Which of the following is an example of a physical change? burning paper digesting food o ripping paper​

Answer:

h = 493 m

Explanation:

Given that,

A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 2.9 s later.

The total distance covered by the rock is 2d. We know that,

Speed = distance/time

\(v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{340\times 2.9}{2}\\\\d=493\ m\)

So, the height of the cliff is equal to 493 m.

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

10. more

I HOPE THESE ARE CORRECT AND IT HELPS

The center of mass of the rod is located at x = 0.5 on the -axis.

To find the mass of the rod, we need to integrate the density function over the length of the rod. We are given that the density of the rod is 3 sin(x) mass per unit length, and the length of the rod is from 0 to pi on the -axis. Therefore, the mass of the rod is:

M = ∫0π (3 sin(x)) dx

Using the integration formula for sin(x), we get:

M = [-3 cos(x)]0π
M = 3(cos(0) - cos(pi))
M = 6

So, the mass of the rod is 6 units.

Next, to find the center of mass of the rod, we need to find the position of the center of mass along the -axis. The position of the center of mass is given by the formula:

x_c = (1/M) ∫0π (x dm)

where x is the position of an infinitesimal element of the rod, and dm is the mass of that element. We can express dm as the product of the density function and the length element dx:

dm = ρ(x) dx = 3 sin(x) dx

Substituting dm into the formula for x_c, we get:

x_c = (1/M) ∫0π (x ρ(x) dx)

x_c = (1/6) ∫0π (x 3 sin(x) dx)

Using integration by parts with u = x and dv = 3 sin(x) dx, we get:

x_c = (1/6) [-x 3 cos(x) + 3 sin(x)]0π

x_c = (1/6) (0 - 0 + 3)

x_c = 0.5

Therefore, the center of mass of the rod is located at x = 0.5 on the -axis.

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We are given the following information

Mass of car = 950 kg

Initial speed of car = 16.0 m/s

Final speed of car = 9.50 m/s

Time = 1.20 s

The average force exerted on the car during braking can be found using Newton's 2nd law of motion

Where m is the mass of the car and a is the acceleration of the car.

The acceleration of the car is given by

The negative sign indicates deacceleration since the car is stopping.

So, the force is

Therefore, an average force of 5145.865 N was exerted on your car during braking.

The distance traveled by the car while braking can be found as

Let us substitute the given values

Therefore, the car traveled a distance of 15.3 m while braking.

The type of figurative language present in the example "in leaves no step had trodden black" is personification.

Personification is a figure of speech in which human qualities or actions are attributed to non-human entities. In the given example, the phrase "leaves no step had trodden black" personifies the leaves by suggesting that they have the ability to tread or walk.

By attributing the human action of stepping to the leaves, the poet gives them a sense of agency and animates them. This personification enhances the imagery of the poem, creating a vivid picture of untouched and unexplored surroundings, where the leaves have remained undisturbed.

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The figurative language present in the phrase 'In leaves no step had trodden black' is imagery. This phrase creates a visual image for the reader, contrasting with symbol, metaphor, and personification.

The phrase 'In leaves no step had trodden black' from your question is an example of imagery. Imagery refers to language that appeals to one or more of the five senses. Here, this phrase creates a visual image of untouched leaves, helping the reader to more vividly imagine the scene being described. It does not suit the definitions of symbol, metaphor, or personification. A symbol represents itself and something else simultaneously, a metaphor makes a direct comparison between two unlike things, and personification assigns human traits to non-human entities or concepts. In this case, the leaves are not representing something else (symbol), not being compared directly with something else (metaphor), and not given human traits (personification).

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(a) The work function  is 1.98 x 10⁻¹⁹ J.

(b) The threshold frequency is 3 x 10¹⁴ Hz.

(c) The maximum kinetic energy of the most energetic electron is 3.32 x 10⁻¹⁹ J.

(d) Photo-electron would be ejected.

(e) The only constant parameter would be speed of the photon.

(f) The slope of the graph is 6.67 x 10⁻³⁴ J.s

(a) The work function of the experimental photo-missive material is calculated as follows;

Ф = hf₀

where;

Ф = hf₀

Ф = 6.626 x 10⁻³⁴ x 3 x 10¹⁴

Ф = 1.98 x 10⁻¹⁹ J

(b) The threshold frequency of the experimental photo-missive material is the frequency at which the kinetic energy is zero = 3 x 10¹⁴ Hz.

(c) The maximum kinetic energy of the most energetic electron is calculated as;

K.E = E - Φ

K.E = ( 6.626 x 10⁻³⁴ x 8 x 10¹⁴) -  1.98 x 10⁻¹⁹ J

K.E =  3.32 x 10⁻¹⁹ J

(d) The frequency of the photon with a wavelength of 500 nm is calculated as;

f = c/λ

where;

f = ( 3 x 10⁸ ) / ( 500 x 10⁻⁹ )

f = 6 x 10¹⁴

Since the frequency of the incoming photon is greater than  the threshold frequency, photo-electron would be ejected.

(e) If you use a different experimental photo-missive material the only parameter that would be the same on the graph is speed of photon.

(f) The slope of the graph is calculated as;

m = (2.5 eV - 0 eV) / [(9 - 3) x 10¹⁴]

m = (2.5 ev) / (6 x 10¹⁴)

m = (2.5 x 1.6 x 10⁻¹⁹ ) / (6 x 10¹⁴ )

m = 6.67 x 10⁻³⁴ J.s

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The maximum number of comparisons that may be needed to find an element in a 1000-element balanced binary search tree is 10.

To find an element in a balanced binary search tree, the maximum number of comparisons required is equal to the height of the tree.

In a balanced binary search tree with 1000 elements, the height can be calculated using the formula:

height = log2(n),

where n is the number of elements in the tree.

Therefore, the height of a balanced binary search tree with 1000 elements is approximately log2(1000) = 9.97.

Since the height of a tree must be an integer, we can round up the height to the nearest whole number, which in this case is 10.

Hence, the maximum number of comparisons that may be needed to find an element in a 1000-element balanced binary search tree is 10.

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Answer:

D is the answer wave behavior

Interference, refraction, diffraction, and dispersion are all aspects of wave behavior. (D).  That is, particles don't do these things.

Remote-control toys make use of pulse code modulation.

Pulse-width modulation (PWM), pulse-position modulation (PPM), and more recently spread-spectrum technology are used in typical radio control systems for radio-controlled models, and servomechanisms are used to activate the various control surfaces. Digital modulation is used in all current infrared remote control designs. Amplitude Shift Keying (ASK) and Frequency Shift Keying (FSK) are two fundamental types of digital modulation. (FSK).

Electronic communication is where amplitude modulation is most commonly utilised. This method is being utilised in numerous communication channels, including computer modems, citizens band radio, VHF aviation radio, and portable two-way radios.

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A 62.0 kg box hangs from a rope. what is the tension in the rope. When the box is at rest, the tension in the rope is equal to the weight of the box When the box moves up at a steady speed of 4.90 m/s, the net force acting on the box is zero.

When the box is at rest, the tension in the rope is equal to the weight of the box. The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity (g = 9.8 m/s^2). Therefore, the tension in the rope is given by:

Tension = Weight = mass * g = 62.0 kg * 9.8 m/s^2 = 607.6 N

Part B: When the box moves up at a steady speed of 4.90 m/s, the net force acting on the box is zero. The tension in the rope must balance the weight of the box. Therefore, the tension in the rope is equal to the weight of the box:

Tension = Weight = 62.0 kg * 9.8 m/s^2 = 607.6 N

Part C: When the box has a vertical velocity (Vy) of 4.50 m/s and is speeding up at 5.50 m/s^2, the net force acting on the box is the sum of the tension in the rope and the force due to the acceleration. The net force can be calculated using Newton’s second law (F = m * a):

Net Force = m * a = 62.0 kg * 5.50 m/s^2 = 341 N

To find the tension in the rope, we need to add the weight of the box to the net force:

Tension = Weight + Net Force = 62.0 kg * 9.8 m/s^2 + 341 N = 972.8 N

Part D: When the box has a vertical velocity (Vy) of 4.50 m/s and is slowing down at 5.50 m/s^2, the net force acting on the box is the difference between the tension in the rope and the force due to deceleration. The net force can be calculated using Newton’s second law:

Net Force = m * a = 62.0 kg * (-5.50 m/s^2) = -341 N (negative because it opposes the motion)

To find tOme tension in the rope, we subtract the net force from the weight of the box:

Kj Tension = Weight – Net Force = 62.0 kg * 9.8 m/s^2 – 341 N = 607.6 N

Therefore, the tension in the rope is 607.6 N in all cases, except for when the box is slowing down, where it is 972.8 N.

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Answer:

57 m

Explanation:

Given:

v₀ = 25 m/s

v = 13 m/s

t = 3 s

FInd: Δx

Δx = ½ (v + v₀) t

Δx = ½ (13 m/s + 25 m/s) (3 s)

Δx = 57 m

You were approximately 5.72 meters away from the second baseman. The vertical drop or distance between point A and point B was approximately 0.4576 meters.

To determine the distance between you (the shortstop) and the second baseman, we can use the formula for horizontal distance (d) traveled by an object moving at a constant horizontal velocity:

d = v * t

where:

- d is the horizontal distance traveled,

- v is the horizontal velocity of the ball,

- t is the time taken.

Given that the horizontal velocity (v) is 13 m/s and the time (t) is 0.44 s, we can calculate the horizontal distance (d) as follows:

d = 13 m/s * 0.44 s = 5.72 meters

So, you were approximately 5.72 meters away from the second baseman.

To find the vertical drop or the distance between point A and point B, we need to calculate the vertical component of the ball's motion. Since the ball is thrown horizontally, it will experience a constant vertical acceleration due to gravity.

The formula to calculate the distance (d) traveled vertically in free fall is:

d = 1/2 * g * t²

where:

- d is the vertical distance traveled,

- g is the acceleration due to gravity (approximately 9.8 m/s²),

- t is the time taken.

Given that the time (t) is 0.44 s, we can calculate the vertical distance (d) as follows:

d = 1/2 * 9.8 m/s² * (0.44 s)² = 0.4576 meters

So, the vertical drop or the distance between point A and point B is approximately 0.4576 meters.

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A. The maximum theoretical efficiency is 67.2 %.

B. The hot reservoir is 56.7 kW.

Solution:

A. Max. theoretical (Carnot) efficiency = 1-T(cold)/T(hot)

or 1 - (430+273)/(1870+273) =   0.672  = 67.2 %.

B. mechanical power delivered = 0.42*1.35*10^5 = 56700 J/s = 56700  W = 56.7 kW.

A Carnot engine operating between two given temperatures exhibits the highest possible efficiency among heat engines operating between these two temperatures. The Carnot engine has the highest efficiency among heat engines operating between two temperatures. Diesel is more flammable than gasoline.

Second, petrol doesn't burn completely, so very little fuel is wasted. Because of this, diesel engines are more efficient than gasoline engines. The Carnot cycle achieves maximum efficiency because all heat is transferred to the working fluid at the highest temperature.

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If one travels away from the center of a mid-ocean rift (ridge) zone, the rock on either side of the rift zone gets progressively younger.

Hence, the correct option is A.

When one travels away from the center of a mid-ocean rift zone (ridge), the rock on either side of the rift zone gets progressively younger.

This is because at the mid-ocean ridge, new oceanic crust is formed through volcanic activity. As tectonic plates move away from the ridge, new magma rises to fill the gap, solidifies, and forms new rock.

Therefore, the rocks farther away from the ridge are younger compared to the rocks closer to the ridge.

Hence, the correct option is A.

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Answer:

E. An object’s velocity changes at a constant rate, and its acceleration remains constant.

Explanation:

When an object is in freefall, it implies that the object is falling freely under gravity. If it falls towards the earth surface, the fall is in the direction of the Earth's gravitational force.

At the point of release of the object, its initial velocity is zero because it is at rest. But when released, its velocity increases at a constant rate until it is acted upon by an external force. But its acceleration remains constant, acceleration due to gravity.

the fractional intensity i/imax at the point on the screen 4.10 mm from the center of the principal maximum is approximately 0.123.

The given parameters are:Width of the slit, d = 0.340 mm

Wavelength of the light, λ = 546.1 nm

Distance from the slit to the screen, L = 90 cm

Distance of the point on the screen from the center of the principal maximum, y = 4.10 mm

The distance between the center of the principal maximum and the first minima is given by:

ym = (m * λ * L) / d

Where m is the order of the minima

From the above equation, we can calculate the order of the minima closest to the given point on the screen as:

m = (y * d) / (λ * L) = (4.10 × 10^(-3) × 0.340 × 10^(-3)) / (546.1 × 10^(-9) × 90 × 10^(-2)) ≈ 1

The intensity at a point on the screen at distance y from the center of the principal maximum is given by the equation:

i / imax = [sin(πa sinθ / λ) / (πa sinθ / λ)]^2

where a is the width of the slit and θ is the angle between the line joining the point on the screen and the center of the principal maximum, and a line perpendicular to the slit at the point where the diffracted beam passes through the slit.θ can be approximated as:

θ ≈ (m * λ) / d = (1 × 546.1 × 10^(-9)) / 0.340 × 10^(-3) ≈ 1 × 10^(-3) radians≈ (180 / π) × (1 × 10^(-3)) degrees = 0.057296 degrees

Putting the values of θ and a in the equation for intensity, we get:

i / imax = [sin(πa sinθ / λ) / (πa sinθ / λ)]^2≈ [sin(π × 0.340 × 10^(-3) × (1 × 10^(-3)) / (546.1 × 10^(-9))) / (π × 0.340 × 10^(-3) × (1 × 10^(-3)) / (546.1 × 10^(-9)))]^2≈ 0.123

Thus, the fractional intensity i/imax at the point on the screen 4.10 mm from the center of the principal maximum is approximately 0.123.

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Answer:

a) F = 5.14 10⁻⁸ Hz,  f = 4.76 10-8 Hz,  b)   v = 2.29 m / s,   f = 42.5 Hz

Explanation:

a)This problem has two parts.

For the calculations relative to the planet Venus, we use that the period and the frequency are related

            f = 1 / T

frequency of the orbit around the Sun

Let's reduce the period to the SI system

           T = 225 days (24h / 1days) (3600 s / 1h) = 1.94 10⁷ s

           F = 1 / 1.94 10⁷

           F = 5.14 10⁻⁸ Hz

rotation frequency

            T = 243 d = 2.1 107 s

             f = 1 / T

             f = 1 / 2.1 107

            f = 4.76 10-8 Hz

b) give the data of some marine waves

the speed of the wave can be found with kinematics

            v = x / t

            v = 50.0 / 21.8

            v = 2.29 m / s

If the wavelength is L = 9.28m

this distance is the distance between two consecutive ridges or valleys

             λ / 2 = L

             λ = 2L

             λ = 2 9.28

             λ = 18.56 m

the speed of the wave is

             v = λ f

             f = v /λ

             f = 2.29 / 18.56

             f = 42.5 Hz

The final velocity of the model rocket when the engine cuts out is \(32.3 m/s.\)

We may solve for the model rocket's final velocity using the formula: work done (W) = change in kinetic energy (KE). We must first identify the work the model rocket engine has done. It is stated that 1500 J of work are applied.
\(W = 1500 J\)
Next, we need to find the change in kinetic energy of the model rocket. We know that it starts from rest and travels 35 meters with a mass of 0.80 kg.
\(ΔKE = 1/2 mv^2 - 0\)
\(ΔKE = 1/2 (0.80 kg) v^2\)
\(ΔKE = 0.40v^2\)
Now we can substitute these values into the equation:
\(1500 J = 0.40v^2\)
Solving for v, we get:
\(v = \sqrt{1500 J / 0.40(0.80 kg}\)
\(v = 32.3 m/s\)
Therefore, the model rocket is going 32.3 m/s when the engine cuts out.

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The magnification is doubled when 10x oculars are used instead of 5x in our microscope with the same objectives.

When multiple lenses are lined together, the overall magnification can be calculated by multiplying the individual magnifications of each lens.

M = M1 × M2 × M3 × ... × Mn

where M is the overall magnification and M1, M2, M3, ..., Mn are the magnifications of the individual lenses.

Let M be the magnification of the objective, then the overall magnification,

when 5x ocular is used,

M1 = M × 5

M1 = 5M

when 10x ocular is used

M2 =  M × 10

M2 = 10M

Therefore, the magnification is doubled when 10x ocular is used instead of 5x in our microscope with the same objectives.

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Answer:

The mechanical advantage of a machine is always less than its velocity ratio.It is because mechanical advantage decreases due to the friction and weight of moving parts of the machine, but the velocity ratio remains constant.

Answer:

The velocity at which it hit the ground is  72.6 meters.

Explanation:

So,

h = 269 meters

u = 0 m/s ( 0 for a dropping object )

g = 9.8 m/s²

By Substitute the kinematic equation

  v² = u² + 2gh

  v² = 0 x 0 + 2 x 9.8 x 269

  v² = 0 + 5272.4 => 5272.4

  v =  √5272.4

  v = 72.6

Therefore the object is dropped from a height of 269 meters, it will hit the ground with a velocity of 72.6 meters per second.

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Answer:

No, The Moon, on the other hand, rotates once around its every 28 days, and once around the Earth in that same 28 days. The result of this combination is that the same side of the Moon is always facing the Earth.

Answer:

Give person above me brainliest

Explanation:

The average speed of April, given the data from the question is 13.33 m/s

Speed is the distance travelled per unit. Mathematically, it can be expressed as:

Speed = distance / time

Average speed is simply defined as the total distance travelled divided by the total time taken to cover the distance .

Average speed = Total distance travelled / total time

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

Average speed = Total distance travelled / total time

Average speed = 1600 / 120

Average speed = 13.33 m/s

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Complete question

As shown in the figure below, April enters a race. She runs leftward 100 m to her horse, then she rides 500m to her truck, then she drives 1000m in a total time of 120s.What is her average velocity over the 120s period? What is her average speed over the 120s period?

Answer:

-13 | Velocity

13 | Speed

To explain the tangent plane to the surface, `z = 53 − x² − 2y²` at the point `(3, 2, 6)`, let us first determine the partial derivatives of `z` with respect to `x` and `y`.

Partial derivative of `z` with respect to `x`, `∂z/∂x = -2x`Partial derivative of `z` with respect to `y`, `∂z/∂y = -4y`Now, let's find the gradient vector `grad z` at `(3, 2, 6)` and the value of `z` at `(3, 2)`.gradient vector `grad z = (-2x, -4y, 1)`gradient vector `grad z = (-6, -8, 1)` at `(3, 2, 6)`.Value of `z` at `(3, 2)` is given by `z = 53 - 3² - 2(2)² = 39`.

Therefore, the equation of the tangent plane to the surface `

z = 53 − x² − 2y²` at `(3, 2, 6)` is:

`z - 6 = -6(x - 3) - 8(y - 2)`

which can be written as:`6x + 8y + z = 50`Thus, the equation of the tangent plane to the surface `z = 53 − x² − 2y²` at the point `(3, 2, 6)` is `6x + 8y + z = 50`.

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The work done by force f on the proton is \(- 6.912 x 10^-34 J\).

The work done by force (f with arrow) when a proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.3 m) is given by:Work = Change in potential energy of the protonThis is because the force is conservative (i.e., it doesn't depend on the path taken by the proton).We know that the proton moves only along the vertical direction, therefore, its displacement is:∆x = 0and

∆y = 0.3 mThe potential energy of a proton is given by:

Potential energy = - q * V where

q = charge of proton (\(1.6 x 10^-19\) C)

V = potential difference between two points at which proton movesThe potential difference between the two points is given as follows:V = VB - VAwhere

VB = potential at point (0.10 m, 0.3 m)VA

= potential at point (0.10 m, 0)We can take

VA = 0 since we are considering it as reference level of potential energy.

The potential at point (0.10 m, 0.3 m) can be calculated as follows:We know that the electric field due to a point charge is given by:E = \(k * q / r^2\) where

k = Coulomb's constant (\(9 x 10^9 N.m^2.C^-2\))

q = charge of point charger

r = distance between point charge and point of interestThus, the electric field at point (0.10 m, 0.3 m) due to the charge is: E = \(k * q / r^2\) where

q = proton charge (\(1.6 x 10^-19\) C)

r = distance between proton and charge

= 0.1 mThus,

E = (\(9 x 10^9\)) * \((1.6 x 10^-19) / (0.1^2)\)

= \(1.44 x 10^-14\) N/CThe potential at point (0.10 m, 0.3 m) due to the charge is:

V = Edwhere

Ed = distance moved by proton

= ∆y

= 0.3 mThus,
\(V = 1.44 x 10^-14 * 0.3\)

= \(4.32 x 10^-15\) VTherefore, the potential difference between the two points is:

V = VB - VA

= \(4.32 x 10^-15 - 0\)

= 4.32 x 10^-15 JThus, the work done by force f on the proton is:

Work = Change in potential energy of the proton

= - q * (VB - VA)

\(= - (1.6 x 10^-19) * (4.32 x 10^-15)\)

\(= - 6.912 x 10^-34 J\).

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In the event of an accident or a car crash, road safety barriers and fences prevent automobiles from running off the road.

Newton's Three Laws of Physics can help explain what these safety barriers are and how they help to save lives during car accidents :

I.  Unless acted upon by an imbalanced force, an object at rest will remain at rest, and an object at constant velocity will remain at constant velocity.

II.  If an imbalanced force occurs, a mass will experience acceleration proportional to its magnitude.

III.  When you apply a force to an object, you will feel a force that is equal    in magnitude but opposite in direction.

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First, for this problem, we will need to know the formula for time dilation

Where t is the time in the reference frame on earth

t0 is the time on the ship

v is the velocity relative to lightspeed

and c is light speed in a vacuum

For this problem, we will need to convert .93x10^8 to a fraction of light speed.

.93x10^8 = .31c

Now, we can plug in the numbers we get from the given

t0 = 6 hours since the ship has been in the air for 6 hours

t = 6.31089

For the answer to be valid, we will convert .31089 to minutes by multiplying it by 60 (60 minutes in a hour)

18.654 minutes

Explanation:

The kinetic energy of an object is given in terms of its mass and object as :

\(K=\dfrac{1}{2}mv^2\)

An object with more mass has more kinetic energy. An object with less  mass, if both objects are moving at the same speed, in opposite directions, will have less kinetic energy. The square of v will give positive number.