If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.

Answers

Answer 1

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.


Related Questions

Under normal circumstances: _________
a. Fetal Hb binds to oxygen more tightly than Mb binds.
b. Fetal Hb binds oxygen more tightly in the absence of 2,3-BPG.
c. Fetal Hb does not bind to oxygen.
d. Adult Hb has the lowest affinity for oxygen of the 3.
e. More than one of these statements is correct.

Answers

Answer:

Fetal Hb binds oxygen more tightly than adult Hb (not option a)

What Are the type's of Tidal turbines?

Answers

Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put​

Answers

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball

Answers

Answer:

  t = 0.437 s

Explanation:

Sound is a wave so its speed is constant

         v = x / t

         t = x / v

         

indicates that the distance is x = 150 m

         t = 150/343

         t = 0.437 s

this is the time it takes to hear the hit

To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s

find the equivalent resistance of this circuit

Answers

Answer:

Req = 564 Ω

Explanation:

The equivalent resistance between R1 and R2:

1/R =1/R1 + 1/R2

1/R =1/960 + 1/640

1/R = 1/384

R = 384

Now, the equivalent resistance between R and R3:

Req = 384 + 180

Req = 564 Ω

Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?

Answers

Answer:

[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

Explanation:

The pressure at a depth of a fluid is

       P = ρ g y

where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.

In this case the pressure for fluid A is

      Pa = ρₐ g y

the pressure for fluid B is

      P_b = ρ_b g y

depth y not changes as the gauge is stationary

if we look for the relationship between these pressures

       [tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]

       

        [tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]

therefore we see that the pressure measured for fluid B is different from the pressure of fluid A

if  ρₐ < ρ_b B the pressure P_b is greater than the initial reading

   ρₐ>  ρ_b the pressure in B decreases with respect to the reading in liquid A

I need help with this please!!!!

Answers

Answer:

1.84 hours

I hope it's helps you

find the upward force in Newton when each of these is under water(density of 1g/cm3) a lump of iron of volume 2000cm3​

Answers

Answer:

Upthrust = 19.6 N

Explanation:

When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.

Upthrust = density of liquid x gravitational force x volume of object

i.e U = ρ x g x vol

Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]

So that;

U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])

   = 19.6 Kg m/[tex]s^{2}[/tex]

U = 19.6 Newtons

The upthrust on the iron is 19.6 N.

which characteristic of nuclear fission makes it hazardous?

Answers

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

A student has a large number of coins of different diameters, all made of the same metal. She wishes to find the density of the metal by a method involving placing the coins in water.

a) State the formula needed to calculate the density.

b) Describe how the measurements of the required quantities are carried out.​

Answers

Answer:

a)density = mass /volume

b)to find volume put water into a container .measure the level of water , put the coins into the beaker containing water , measure the level of water again, subtract the new volume withe the first one . the result is the volume of coins

state the laws of reflection​

Answers

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

explain what would happen if the cell was disconnected from the circuit ( please help me)​

Answers

Most of the time, "cell" means a battery, and if it gets disconnected, all voltage and current in the circuit goes away.  

I know this is a lame answer, but it kinda depends on what the cell was doing in the circuit, and what else is in the circuit besides the cell, and you haven't told us anything about these details, so that's really all we can guess.

Although your question is incomplete a general answer within the concept of your question is provided:    when the cell is disconnected from the Circuit the flow of voltage across the circuit will be halted.

A cell/battery is often used as a voltage source in electrical circuits, since the cell is a power source, when the cell is connected to a circuit the cell discharges some of its voltage to the components of the circuit ( such as capacitors ) to keep the circuit functional. so when the cell is disconnected from the circuit the flow of voltage is halted

Hence we can conclude that without a voltage source ( cell ) in a circuit there will be no flow of voltage across the components of the circuit.

learn more : https://brainly.com/question/16598952

In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.

Answers

Answer:

λ₂ = 357.3 nm

Explanation:

The expression for double-slit interference is

          d sin θ = m λ                 constructive interference

          d sin θ = (m + ½) λ        destructive interference.

The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits

         d sin θ = m λ₁

       

now ask for destructive interference for m = 4

        d sin θ = (m + ½) λ₂

we match these two expressions

         m λ₁ = (m + ½) λ₂

         λ₂ = ( m / m + ½) λλ₁  

let's calculate

         λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]

        λ₂ = 357.3 nm

can you guys pls also solve for average speed.

Answers

Answer:

d_t = 3.05km

v_a = 4.3km/h

Explanation:

42mins*(2/3) = 28mins

42mins-28mins = 14mins

d = v*t

d_1 = (4km/h)*(1h/60mins)*(28mins)

d_1 = 1.87km

d_2 = (5km/h)*(1h/60mins)*(14mins)

d_2 = 1.17km

d_t = d_1+d_2

d_t = 1.87km+1.17km

d_t = 3.05km

v_a = (v_1+v_2)/2

v_a = [(2*4km/h)+5km/h)]/3

v_a = 4.3km/h

If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.

Answers

Answer:

W = 641.52 J

Explanation:

The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:

[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]

where,

W = Work Done = ?

m = mass = 6 kg

v = speed = 4.2 m/s

g = acceleration dueto gravity = 9.81 m/s²

h = height = 10 m

Therefore,

[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]

W = 52.92 J + 588.6 J

W = 641.52 J

Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?

Answers

Answer:

Explanation:

Given force between 2 currents carrying

wires = F₀

Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L

where I₁=Current in wire 1

           I₂= Current in wire 2

           L= Length of the wire

when one current is doubled and the other is halved

I₁= 2 I₁

I₂=    I₂/2

F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L

The question is in the photo.​

Answers

Answer:

heyaa thereeee

see temperature is rising in interval of

0 to 4 minutes

8 to 10 minutes

but 8 to 10 is NOT in options

so answer is option a) 0 to 4 minutes

:))))

A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?

Answers

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

[tex] A = l*w [/tex]

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]                            

And the largest area is:

[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Answer:

the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m

the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

[tex]A_l[/tex]= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

[tex]A_s[/tex]= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?​

Answers

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation

[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]

Solving for [tex]v_{0y}[/tex],

[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]

[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]

b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as

[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]

or

[tex]v_{0x} = 150\:\text{m/s}[/tex]

How do you know that a liquid exerts pressure?​

Answers

Answer:

The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.

name a device that converts mechanical energy into electrical energy.

Answers

Answer:

Electric generator is the device that converts mechanical energy into electrical energy

Convert 385k to temperature of

Answers

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons

pls write the formula too ​

Answers

Element is

[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]

Atomic number=13Mass number=27

[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]

Express 6revolutions to radians

Answers

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

the specific heat capacity of a substance is 500J/kg/oC. Find the heat required to rise the temperature of 10 quintial of the substance by 3 degree celcius

Answers

Heat capacity=c=5000J/kg°CMass=10quintal=1000kg=mTemeperature=T=3°CHeat=Q

[tex]\boxed{\sf Q=mc\Delta T}[/tex]

[tex]\\ \sf\longmapsto Q=1000(5000)(3)[/tex]

[tex]\\ \sf\longmapsto Q=15000000J[/tex]

[tex]\\ \sf\longmapsto Q=1.5\times 10^7J[/tex]

2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor​

Answers

Check attached photo

Check attached photo

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube

Answers

Answer:

[tex]v=0.9833\ c[/tex]

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

[tex]$\text{Density} = \frac{m}{lwh}$[/tex]

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 [tex]kg/m^3[/tex]

So,

[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]

[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]

l = 0.024 m

Then for relativistic length contraction,

[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]

[tex]$\frac{v^2}{c^2}= 0.967$[/tex]

[tex]$\frac{v}{c}=0.9833$[/tex]

[tex]v=0.9833\ c[/tex]

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially

Answers

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?

Answers

Answer:

The charges are + 74.3 μC and - 74.3 μC

Explanation:

Let the charges be q and q'.

Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by

F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m

When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.

They now repel each other.

So, the magnitude of the force of repulsion is given by

F' = k[(q + q')/2][(q + q')/2]/r²

F' = k[(q + q')²/4r²

Since the magnitude of the force of attraction and repulsion are the same, we have that

F = F'

kqq'/r² = k[(q + q')²/4r²

qq' = (q + q')²/4

(q + q')² = 4qq'

q² + 2qq' + q'² = 4qq'

q² + 2qq' - 4qq' + q'² = 0

q² - 2qq' + q'² = 0

(q - q')² = 0

q - q' = 0

q = q'

Substituting q = q' into F, we have

F = kqq'/r²

F = kq²/r²

making q subject of the formula, we have

q² = Fr²/k

q = √(Fr²/k)

q = r√(F/k)

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

q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)

q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)

q = 0.189 m(0.3923 × 10⁻³ C/m)

q = 0.0743 × 10⁻³ C

q = 74.3 × 10⁻³ × 10⁻³ C

q = 74.3 × 10⁻⁶ C

q = 74.3 μC

Since q and q' initially attract, it implies that they initially had opposite charges.

So, q = 74.3 μC and q' = -74.3 μC

So, the charges are + 74.3 μC and - 74.3 μC

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground

Answers

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J

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