Which statements accurately describe sound waves? Check all that apply.
Sound waves are transverse waves.
Sound waves require a medium to transfer energy.
Sound waves start with a vibration.
Energy from a vibration creates a mechanical wave.
Sound can be heard as soon as an object begins vibrating.

Answer:

megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter

Explanation:

plz mark brainlest

Answer:

Explanation:

1 ) Let the initial horizontal velocity of car be v .

For vertical displacement

vertical displacement h = 21.3 - 2.3 = 19 m

Time taken to fall by 19 m be t

19 = 1/2 x 9.8 t² ( initial downward velocity is zero )

t = 1.97 s

This is also the time taken to cover horizontal displacement of 54 m which is width of river .

horizontal speed v = 54 / 1.97 m /s

v = 27.41 m /s

2 )

At the time of landing on other side , car will have both vertical and horizontal speed .

vertical speed

v = u + gt

= 0 + 9.8 x 1.97 = 19.31 m /s

horizontal speed will remain same as  the initial speed = 27.41 m /s

Resultant speed = √ ( 27.41² + 19.31² )

= √ ( 751.3 + 372.87)

= 33.52 m /s

Answer:

47947.52 J.

Explanation:

From the question,

Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1

Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.

Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.

Substitute these values into equation 1 above

Q = 640×4.816(48)

Q = 147947.52 J.

Hence the amount of heat given off is 47947.52 J.

Answer:

True

Explanation:

Answer:

density = mass/volume

so . . .  

density = (36 g)/(15 cm³) = 2.4 g/cm³

Explanation:

Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.

Balloon will burst, due to more heat and in another flask air gets heated.

"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."

"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."

Know more about heat here

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Answer:

Compared to the first student, the second student did twice as much work as the first student.

Explanation:

The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance).  The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).

Answer:

b

Explanation:

because of bonds and the bowl is used for heating purpose

Answer:

I think b not sure but I hope its right and is that the nwea test??

Answer:

Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.

The answer:

(a) 88N, northward.

(b) 78.4‬, southward.

Explanation:

(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)

Normal force = mass X acceleration due to gravity

Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case  X acceleration due to gravity)

Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N

While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)

Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N

With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.

The answer is 88N, northward.

(b)  Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.

Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136‬ N

We can conclude that the reaction force (136‬ N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.

This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:

kinetic force = (coefficient of kinetic friction) X  (mass of packing case  X acceleration due to gravity)

= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4‬N

Answer:

The frequency must be: [tex]725\,\,10^{12} \,Hz[/tex]

Explanation:

If the work function of the metal ([tex]\phi[/tex]) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:

[tex]KE= h*f-\phi[/tex]

considering for the minimum KE = 0, and using the Plank constant h in eV s as: 4.14 * 10 ^(-15) eV s, to solve for the frequency:

[tex]h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz[/tex]

Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.

Explanation:

Answer:

A is the answer okkkkkkkkkkkkkkkk

Answer:

15m/s

Explanation:

[tex]20m = \frac{1}{2} g {t}^{2} \\ t = 2s \\ t0 = 2s - 1s = 1s \\ vt + \frac{1}{2} g {t}^{2} = 20m \\ v \times 1s + \frac{1}{2} \times 10 \times 1 = 20 \\ v = 15[/tex]

Answer:

D. 3 : 1

Explanation:

Let suppose that A and B are particles. From statement we know that [tex]K_{A} = K_{B }[/tex] (same kinetic energy) and [tex]m_{A} = 9\cdot m_{B}[/tex]. Then,

[tex]K_{A} = K_{B}[/tex]

[tex]\frac{1}{2}\cdot m_{A}\cdot v_{A}^{2} = \frac{1}{2}\cdot m_{B}\cdot v_{B}^{2}[/tex]

[tex]m_{A} \cdot v_{A}^{2} = m_{B}\cdot v_{B}^{2}[/tex]

[tex]\frac{v_{B}}{v_{A}} = \sqrt{\frac{m_{A}}{m_{B}} }[/tex]

[tex]\frac{v_{B}}{v_{A}} = 3[/tex]

And the ratio of the momentum of A to the momentum of B is:

[tex]r = \frac{m_{A}\cdot v_{A}}{m_{B}\cdot v_{B}}[/tex]

[tex]r = 9\times \frac{1}{3}[/tex]

[tex]r = 3[/tex]

Hence, the correct answer is D.

Answer:

Explanation:

The magnitude of the force is known as the resultant.

R = √Fx²+Fy²

Fx = 80cos 20 + 40cos70

Fx = 80(0.9397)+40(0.3420)

Fx = 75.176 + 13.68

Fx = 88.856N

Fy = 80sin 20 + 40sin70

Fy = 80(0.3420)+40(0.9397)

Fy = 27.36 + 37.588

Fy = 64.948N

R = √88.586²+64.948²

R = √7,847.48+4,218.24

R = √12,065.72

R = 109.5

R = 110N

Hence the magnitude of the forces is 110N

Answer:

16.6 m

Explanation:

Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.

Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.

At catch up, D = d'

d + vt = v't

d = v't - vt

d = (v' - v)t

Substituting the values of the variables into d, we have

d = (2.75 m/s - 1.17 m/s)10.5 s

d = (1.58 m/s)10.5 s

d = 16.59 m

d ≅ 16.6 m

So, the other person was 16.6 m ahead of you when you started running.

Answer:

Circuit B

Explanation:

To know the correct answer to the question, we shall determine the reading of the ammeter i.e the current in each circuit. This can be obtained as follow:

For circuit A:

Resistance (R) = 5 ohms

Voltage (V) = 1.5 V

Current (I) =?

V = IR

1.5 = I × 5

Divide both side by 5

I = 1.5 / 5

I = 0.3 A

For circuit B:

We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:

Resistance 1 (R₁) = 5 ohms

Resistance 2 (R₂) = 5 ohms

Total resistance (Rₜ) =?

Since they are in parallel connections, the total resistance can be obtained as illustrated below:

Rₜ = R₁ × R₂ / R₁ + R₂

Rₜ = 5 × 5 / 5 + 5

Rₜ = 25 / 10

Rₜ = 2.5 ohms

Finally, we shall determine the current in the circuit.

Resistance (R) = 2.5 ohms

Voltage (V) = 1.5 V

Current =?

V = IR

1.5 = I × 2.5

Divide both side by 2.5

I = 1.5 / 2.5

I = 0.6 A

For circuit C:

We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:

Resistance 1 (R₁) = 5 ohms

Resistance 2 (R₂) = 5 ohms

Total resistance (Rₜ) =?

Since they are in series connections, the total resistance can be obtained as illustrated below:

Rₜ = R₁ + R₂

Rₜ = 5 + 5

Rₜ = 10 ohms

Finally, we shall determine the current in the circuit.

Resistance (R) = 10 ohms

Voltage (V) = 1.5 V

Current =?

V = IR

1.5 = I × 10

Divide both side by 10

I = 1.5 / 10

I = 0.15 A

SUMMARY:

Circuit >>>>>>> Current

A >>>>>>>>>>>> 0.3 A

B >>>>>>>>>>>> 0.6 A

C >>>>>>>>>>>> 0.15 A

from the above calculations, circuit B has the highest ammeter reading.

Answer:

A resultant vector is a combination or, in simpler words, can be defined as the sum of two or more vectors which has its own magnitude and direction.

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)

∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - [tex]a_{Net}[/tex] × t

∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex]  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

Answer:

a)  i = 0.746 A, b)   # _electron = 2.44 10¹⁹ electrons,  c)   E = 1.87 10⁴ J

Explanation:

a) The definition of current is the charge per unit of time

           i = Q / t

           i = 1.56 / 2.09

           i = 0.746 A

b) Let's look for the cargo in passing at this time

           i = Q / t

           Q = i t

            Q = 0.746 5.24

            Q = 3.904 C

an electron has a charge e = -1.6 10⁻¹⁹ C, let's use a direct proportions rule

            # _electron = 3.904 C (1 electron / 1.6 10⁻¹⁹)

            # _electron = 2.44 10¹⁹ electrons

the number of electrons has to be an integer

c) In this part you are asked to calculate the power

            P = V i

            P = 12 0.746

            P = 8.952 W

            P = E/t

            E = P t

            E = 8.952 2.09

            E = 1.87 10⁴ J

Answer:

Explanation:

When two charges of equal magnitude and sign approach each other, they interact through Coulomb's law

          F = [tex]k \frac{q_ 1q_2}{z^2 }[/tex]k q1 q2 / r2

In you case the house are of equal magnitude and sign

          q1 = q2 = q

          F = k q2 / r2

Let's analyze this expression, the charge is repulsive on each charge, when they are on a collision course as they approach they feel an electric field opposite to their direction of movement, this field decreases its speed, the closer they get, the greater the repulsive force. , up to the point where this force is equal to or greater than the impulse, therefore the point where the velocity reaches zero, for this reason the particles do not actually touch

Answer:

a. Power = 1000 Watts or 1 Kilowatts.

b. Current = 4 Amperes.

Explanation:

Given the following data;

Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules

Voltage = 250V

Time = 30 minutes to seconds = 30 * 60 = 1800 seconds

To find the power rating;

Power = energy/time

Substituting into the equation, we have;

Power = 1800000/1800

Power = 1000 Watts or 1 Kilowatts.

b. To find the current taken from the supply;

Power = current * voltage

1000 = current * 250

Current = 1000/250

Current = 4 Amperes.

Answer:

Explanation:

Yes , it is possible .

When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .

Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .

V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .

Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

A) i) Dynamic error ≈ 3.1%

   ii) phase shift ≈ -12°

B) 79971.89 rad/s

Explanation:

Given data :

Damping ratio = 0.5

natural frequency = 18,000 Hz

a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz

i) Dynamic error

This can be calculated using magnitude ratio formula attached below is the solution

dynamic error ≈ 3.1%

ii)  phase shift

This phase shift can be calculated using frequency dependent phase shift formula

phase shift ≈ -12°

B) Determine resonance frequency

  Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s

C) The maximum magnitude ratio that the system can achieve

Answer:

a) 0, ±1.65  b)  ± 0.825m

Explanation:

This is a sound interference exercise, it can be described by the path difference between the two waves

for the case of constructive interference

          Δr = 2n λ/ 2                   n = 0, 1, 2 ...

for the case of destructive interference

          Δr = (2n + 1) λ/ 2

the speed of sound is related to the wavelength

          v = λ f

          λ = v / f

          λ= 340/206

          λ = 1.65 m

let's set a reference system in the center between the two speakers

a) let's find the distances for constructive interference

            Δr = 2n 1.65 / 2

            Δr = 1.65 n

* the first interference occurs at n = 0

          Δr = 0

therefore the interference in the center is maximum

* n = 1

          Dr = 1.65 m

the second inference occurs at 1.65 m from the center, therefore there is a right wing and a left wing,

We do not have any more interference between the speakers because

n = 2 Δr = 3.3m this distance can be from the speaker

b) let's look for the destructive interference points

           Δr = (2n + 1) 1.65 / 2

           Δr = (2n + 1) 0.825

m = 0       Δr = 0.825m

m = 1        Δr = 2,475m

We can see that we only have the first destructive interference, one on each side.

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

21560 J work must be done to raise an 1100 kg car 2m above the ground.

Work = mass * gravity * height

= 1100 * 9.8 * 2

= 21560 J

In precis, work is carried out whilst pressure acts upon an item to purpose a displacement. 3 portions must be regarded in the way to calculate the quantity of work. The ones 3 portions are force, displacement, and the perspective between the pressure and the displacement.

Paints carried out are elaborated in this type of manner that includes both forces exerted on the body and the whole displacement of the body. This block is preceded by a steady force F. The purpose of this pressure is to move the body a certain distance d in an immediate route in the route of the pressure.

Learn more about work here: https://brainly.com/question/1382377

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Answer:

Speed = 10.24 m/s.

Explanation:

Given the following data;

Distance = 100m

Time = 9.77

To find her speed;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

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

Substituting into the equation, we have;

[tex]Speed = \frac{100}{9.77}[/tex]

Speed = 10.24 meter per seconds.

Answer:

Check Newton's Rings:

d = height of air film

s = distance from center to ring being considered

R = radius of circle considered

The approximate formula is:

d = s^2 / (2 R)   or s = (2 R d)^1/2

If we just use 4000 mi for R and 1/4 mi for d the height

we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi

Answer:

2.5 cm

Explanation:

Using the relation :

Refractive index = Real Depth / Apparent depth

Refractive index = 1.6

Real depth = 4cm

Virtual depth = apparent depth = x

1.6 = 4cm / x

1.6x = 4

x = 4 / 1.6

x = 2.5

Hence, virtual depth = 2.5cm