tex] v_{f} = v_{0} + at [/tex]  

[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)

So first, we need to find the acceleration:

[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex]    (2)  

By entering equation (2) into (1) we have:

[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]

Since it starts from rest (v₀ = 0):  

[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]

Finally, we can find the revolutions:  

[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

If the peak wavelength of a star at rest is 615 nm, then what peak wavelength is observed when the star is traveling 2,500,000 m/s toward the Earth.

Answers

Answer:

1612903nm

Explanation:

Doppler effect can be regarded as the change in frequency of a wave with respect toobserver that move relative to the wave source

We can expressed as

(λo - λs)/λs = v/s

λo= peak wavelength

λs= peak wavelength observed

C= speed of light

(λo -615×10^9 )/615×10^9 = 2,500,000/(3×10^8)

1.5375=3×10^8λo -184.5

1.5375+184.5=3×10^8λo

186=3×10^8λo

λo=1612903nm

Therefore the peak wavelength that is observed when the star is traveling away from the eart to the velocity given is 1612903nm

branches of sicence​

Answers

Answer: Natural science can be divided into two main branches

Explanation:

life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.

how does tom and jerry movie character influence your attitude​

Answers

Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

Describe why you are doing the experimen?​

Answers

What is the experiment ?

Answer:

An experiment is a procedure carried out to support, refute, or validate a hypothesis. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.

Explanation:

A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?

Answers

Answer:

t = 1.32 s

Explanation:

We are given;. Frequency of C4 note; F_c = 262 Hz

In conversions, we know that 1 Hz = 1 cycle/s

Thus, F_c = 262 cycles/s

Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.

346 air pressure maxima denotes that the air pressure maxima is 346 cycles.

Thus, time will be;

t = 346 cycles/262 cycles/s

t = 1.32 s

The time taken for the musical note to pass the stationary listener is 1.32 s.

The given parameters:

frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346

The frequency of a sound wave is defined as the number of cycles completed per second by the wave.

[tex]F = \frac{n}{t}[/tex]

where;

t is the time to compete the maximum cycle

The time taken for the musical note to pass the stationary listener is calculated as follows;

[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]

Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.

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