What do ocean waves and sound waves have in common? A. Both are transverse waves. B. both are longitudinal waves C. Both are electromagnetic waves. D. Both exhibit the same particle-to-particle interaction. E. Both are mechanical waves.

Answer: the mass is 1.60 kg and the acceleration on the rocket is...

58.4 m/s with a Power of Two

Explanation:

A model rocket lifting off from the launch pad is a good example of this principle. Just prior to engine ignition, the velocity of the rocket is zero and the rocket is at rest. If the rocket is sitting on its fins, the weight of the rocket is balanced by the re-action of the earth to the weight as described by Newton's third law of motion. There is no net force on the object, and the rocket would remain at rest indefinitely. When the engine is ignited, the thrust of the engine creates an additional force opposed to the weight. As long as the thrust is less than the weight, the combination of the thrust and the re-action force through the fins balance the weight and there is no net external force and the rocket stays on the pad. When the thrust is equal to the weight, there is no longer any re-action force through the fins, but the net force on the rocket is still zero. When the thrust is greater than the weight, there is a net external force equal to the thrust minus the weight, and the rocket begins to rise. The velocity of the rocket increases from zero to some positive value under the acceleration produced by the net external force. But as the rocket velocity increases, it encounters air resistance, or drag, which opposes the motion and increases as the square of the velocity. The thrust of the rocket must be greater than the weight plus the drag for the rocket to continue accelerating. If the thrust becomes equal to the weight plus the drag, the rocket will continue to climb at a fixed velocity, but it will not accelerate.

Answer:

3050.6 Litre .

Explanation:

Total time of heart beat = Total time of race  = 2 hrs , 39 minutes and 54 seconds

= 2 x 60 + 39 + 54/60 min

= 120 + 39 + .9 min

= 159.9 min

rate of heart beat = 170 per min

Total no of heart beat during race = 170 x 159.9

volume of blood per kg per beat = 2.5 mL per kg of weight

body weight  = 99 pounds = .4535 x 99  kg = 44.89 kg

volume of blood  per beat = 2.5 mL x 44.89 mL

= 112.225 mL .

Total required volume of blood =  112.225  x 170 x 159.9 mL

= 3050612 mL

= 3050.6 L.

Answer:

1. The density of the cube is 1.03 g/mL.

2. Dish soap

Explanation:

1. Determination of the density of the cube.

From the question given above, the following data were obtained:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:

Density (D) = mass (m) / volume (V)

D = m / V

With the above formula, we can obtain the density of the cube as follow:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

D = m / V

D = 21.7 / 21

D = 1.03 g/mL

Thus, the density of the cube is 1.03 g/mL.

2. Determination of the layer of density the cube will settle in.

From the question given above,

Subtance >>>>>>>> Density

Vegetable oil >>>>> 0.91 g/mL

Grape juice >>>>>> 0.97 m/L

Water >>>>>>>>>>> 1 g/mL

Dish soap >>>>>>>> 1.03 g/mL

Maple syrup >>>>>> 1.37 g/mL

Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.

Answer:

the correct answer to increase the thickness of the windows

Explanation:

Thermal transfer by conduction is

           P = k A   [tex]( \frac{ T_h - T_c}{L} )[/tex]

where P is the heat transfer rate, A is the area and L is the thickness

In the given case we have that the window is made up of two materials, two layers of transparent glass and a layer of air between them, let us use the index 1 for the glass and the index 2 for the air, the equation remains

           P =[tex]k_1 A_1 \frac{ \Delta T }{2 \ e_1} + k_2 A_2 \frac{ \Delta T}{e_2}[/tex]

where the number two (2) is due to the two layers of glass

the thermal conductivity of the glass is k₁ = 0.75 W/m K for a glass of

e₁ = 4 mm, this is more used for windows

the thermal conductivity of the air is k₂ = 0.024 W / m K for a temperature of T = 0ºC

In a window the area and the temperature difference are constant for each part for a given configuration

           P = [tex]( \frac{k_1}{2 \ e_1} + \frac{k_2 }{ e_2} )[/tex]   A ΔT

To decrease the speed of thermal transfer we can decrease the area, but this also reduces the lighting in the room, which brings other costs.

We can also increase the thickness of the materials, as we see the thermal conductivity of the air is very less than that of glass

                k₂  « k₁

increasing the thickness of the air layer would have the greatest effect is the decrease in heat transfer in window light

consequently the correct answer to increase the thickness of the windows

Answer:

if its winter

Explanation:

then we know it will me cold

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

Substitute the known values into the equation.

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

Answer:

Explanation:

a )

in the regions r < R₁

charge q inside sphere of radius R₁ = 0

Applying gauss's law for electric field E at distance r <R₁

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = q / ε₀ = 0 / ε₀

E = 0

Applying gauss's law for electric field E at distance R₁ < r < R₂ .

charge q inside sphere of radius R₁ = q₁

Applying gauss's law for electric field E at distance R₁ < r < R₂

electric flux through Gaussian  surface of radius r = 4π r² E

4π r² E = q₁ / ε₀

E = q₁ / 4πε₀

in the regions r> R₂

charge q inside sphere of radius R₂ = (q₁ + q₂)

Applying gauss's law for electric field E at distance r > R₂

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = (q₁ + q₂) / ε₀

E =  (q₁ + q₂) /4π ε₀

b )

For electric flux to be zero at r > R₂

(q₁ + q₂) /4π ε₀  = 0

q₁ + q₂ = 0

q₁ / q₂ = - 1 .

(a) The value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex]

(b)The ratio of the charges q1/q2 will be -1.

The charge contained divided by the permittivity equals the total electric flux out of a closed surface, according to Gauss Law.

The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane parallel to the field.

Let's analyse regions r < R₁

charge q inside sphere will be zero.

According to gauss's law for electric field E at distance r <R₁

electric flux for the gaussian surface  = 4π r² E

[tex]4\pi r^2E=\frac{Q}{\epsilon_0} =0[/tex]

By the applications of  gauss's law for electric field E at distance R₁ < r < R₂ .

q is charge inside sphere having radius R₁ = q₁

By the applications of gauss's law for electric field E at distance R₁ < r < R₂

electric flux obtained due to Gaussian surface of radius r = 4π r² E

[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1}{\epsilon_0}[/tex]

For the regions r> R₂

q is the charge inside sphere of radius hyaving  R₂ = (q₁ + q₂)

By the application of gauss's law for electric field E which is at distance r > R₂

Electric flux through Gaussian surface of radius r = 4π r² E

[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1+q_2}{\epsilon_0}[/tex]

Hence the value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex].

(b) To finding the ratio of charges the value of  electric flux must be zero at r > R₂

[tex]\frac{(q_1+ q_2) }{4\pi\epsilon_0} =0\\\\q_1 + q_2= 0\\\\\frac{q_1}{q_2} =-1[/tex]

Hence the ratio of the charges q1/q2 will be -1.

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Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of  119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

Answer:

0.9Ns

Explanation:

Impulse formula is expressed as;

Impulse = Ft = m(v-u)

Impulse = m(v-u)

m is the mass of football = 0.45kg

v is the final velocity = 22m/s

u is the initial velocity = 0m/s

Impulse = 0.45(22-0)

Impulse = 0.45 * 22

Impulse =  0.9Ns

Hence the magnitude of the impulse imparted to the receiver by the ball is 0.9Ns

Answer:

Electric field points in the direction of the force experienced by a positive charge. Magnetic field points in the direction of the force experienced by a north pole.

Answer:

V = 23.6062 cm³

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Gas Laws

Explanation:

Step 1: Define

Mass m = 455.6 g

Density D = 19.3 g/cm³

Step 2: Solve for V

Answer:

  x = 10.53 m

Explanation:

Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive

Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0

           y =y₀ + v₀ t - ½ g t²

           0 = y₀ - ½ g t²

           t = [tex]\sqrt{ \frac{2 y_o}{g} }[/tex]

we calculate

           t = √(2 3.22 / 9.8)

           t = 0.81 s

the horse goes at a constant speed

           x = [tex]v_{c}[/tex] t

           x = 13 0.81

           x = 10.53 m

this is the distance where the horse should be when in cowboy it is left Cartesian

The position of the horse from the cowboy must be 10.53 m apart.

The time taken by the cowboy to fall must be equal to the time that the horse takes to arrive at the point where the cowboy falls.

We can calculate the time taken by the cowboy to reach the saddle by applying the second equation of motion

h = ut + (1/2)gt²

here h =3.22m , u = 0m/s as he starts from rest

[tex]t=\sqrt[]{\frac{2h}{g} }\\ t=\sqrt{\frac{2*3.22}{9.8} } [/tex]

t = 0.81 s

the horse goes at a constant speed

x = vt , given that v = 13 m/s

x = 13 × 0.81

x = 10.53 m is the distance where the horse should be when the cowboy jumps.

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Answer:

The force required to get it there is 200000 N.

Explanation:

The force can be calculated by the second Newton's law:

[tex] F = ma [/tex]

Where:

m: is the mass = 25000 kg

a: is the acceleration

The acceleration is given by:

[tex] a = \frac{v}{t} [/tex]

Where:

v: is the velocity = 80 m/s

t: is the time = 10 s

[tex] a = \frac{v}{t} = \frac{80 m/s}{10 s} = 8 m/s^{2} [/tex]

Hence, the force is:

[tex] F = ma = 25000 kg*8 m/s^{2} = 200000 N [/tex]

Therefore, the force required to get it there is 200000 N.

I hope it helps you!

Answer:

impulse = change in linear momentum

3.75kgm/s

Explanation:

initial momentum is 0 because at the start golf ball is at rest

and our final momentum 0.050kg × 75m/s = 3.75kgm/s

momentum final - momentum initial

3.75kgm/s  -  0

3.75kgm/s

Answer:

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

    4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

Substitute the known values into the equation and solve for Δx.

The high jumper can jump to a max height of 4.80 m.

Answer:

   F = 1494.52 N,   θ = 44º

Explanation:

For the sum of vectors by the parallelogram method, see attached, the vectors are drawn, the parallelogram is completed and a vector is drawn from the origin of the two vectors to the end point of the rectangle, this is the resulting vector.

The attachment shows this roughly.

For the Cartesian coordinate method, each vector is decomposed into its components, they are added algebraically and then the resulting vector is composed in the form of a module and angles

we use trigonometry to decompose the vectors.

The coordinate system can be seen in the attachment

           sin θ = y / R

           cos θ = x / R

            y = R sin θ

            x = R cos θ

Vector 1

module F₁ and angle β₁ = 50

            sin 50 = [tex]\frac{F_{1y} }{F_1}[/tex]

            cos 50 = [tex]\frac{F_{1x} }{F_1}[/tex]

            [tex]F_{1y}[/tex] = F₁ sin 50

            F₁ₓ = F₁ cos 50

            F_{1y} = 600 sin 50 = 459.63 N

            F₁ₓ = 600 cos 50 = 385.67 N

Vector 2

modulus F₂ = 900N, angle β₂ = 40

            F_{2y} = 900 sin 40 = 578.51 N

            F₂ₓ = 900 cos 40 = 689.44 N

we find the resultant of each component

           F_{y} =F_{1y} + F_{2y}

           F_{y}  = 459.63 + 578.51

           F_{y}  = 1038.14 N

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 385.67 + 689.44

             Fₓ = 1075.11 N

We use the Pythagorean theorem to find the modulus of the resultant

            F = Fₓ² + [tex]F_{y}^2[/tex]

            F = √(1075.11² + 1038.14²)

            F = 1494.52 N

we use trigonometry for the angle

            tan θ = F_y / Fₓ

            θ = tan⁻¹ (F_y / Fₓ)

            θ = tan⁻¹ (1038.14 / 1075.11)

            θ = 44º

Answer:

For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed

Explanation:

Answer:

18000

Explanation:

the is to raise how much work is required is 18000

Work done to raise an object of mass 1200 kg to the height of 15 meters will be equal to 176,400 J.

In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.

The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.

As per the data given in the question,

Mass of the object, m = 1200 kg

Displacement, d = 15 meters

Then, use the equation of work done,

W = fd

W = (m×a)d

Substitute the values in above equation,

W = 1200 × 9.8 × 15

W = 176,400 J.

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Answer:

to answer it just get the distance traveled (m) and divide it by the time it took

Explanation:

so for the first one just do 100 divided by 47 which would equal 2.12 m/s

Answer:

W = 65.04 μW

Explanation:

For this exercise we will use the definition of sound intensity which is the power per unit area

           I = W / A

In this case, they indicate that for the intensity of the first student is I1 = 0.58 W / m², and the intensity of the second student is

            I₂ = 1.18 I₁

            I₂ = 1.18 0.58

            I₂ = 0.6844 W / m²

Ask the power

              W = I A

the diameter of the size is 1.1 cm and the radius is r = 0.55 cm = 0.55 10⁻² m, the approximate eardrum area as a circle

             A = π r²

we substitute

             W = I π r²

let's calculate

             W = 0.6844 π (0.55 10⁻²)²

             W = 6.504 10⁻⁵ W

they ask us to reduce to microwatts

              W = 6.504 10⁻⁵ W (106 μW / 1W)

              W = 65.04 μW

the power is constant, what changes is the intensity that depends on the sensor area, therefore the two students review the same power

Acoustic power, in microwatt W is 65.04 μW

We know that;

I = W / a

In case 1;

I1 = 0.58 W / m²

So, I₂ = 1.18 I₁

I₂ = 1.18 × 0.58

I₂ = 0.6844 W / m²

Diameter of the size = 1.1 cm

So,

Radius r = 1.1/2 = 0.55 cm

Area A = πr²W = Iπr²

W = 0.6844(3.4)(0.55)²

W = 6.504 × 10⁻⁵ W

Acoustic power, in microwatt W = 65.04 μW

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Answer:

always same

Explanation:

velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity

Answer:

2.40 Km/hr.

Explanation:

From the question given above, the following data were obtained:

Mass of truck (m₁) = 3500 kg

Velocity of truck (u₁) = 7 Km/hr

Mass of car (m₂) = 1200 kg

Velocity of car (u₂) = 11 km/hr

Post collision velocity (v) =?

The post collision velocity of the truck and car can be obtained as illustrated below:

m₁u₁ – m₂u₂ = v(m₁ + m₂)

(3500 × 7) – (1200 × 11) = v(3500 + 1200)

24500 – 13200 = v × 4700

11300 = v × 4700

Divide both side by 4700

v = 11300 / 4700

v = 2.40 Km/hr.

Therefore, the post collision velocity of the truck and car is 2.40 Km/hr.

Answer:

a) the required time interval is 0.25 s

b) the distance traveled while accelerating is 0.075 m

Explanation:

Given that;

Initial speed of the squid u = 0.15 m/s

final speed of the squid v = 0.45 m/s

acceleration a = 1.2 m/s²

Time period T = ?

using equation of motion

v = u + at

solve for t

at = v - u

t = v - u / a

we substitute

t = ( 0.45 - 0.15 ) / 1.2

t = 0.3 / 1.2

t = 0.25 s

Therefore, the required time interval is 0.25 s

b)

the other possible question is;  Determine the distance the squid has traveled while accelerating.

using the following expression

v² = u² + 2as

we solve for distance s

2as = v²- u²

s = v²- u² / 2a

we substitute

s = ((0.45)²- 0.15)²) / 2×1.2

s = (0.2025 - 0.0225) / 2.4

s = 0.18 / 2.4

s = 0.075 m

Therefore,  the distance traveled while accelerating is 0.075 m

Answer:

This is Introvert ting

Explanation:

u know when people are shy

i think they r known as Introvert

i hope this helps

Answer:

V₁ = 1.75 m³

Explanation:

Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:

[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

where,

P₁ = Initial Pressure of the Gas = 4 KPa

V₁ = Initial Volume of the Gas = ?

P₂ = Final Pressure of the Gas = 2 KPa

V₂ = Final Volume of the Gas = 3.5 m³

Therefore,

[tex](4\ KPa)V_{1} = (2\ KPa)(3.5\ m^{3})\\\\V_{1}=\frac{2\ KPa}{4\ KPa}(3.5\ m^{3})\\\\[/tex]

V₁ = 1.75 m³

Answer:

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

Explanation:

An atmosphere equals 14.695 pounds per square inch. We find the equivalent of given pressure in atmospheres by means of simple rule of three:

[tex]x = 25.5\,\frac{lb}{in^{2}} \times \frac{1\,atm}{14.695\,\frac{lb}{in^{2}} }[/tex]

[tex]x = 1.735\,atm[/tex]

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

Explanation:

third and fifth option along with the first one are right

The volume of snow on the lawn, in cubic m, is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

It is defined as a three-dimensional space enclosed by an object or thing.

It is given that:

The lawn outside your neighbor's house has an approximate area of 175 m². One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm.

As we know, unit conversion can be defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.

25.5 cm =  0.255 m

Volume = area×depth

Volume = 175×0.255

Volume = 44.63 cubic meters

Thus, the volume of snow on the lawn, in cubic m is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

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Answer:

The effects that certain frequencies of EMR have when absorbed by matter is explained below in complete detail.

Explanation:

Electromagnetic radiation of distinct frequencies associates with material adversely. ... Gamma rays, though commonly of somewhat greater frequency than X rays have the equivalent creation. When the power of gamma rays is consumed in material, its influence is practically indistinguishable from the outcome generated by X rays.