if you wish to reconstruct a complex wave from its fourier spectrum, what additional information must you obtain, other than the amplitudes of the harmonics?

Answers

Answer 1

In order to reconstruct a complex wave from its fourier spectrum, we must obtain the phase of each harmonic other than amplitudes of the harmonics.

The phase of each harmonic is needed because it tells you exactly where in time the harmonic will occur.

For example, if a wave has a frequency of 100 Hz, and you know that the phase of each harmonic is regularly spaced by 1 radian (180 degrees), then you can use this information to determine when each harmonic will occur. You can do this by dividing 360 degrees by the frequency and multiplying by 2 pi.

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Related Questions

you and a friend are on a swing set and her swing is slightly longer than yours. if you both start swinging at the same time, from the same height, where will she be after you have completed one complete swing back and forth?

Answers

She might be barely lower than you however shifting upward in the direction of you.

Oscillation is the repetitive or periodic variation, generally in time, of a few measures approximately a critical price or among or greater distinct states. familiar examples of oscillation consist of a swinging pendulum and alternating modern-day.

A pendulum is a weight suspended from a pivot so that it can swing freely. whilst a pendulum is displaced sideways from its resting, equilibrium position, it's miles issue to a restoring force due to gravity so one can accelerate it returned toward the equilibrium function.

Oscillation is defined as the method of repeating versions of any quantity or degree about its equilibrium price in time. Oscillation can also be defined as a periodic variant of a rely on among values or approximately its crucial price.

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in a double-slit experiment, the slit separation d is 2.00 times the slit width w. how many bright interference fringes are in the central diffraction envelope

Answers

The number of bright interference fringes in the central diffraction envelope is 3.

To determine the number of bright interference, we need to understand the equation of first minima in the diffraction pattern, and the equation of angular locations of the double slit interference pattern.

For the equation of first minima in the diffraction pattern is:

[tex]W[/tex]·[tex]Sin[/tex]θ = [tex]m_{1}[/tex]·λ

For the equation of angular locations of the double slit interference pattern is:

[tex]d[/tex]·[tex]Sin[/tex]θ = [tex]m_{2}[/tex]·λ..... (1)

Here, W is single slit width while d is slit separation

Next, we need to determine the number of bright interference fringes in the central diffraction envelope.

For the first minima, [tex]m_{1} = 1[/tex], then rewrite the equation (1) as follows.

=[tex]a[/tex]·[tex]Sin[/tex]θ = [tex]m_{1}[/tex]·λ

=[tex]a[/tex]·[tex]Sin[/tex]θ = [tex]1[/tex]·λ

=[tex]a[/tex]·[tex]Sin[/tex]θ = λ..... (2)

Then, from the equations (1) and (2)

[tex]=m_{2}=\frac{d}{w} \\=m_{2}=\frac{2w}{w}\\=m_{2}=2[/tex]

Therefore, there are 3 bright fingers, 1 at the centre and 2 in each side.

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what is the heaviest the book he can hold vertically before it slips out of his fingers? the coefficient of static friction between his fingers and the book cover is 0.88

Answers

Mass of the heaviest book is 0.733 kg.

Heaviest book has weight

= (6 + 6) * 0.6 N

= 7.2 N

If m = mass of the heaviest book in kg

m = 7.2/9.81 kg

   = 0.733 kg.

Children often bring a well-established "life-view" of friction with them because of their experiences with slippery surfaces like frozen ponds (low friction) and "gripping" surfaces like deep pile carpets (high friction) and the effects they have on movement. This way of existence needs to be expanded and understood in the perspective of science.

The force that modifies movement as a result of surface/surface interaction is known scientifically as friction (all changes of movement require the action of a force). When using diagrams to portray forces in action, the direction of the frictional force should be depicted as being opposed to that of the movement.

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what is the average flow rate in cm3/s of gasoline to the engine of a car traveling at 104 km/h if it averages 8.0 km/l? (you do not need to enter any units.)

Answers

The average flow rate is 3.611 cm^3/s

The speed V is = 104Km/h

The velocity of vehicle z is = 8km/l

Qavg = V/Z

         = 104/8 =13

         = 13 (1000)/1(3600)

Qavg = 3.611 cm^3/s

The physical parameters flow rate and velocity are linked yet very distinct. Consider a river's flow rate to help you understand the difference. The flow rate of the river increases as water velocity increases. However, the size of the river also affects the flow rate. The Amazon River in Brazil, for instance, carries much more water than a swift alpine stream. When A is the cross-sectional area and v is the average velocity, the flow rate Q and velocity v are precisely related. This equation seems to make sense. According to the relationship, the size of a river, pipe, or other body of water as well as the magnitude of the average velocity (hereinafter referred to as the speed) directly affect the flow rate.

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A car moves at a velocity of 8.5m/s. It then accelerates at a constant rage of 2.5m/s/s for a total time of 5 second. How fast is the car moving at the end of the 5 second

Answers

The car would be moving with a velocity of 21 m/s at the end of the five seconds.

We know that the first equation of motion is ⇒ v = u + at (i)

Here, the car moves with an initial velocity, u =  8.5 m/s (ii)

The car accelerates at a constant rate, a = 2.5 m/s² (iii)

The total time during this process, t = 5 seconds (iv)

Putting the values of the initial velocity, acceleration, and time (ii, iii, and iv) in equation (i), we getv = 8.5 + (2.5)(5)

v = 21 m/s

Therefore, the car moves with a final velocity of 21 m/s at the end of the five seconds.

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A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with the consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?

help someone help me​

Answers

Hello..!

Subject: Thermodynamics

The first law of thermodynamics relates work and transferred heat exchanged in a system through a new thermodynamic variable, internal energy. This energy is neither created nor destroyed, only transformed.

We can think of gas as a thermodynamic system, all because gases can work and absorb heat, and then they can turn all that into energy.

The formula for the change in energy is given by the first law of thermodynamics expressed as:

[tex] \: \: \: \: \: \: \: \: \: \: {\boxed{\boxed{ \sf\large \rm \Delta U = Q - W }}}[/tex]

Being:

ΔU = change in energyQ = added heatW = Work done

Problem:

A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?

Data:

ΔU = ¿? (Meet)Q = 1000 JW = 600 J

Now adding the data in the formula to find the energy change:

[tex] \sf\large \rm \Delta U = 1000J - 600J[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{\boxed{\boxed{\large \rm \Delta U = 400 J}}}[/tex]

[tex]\begin{gathered}\rule{7cm}{0.01mm}\\\texttt{Good studies! :D}\\\rule{7cm}{0.01mm}\end{gathered}[/tex]

Which letter represents the normal force acting on the box?
A
B
C
D
What is the answer

Answers

Answer: (C)

Explanation:  (B) is the weight of the box acting downwards

(A) is the frictional force acting on the box because of moving forward against the plane

(D) is the force that moves the box forward

( C) is perpendicular to the box which means it is at 90 degrees with the box. Normal usually means at 90 degrees

Please someone help!! I don’t know how to figure this out!

Answers

Answer:

relax

Explanation:

your answer is right there in front of you force = weight

area of feet = area

while your pressure = force per unit area i.e

force/area = 600/0.03

= 2 × 10⁵

The change in velocity v of an object is zero over a short time interval t. Which of the following is true? Assume quantities are instantaneous unless stated otherwise.

Answers

Answer:

2. The object must have zero average acceleration over the interval.

Explanation:

Notes:

-Velocity can have zero change if it is either at rest or moving at a constant velocity. This also means there is zero acceleration (acceleration is any change in velocity).

-Velocity is also speed and direction, so if it changes direction (for example: moving backward or in a circle) it is not constant even if it has a constant speed. It also means the object is accelerating.

1: Nothing can be determined without additional information.

Incorrect, because it is solvable by the process of elimination.

2: The object must have zero average acceleration over the interval.

Correct, because zero acceleration equals zero change in velocity.

3: The object must be changing position.

Incorrect, because you don't know whether or not the object is accelerating.

4:  The object must have zero average velocity over the interval.

Incorrect, because this implies it is at rest. To have zero change, it can be at rest OR moving at a constant velocity.

5:  The object must have constant velocity over the interval.

Incorrect, because zero change in velocity can mean either be at rest OR moving at a constant velocity.

6: The object must begin and end at the same position.

Incorrect, because to begin and end at the same position implies that it was at rest or it changed direction while moving to end at the same position. (see Notes above for explanation)

7: The object must have constant acceleration over the interval.

Incorrect, because you can't have zero change in velocity if there is acceleration (Acceleration is any change in velocity).

8: The object must be at rest.

Incorrect, because to have zero change, it can be at rest OR moving at a constant velocity.

when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about:

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When an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

Luminosity and apparent brightness are about brightness, but from a different point of view. The difference between luminosity and apparent brightness is that luminosity tells us exactly how bright a star really is while apparent brightness only tells us its brightness seen from Earth.

L / Lsun = ( d / dsun )² = b / bsun

L = Luminosity of a star

Lsun = Luminosity of sun

d = Distance of star

dsun = Distance of sun

b = Brightness of star

bsun = Brightness of sun

Therefore, when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

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Light passes straight through two vertical polarizers. The axes of the filters are aligned parallel to each other. How does the light that passes through the second filter compare to the light that passed through the first filter?.

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The direction of the polarized light after passing the first and the second filter are the same, however the intensity of the light after passing the first and second polarizer are half and fourth of its original intensity, respectively.

Light polarization is a filter of electromagnetic waves such that it propagates into one transmission axis only.

When light passes through a polarizer, its intensity will decrease by half.

Suppose the intensity of unpolarized light is  I₀,  after passing the first polarizer, its intensity becomes:

I₁ = 1/2 . I₀

After passing the second polarizer, the intensity will further decrease by half, or:

I₂ = 1/2 . I₁

I₂ = 1/2 .1/2 . I₀ = 1/4. I₀

Since both polarizers are vertical filters, then the direction of the polarized light after passing the first and the second filter are the same.

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A person standing on a hill throws a ball horizontally with a velocity of 12m / s The ball lands 10.1 m away from the hill. How tall is the hill? Assume no air resistance

A.3.5
B.6.9
C.0.84
D.0.42

Answers

The height of a hill from which the person throws a ball horizontally with a velocity of 12 m/s and the ball lands 10.1 m away from the hill is 3.47 m.

What is the height of a hill from which the person throws a ball horizontally with a velocity of 12 m/s and the ball lands 10.1 m away from the hill?

The height of a hill from which the person throws a ball horizontally with a velocity of 12 m/s and the ball lands 10.1 m away from the hill is calculated as follows;

The ball was projected horizontally from the top of a hill, therefore, there is no vertical component of the velocity of projection and the object is considered to be undergoing free fall under gravity from a height, h.

The height from which an object falls freely under gravity is calculated with the formula below:

H = gt²/2

where g is the acceleration due to gravity = 9.81 m/s²

t is time.

the time taken for the fall = horizontal distance / velocity

t = 10.1 / 12

t = 0.841 s

Therefore;

h = 9.81 * (0.841)² / 2

h = 3.47 m

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the vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. the potential difference between the plates is 25.0 v. typical dimensions are about 3.9 cm on a side, with a separation of about 5.4 mm. the plates are close enough that we can ignore fringing at the ends. part a under these conditions, how much charge is on each plate?

Answers

If we keep oscilloscope plates close with potential difference 25 V , the  Charge on each plate will be 4.68 *10^-11 C.

How to calculate charge ?Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge.The number of charges (electrons) that go from a higher potential to a lower potential is referred to as quantity of charge. It refers to the total amount of electricity flowing via a conductor.

The capacitance of a parallel plate capacitor is given by:

C = ∈₀A / d

where

∈₀ = 8.85 *10^-12 F/m is the vacuum permittivity

A is the area of each plate

d is their separation

For the capacitor in the problem:

A= 0.033m² = 0.0011m²

d = 5.2 mm = 0.0052 m

Substituting,

C = (8.85 *10^-12) (0.0011m)/ 0.0052 = 1.87 *10^-12 F

The capacity is related to the charge on the plate by:

Q=CV

where

V = 25.0 V is the potential difference between the plates

C = 1.87 *10^-12 F

Substituting,

Q = (25.0) (1.87 *10^-12) = 4.68 *10^-11 C

Charge on each plate = 4.68 *10^-11 C

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A 5 kg block is pulled across the ground to the right by a tension force of 40 N with a frictional force of 8 N opposing the motion. What is the acceleration of the object?

Answers

The Acceleration of the object = 6.4 m/s²

Mass of block (m) = 5 kg

Action force on block, (F₁) = 40 N

Frictional force opposing the motion (F₂) = 8 N

Acceleration of the object (a) = ?

⇒ Net force = Action force on block - Opposing friction force

⇒ F = F₁ - F₂

⇒ F = 40 - 8

⇒ F = 32 N

Net force of the block (F) = 32 N

Mass of block (m) = 5 kg

F is the Force in N.

m is the Mass in kg.

a is the Acceleration in m/s².

F = ma

⇛ a = F/m

⇛ a = 32/5

⇛ a = 6.4 m/s²

You must use Fnet=ma

Ft-Ff=ma

40-8=5*a

a=6.4m/s^2

the top of a ladder slides down a vertical wall at a rate of 0.15 mys. at the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 mys. how long is the ladder?

Answers

The length of the ladder which is placed 3 meters from the wall is           5 meters

The rate of the change of ladder decreasing in vertical wall = - 0.15 m/s

The distance between the bottom of the ladder and the wall = 3 meter

The rate of change of the bottom of the ladder away from the wall = 0.2 m/s.

The length of the ladder can be found using the Pythagoras theorem,

               x² + y² = L²

where x is the distance between the bottom of the ladder and the wall

           y is the distance from the top of the ladder to the bottom of the wall.

           L is the length of the ladder

Let us differentiate in terms of the rate of change in the above equation,

              2x dx/dt + 2y dy/dt = 0

Now let us substitute the known values,

                 2(3)(0.2) + 2y(-0.15) = 0

                           1.2  - 0.3y = 0

                                 0.3y = 1.2

                                      y = 1.2 / 0.3

                                        = 4

Then, the length of the ladder is

               3² + 4² = L²

                 9 + 16 = L²

                        L = √25

                        L = 5

Therefore, the length of the ladder is 5 meters

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describe the rotation curve of the milky way galaxy and contrast its shape with that of the solar sysstem

Answers

Compared to the orbital speeds of the planets in our solar system, the Milky Way Galaxy's rotation curve is far more flat and steady. While there is less mass as we get farther from the sun in our solar system.

The orbital speed of an astronomical body or object (such as a planet, moon, artificial satellite, spacecraft, or star) in gravitational bound systems is its speed relative to the most massive body's centre of mass, or the barycenter, if one body is significantly more massive than the other bodies in the system put together.

The phrase can be used to describe either the mean orbital speed (i.e., the speed throughout the course of an orbit) or its instantaneous speed at a specific location in the orbit. For objects in closed orbits, the maximum (instantaneous) speed occurs at periapsis (perigee, perihelion, etc.), whereas the smallest speed occurs at apoapsis (apogee, aphelion, etc.). When two-body systems are perfect, things An astronomical body or object (such as a planet, moon, man-made satellite, spacecraft, or star) moves through space at a certain speed called its orbital speed in gravitational bound systems.

When a system closely resembles a two-body system, the object's specific orbital energy, also known as "total energy," and its distance from the central body can be used to calculate the object's instantaneous orbital speed at a specific point in the orbit. Independent of position, the specific orbital energy is constant.

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) a 45 kg box is pulled horizontally across the floor will a force directed horizontally to the left with a magnitude of 200.0n. (a) what is the acceleration of the box if the coefficient of kinetic friction is 0.10? (b) if the box starts from rest, how fast will it be moving at 3.0 s assuming the constant force is applied for the entire 3.0s?

Answers

The acceleration of the box is 3.4m/s² and if the box starts from rest at constant force for 3 seconds, then the speed of the box will be 10.2m/sec

The frictional force (f) = μN

where μ= coefficient of kinetic friction

N = Normal force = mg

The given value for μ = 0.10

∴ f = 0.10 × 45 × 10

f = 45 N

Hence F net = applied force - frictional force

F net = 200 - 45

F net = 155N

Thus, the acceleration of box (a) will be, F = ma

where m = mass of box =45kg

a = F /m

a = 155/45

a = 3.4 m/s²

a) Hence acceleration of the box pulled horizontally is 3.4m/s²

The speed of the box pulled, if initially the box was at rest and constant force is applied for 3 seconds,

v = u + at

v = 0 + 3.4 × 3

v = 10.2 m/s

b) Hence, the speed of the box at constant force is 10.2m/s

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tarzan, who weighs 688 n, swings from a cliff at the end of a vine 18 m long (fig. 8-38). from the top of the cliff to the bottom of the swing, he descends by 3.2 m. the vine will break if the force on it exceeds 950 n. (a) does the vine break? (b) if no, what is the greatest force on it during the swing? if yes, at what angle with the vertical does it break?

Answers

No, the vine didn't breakThe greatest force will be, T = 932.6 NWhat is tension?

Tension is defined as the pulling force transmitted axially using a string, a rope, chain, or similar object, or by each end of a rod, truss member, or similar 3-D object; tension might also be expressed as the action-reaction pair of forces acting at each end of said elements.

Maximum tension:

T - mg = mv²/L

By energy conservation:

mgh = ¹/₂ mv²

v² = 2gh

Now the tension force in the vine at this position is given as:

T = mg +  mv²/L

Now substitute the values in the above equation:

T = 688 N + m(2gh)/L

T = 688 + 2×3.2(688)/18

T = 932.6 N

As the force is less than the limit of 950 N so the vine didn't break.

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Which of the following describes a way that neighborhoods help children build new relationships?
OA neighborhood holds a winter festival with activities for children and families.
O Families of the neighborhood walk around the block on differing weeknights.
O Lawn crews ensure that the neighborhood is properly maintained each week.
O Members of the community form a Neighborhood Watch team.

Answers

The neighborhood can help children build new relationships by holding a winter festival with activities for children and families.

In today's world, where everyone is quite busy in their own world, children have academic pressure and many more.

They tend to detach themselves from people, and hence many relationships ruin.

Building relationship helps the children grow and makes them more sociable. It boots their intelligence and makes them aware of the importance of having a relationship and building bonds with people.

The neighborhood holds a winter festival with activities for children and families is a great way to build new relationships.

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A championship weight lifter did 11,000 J of work on a set of barbells weighing 3680 N. How much gravitational potential energy did the barbells have at the maximum height of the lift? (show the steps you followed to solve the problem - ie. the formula, calculations, and final answer)

Answers

The GPE that the barbells have at max height is ,

3680.h j=11,000 units

where h= maximum height the barbells were lifted

work done by the championship lifter ,W = 11,000 units

weight of the barbells, N = 3680 N

The gravitational potential energy, P.E., the barbells had at their maximum height of lift is given as follows;

P.E. = m × g × h

Where;

m = The mass of the barbells;

g = The acceleration due to gravity = 9.8 m/s^2

h = The maximum height to which the barbells are lifted

m × g = The weight of the barbells = 3680 N

∴ P.E. = 3680 N × h = 3680·h J

we know the law of conservation of energy, according to this the work done by the weight lifter is equals to the maximum gravitational potential energy gained by the barbell is equal to energy at maximium height i.e P.E

therefore, GPE = 3680.h j = W = 11,000j is your answer.

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In which scenario below does the ball have more gravitational potential energy when sitting at the top?
Why?
A. The ball
travels up
the stairs
to a
height of
3 ft.
B. The ball
travels straight
up the column
to a height of 3
ft.

Answers

In both the cases gravitational potential energy is same.

What is gravitational potential?

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s².

Given two cases height of the top is 3 ft so the potential energy is same as potential energy is dependent on height of the ball.

In both the cases gravitational potential energy is same.

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At rest, a car's horn has a frequency of
395 Hz. Car A passes car B on the street
in the same direction. If car A is traveling
at 22.0 m/s and car B is traveling
at 19.5 m/s, what frequency does
car B hear when car A honks?

(Speed of sound = 343 m/s)
(Unit = Hz)

Answers

The frequency heard by car A is determined as 398.4 Hz.

What is the frequency heard by car A?

The frequency heard by car A is determined by applying the following equation.

f = fs(v - v₀) / (v - vs)

where;

v is the speed of sound = 343 m/sv₀ is the speed car B = 19.2 m/svs is the speed of car A = 22 m/sfs is the frequency of car A = 395 Hzf is the frequency of car B = ?

f = 395(343 - 19.2) / (343 - 22)

f = 395(1.0087)

f = 398.4 Hz

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a skater spins with an angular speed of 12.0rad/s with his arms outstretched. he lowers his arms, decreasing his moment of inertia from 41kgm2 to 36kgm2. a. calculate his initial and final rotational kinetic energy. b. does his rotational kinetic energy increase or decrease? c. what causes this change in kinetic energy?

Answers

It has a 2952 J initial rotational energy.

492 kg-m2/s represents its moment of momentum.

13.67 rad/s of angular momentum is conserved.3364J

A) is the final rotational kinetic energy. The rotational energy of it at first is

KE=[tex]\frac{1}{2}\\[/tex]Iω²

=[tex]\frac{1}{2}\\[/tex]

=[tex]2952J[/tex]

B) Her angular momentum is

L=Iω

=(41)(12)

= 492 kg-m²/s

C) Angle momentum is kept constant. Therefore, even if he draws his hands in, nothing will change. She still has 492 kg-m2/s of angular momentum after pulling her arms back. Her moment of inertia has altered, though. L must therefore change in order for its angular velocity to remain constant.

L=Iω

ω=L/I

=492/36

= 13.67 rad/s

D) So her final rotational kinetic energy is

KE=[tex]\frac{1}{2}\\[/tex]Iω²

 =[tex]\frac{1}{2}\\[/tex][tex](36) (13.67)[/tex]

=[tex]3364J[/tex]

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a moving hammer hits a nail and drives it into a wall. if the hammer hits the nail with four times the speed, how much deeper will the nail be driven? (assume that all of ke goes into work)

Answers

According to the work-energy theorem, twice the speed corresponds to 4 times the energy and 4 times the driving distance.

What is energy?

There are different forms of energy on earth. The sun is considered the elementary form of energy on earth. In physics, energy is considered a quantitative property that can be transferred from an object to do work. Therefore, we can define energy as the force required for any  physical activity. So we can define energy in simple words as follows Energy is the ability to do work  According to the laws of conservation of energy, "energy can neither be created nor destroyed, it can only be changed from one form to another". The SI unit of energy is the joule.

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A student conducts an experiment to check how high a basketball can bounce when different volumes of air are filled inside. Which of these steps will most likely help the student get reliable results?(1 point)
Changing the ball for each trial
Doing only one trial with the ball
Doing at least three trials with the ball
Using a different watch to time each trial

Answers

Doing atleast 3 trials with the ball

many astronomers believe that the massive object at the center of the milky way galaxy is a black hole. if so, what must the schwarzschild radius rs of this black hole be?

Answers

The Schwarzschild Radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius or gravitational radius is a physical parameter in the Schwarzschild solution of Einstein's field equations that corresponds to the radius defining the event horizon of a Schwarzschild black hole.

Let's assume that the ring shaped disk is circular and thus, to find the mass of the object at the center of the milky way, we can use the circular orbit equation;

v = √(GM/r)

where;

G is the gravitational constant and has a value of 6.67 x 10^(-11) Nm²/kg²

R is the radius of the orbit,

M is the mass of the larger object

v = velocity = 190km/s = 190000 m/s

Also

1 light year = 9.4605 x 10^(15) m

Thus,7.5 light years = 7.5 x 9.4605 x 10^(15) m

so, M = v²r/G

Putting all the values in the equation,

M = [190000² x 7.5 x 9.4605 x 10^(15)]/6.67 x 10^(-11)

M = 3.84 x 10^(37) kg

Now, mass of the sun generally has a value of 1.9891 x 10^(30) kg

Thus, value of mass of object in solar masses = 3.84 x 10^(37)/1.9891 x 10^(30) = 1.931 x 10^(7) solar masses

The formula for Schwarzschild radius is given as;

Rs = 2GM/c²

Where c is speed of light = 3 x 10^(8) m/s

Thus,

Rs = 2 x 6.67 x 10^(-11) x 3.84 x 10^(37)/(3 x 10^(8))² = 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

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What was the force of gravity on the ball when it was at the highest point in its flight?

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Acceleration of gravity is 9.8

if more massive stars have more hydrogen in their cores than less massive stars, why do they have shorter lifetimes?

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Thermonuclear fusion occurs faster in massive stars so larger stars use up all their fuel in less time.

The more mass a star has the faster it will exhaust its fuel supply and the shorter its lifespan. The most massive stars could burn up and explode in a supernova after just a few million years of nuclear fusion. Massive stars are the largest, hottest, and brightest main-sequence stars and are blue, blue-white, or white in color.

Massive stars run out of hydrogen fuel very quickly and therefore have short lives. This is because the more massive the star the greater the fuel consumption. Even if a high-mass star has more fuel it uses it up very quickly so it does not live as long as a low-mass star. A main-sequence star's mass determines the fundamental properties of its luminosity surface temperature radius and lifetime.

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3) mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. why?

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mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. The planets should have been much farther apart.

The gravitationally bound system of the Sun and the satellites in its orbit is referred to as the Solar System. The gravitational collapse of a massive interstellar molecular cloud gave it birth 4.6 billion years ago. The Sun is the system's primary mass, accounting for 99.86% of its total mass, with Jupiter making up the majority of the remaining mass. Mercury, Venus, Earth, and Mars are the four planets in the inner solar system, and they are all terrestrial planets with rocky and metallic cores. In comparison to the terrestrial planets, the four giant planets of the outer solar system are significantly bigger and more massive. The two biggest, Jupiter and Saturn, are gas giants made primarily of hydrogen and helium; the next two, Uranus and Neptune, are ice giants made primarily of highly volatile substances.

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Two projectiles of mass and are fired at the same speed but in opposite directions from two launch sites separated by a distance . they both reach the same spot in their highest point and strike there. as a result of the impact they stick together and move as a single body afterwards. find the place they will land.

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Two projectiles of mass and are fired at the same speed but in opposite directions from two launch sites separated by a distance, will land at a place x = D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)}

This is calculated using the conservation of linear momentum in the horizontal direction as ,

(vm₁vₓ₁ - m₂vₓ₁ ) ₓî = (m₁ + m₂) vₓ₂ ₓî

vₓ₂ = {(m₁ - m₂) / m₁ + m₂)} × vₓ₁

vₓ₂ = {(m₁ - m₂) / m₁ + m₂)} × v Cos Θ

t max = vₓ₁ / g

=  v Sin Θ / g

x =(D / 2) + vₓ₂t max

= (D / 2) { {(m₁ - m₂) / m₁ + m₂)} × v² Sin Cos Θ / g } ------ (1)

now,

D = 2 vₓ₁vₓ₂ / g

D / 2 = v² Sin Cos Θ / g

From equation (1) we get,

x = D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)}

Hence , D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)} is the place they will land.

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