Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave

Ez Physics question will mark brainliest.

Sound Is An Example Of A:Select One:O A. Rolling Wavesb. Longitudinal WaveO C. Traverse WavesO D. Surface

Answers

Answer 1

Answer:

The answer is B. longitude wave


Related Questions

Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3 ​

Answers

Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)

Answers

Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.

Answers

The horizontal force applied to the lower block is approximately 1,420.85 Newtons

The known parameters are;

The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N

The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N

The length of the string attached to the block, W₂, l = 6 m

The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m

The coefficient of friction between the surfaces, μ = 0.25

Let T represent the tension in the string

The upward force on W₂ due to the string = T × sin(θ)

The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)

The tension in the string, T = N₂ × μ × cos(θ)

∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)

sin(θ) = √(6² - 5²)/6

cos(θ) = 5/6

T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T ≈ 183.27 N

The normal reaction on W₂, N₂ = T/(μ × cos(θ))

∴ N₂ = 183.27/(0.25 × 5/6) = 879.7

N₂ ≈ 879.7 N

The friction force, [tex]F_{f2}[/tex] = N₂ × μ

∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N

The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂

[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N

The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ

∴  [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N

The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = 1,200.925 N + 219.925 N = 1,420.85 N

The horizontal force applied to the lower block, P ≈ 1,420.85 N

A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects

Answers

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?

Answers

Answer:

Charge of the particle is 1 coulomb.

Explanation:

Force, F:

[tex]{ \bf{F=BeV}}[/tex]

F is magnetic force.

B is the magnetic flux density.

e is the charge of the particle.

V is the velocity

[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]

Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.

Answers

We have that the spring constant is mathematically given as

[tex]k=2.37*10^{11}N/m[/tex]

Generally, the equation for angular velocity is mathematically given by

[tex]\omega=\sqrt{k}{m}[/tex]

Where

k=spring constant

And

[tex]\omega =\frac{2\pi}{T}[/tex]

Therefore

[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]

Hence giving spring constant k

[tex]k=m((\frac{2 \pi}{T})^2[/tex]

Generally

Mass of earth [tex]m=5.97*10^{24}[/tex]

Period for on complete resolution of Earth around the Sun

[tex]T=365 days[/tex]

[tex]T=365*24*3600[/tex]

Therefore

[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]

[tex]k=2.37*10^{11}N/m[/tex]

In conclusion

The effective spring constant of this simple harmonic motion is

[tex]k=2.37*10^{11}N/m[/tex]

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The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.

I will give brainliest!

P.S. AL2006 if you see this please help!​

Answers

I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.

"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .

I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.

Answer:

1 -> 2 & 3 -> 4: Isobaric process

4 -> 1: Work done BY the system

2 -> 3: Work done ON the system

W(total): W = 0J

Explanation:

The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.

I hope this helps :))

​Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.

Answers

Answer:

PlROCA

Explanation:

The cells lie odjacent to the sieve tubes​

Answers

Answer:

Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.

Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.

Answers

Answer:

Florida is tropical, with a significant rainy seson

define nortons theorem​

Answers

Answer:

In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.

PLEASE ANSWER IF YOU CAN AND NOT FOR THE SAKE OF GAINING POINTS!

Answers

A is the correct answer because the the plane mirror would move the same speed in the same direction

An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.​

Answers

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

Mention the importance of occupation??​

Answers

Answer:

ln the contemporary time , farming can be considered as comparitively important occupation as it can feed the population , So agriculture is having a greater importance than any other occupation.

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B​

Answers

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor

Answers

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]

Calculating the common potential:

[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]

Calculating the charge after redistribution:

[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]        

[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]

What is an internal resistance?

Answers

Explanation:

some thing inside a resistor

When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity

Answers

The answer is terminal force

Hi Friends!

please help me with these questions!

SUBJECT: Chemistry, Physics,Biology

Answers

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

Is it true that as we gain mass the force of gravity on us decreases

Answers

Answer:

No. As we gain mass the force of gravity on us does not decrease

15 . A scientist who studies the whole environment as a working unit .

Botanist
Chemist
Ecologist
Entomologist

Answers

Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4​

Answers

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

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The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?

Answers

Answer:

[tex]v_1=87.40m/s[/tex]

Explanation:

From the question we are told that:

Mass of arrow [tex]m=52g[/tex]

Mass of rock [tex]m_r=1.50kg[/tex]

Height [tex]h=0.47m[/tex]

Generally the equation for Velocity is mathematically given by

 [tex]v = \sqrt{(2gh)}[/tex]

 [tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]

 [tex]v= 3.035m/s[/tex]

Generally the equation for conservation of momentum is mathematically given by

 [tex]m_1v_1=m_2v_2[/tex]

 [tex]0.052kg * v = 1.5 * 3.03m/s[/tex]

 [tex]v_1=87.40m/s[/tex]

cyclist always bends when moving the direction opposite to the wind. Give reasons​

Answers

When he bends he kinda off his feet and light but if it’s not i’m so sorry this is just my thinking.

Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2

Answers

The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶  W/m².

The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.    

a) Intensity of the electromagnetic wave for the electromagnetic field.

The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:

[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex]   (1)

Where:

c: is the speed of light = 3.00*10⁸ m/s  

E: is the magnitude of the electric field = 1.5 V/m

ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²

Hence, the intensity of the electromagnetic wave (eq 1) is:

[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]                                                                                          

b) Intensity of the electromagnetic wave for the magnetic field

We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:

[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex]   (2)

Where:

B: is the magnitude of the magnetic field = 10⁻¹⁰ T

μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A

Therefore, the intensity of the electromagnetic wave (eq 2) is:

[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]

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A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.

a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.

Answers

(A) The pressure will be greater than 10% overpressure limit.

(B) The final pressure will be "582.915 kPa".

Given:

Volume,

[tex]V = 0.0021 \ m^3[/tex]

Initial pressure,

[tex]P_o= 495 \ kPa[/tex]

Initial temperature,

[tex]T_o = -14^{\circ} C[/tex]

            [tex]= 259 \ K[/tex]

Final temperature,

[tex]T = 32^{\circ} C[/tex]

(B)

Number of moles,

→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]

then,

The final absolute pressure,

→ [tex]P = \frac{nRT}{V}[/tex]

      [tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]

      [tex]=(\frac{T}{T_o} )P_o[/tex]

      [tex]= (\frac{305}{259} )\times 495[/tex]

      [tex]= 582.915 \ kPa[/tex]

Thus the above approach is correct.

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The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?

Answers

When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answers

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

[tex]\phi = BAcos \theta[/tex]

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber

a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm

Answers

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

20

1

+

12

1

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

−16

1

+

12

1

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m​

Answers

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex]  Rounding to the correct number of sig fig's to simplify:

[tex]v=\sqrt{400+2.0*10^2}[/tex] to get

v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

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