Which of the following puts the MOST demand on water supplies? Question 18 options: A) flood B) overpopulation C) drought D) pollution

Answers

Answer 1

option (b) is correct.

Overpopulation puts the most demand on water supplies.The reservoirs of the water are fixed but the human population is ever increasing. The needs of water for drinking, washing,industrial use like medicines, construction etc are thus increasing with every passing day.So, we are continuously running out of the water supplies. Because of the overpopulation. the pollution of water is also increasing, thus rendering the water unfit for drinking.

Thus overpopulation puts the most demand on water supplies.

Answer 2

Answer:

option (b) overpopulation is the  correct option for this question

Explanation:

Overpopulation is the reason  in increasing demand on water supplies.Overpopulation is a bothersome condition where the quantity of existing human populace surpasses the conveying limit of Earth. Overpopulation is brought about by number of variables. Decreased death rate, better restorative offices, consumption of valuable assets are few of the causes which results in overpopulation. Because of the overpopulation. the pollution of water is also increasing, thus rendering the water unfit for drinking



Related Questions

What is the Milgram experiment?

Answers

Answer:

The study was unethical

Explanation:

The study was deceptive to the participants, It failed to protect the participants involved and it denied them the right to withdraw from the experiment.

Airbags will deploy no matter from what angle your car is hit.

Answers

so whats the question?

Answer:

Airbags:

" As the airbags comes in inbuilt form these days in most of the cars, as they fulfill the safety concerns of the passengers and make it sure that non of the individuals gets hurt in any accident faced during the travel."

Angle of Deployment:As the airbags will only deploy in the specific conditions provided, as the car must be speed over about the 25 km/hr and the accident must be more like a head-on-collision making the airbags to deploy. Or else if the car experiences an side wise crash with any other entity then it will never deploy in that case.

Determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. -1.82 × 10-19 j +2.69 × 10-19 j +5.51 × 10-19 j +3.03 × 10-19 j -3.03 × 10-19 j

Answers

The expression for the change in energy for a transition from a state [tex]n_i[/tex] to a state [tex]n_f[/tex] in a hydrogen atom is
[tex]\Delta E=-13.6 Z^2 ( \frac{1}{n_f^2}- \frac{1}{n_i^2} )[eV][/tex]
where Z is the atomic number, which is 1 for hydrogen. Substituting [tex]n_i=3[/tex] and [tex]n_f=2[/tex] we find
[tex]\Delta E = -1.89 eV[/tex]
And to convert it into Joule, we should multiply this number by the elementary charge, obtaining:
[tex]\Delta E = (-1.89 eV)(1.6 \cdot 10^{-19}c)=-3.03 \cdot 10^{-19}C[/tex]

The energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom is about -3.03 × 10⁻¹⁹ J

[tex]\texttt{ }[/tex]

Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

[tex]\large {\boxed {E = h \times f}}[/tex]

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

[tex]\texttt{ }[/tex]

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]

[tex]\large {\boxed {E = qV + \Phi}}[/tex]

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial shell = n₁ = 3

final shell = n₂ = 2

Unknown:

ΔE = ?

Solution:

We will use this following formula to solve this problem:

[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{2^2} - \frac{1}{3^2})[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{4} - \frac{1}{9} )[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times \frac{5}{36}[/tex]

[tex]\Delta E \approx -3.03 \times 10^{-19} \texttt{ J}[/tex]

[tex]\texttt{ }[/tex]

Learn morePhotoelectric Effect : https://brainly.com/question/1408276Statements about the Photoelectric Effect : https://brainly.com/question/9260704Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544

[tex]\texttt{ }[/tex]

Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics

At approximately what frequency do most people perceive the least intense sounds? enter your answer numerically in hertz to two significant figures.

Answers

Final answer:

Most people perceive the least intense sounds at a frequency around 2000 Hz with an intensity level at the threshold of hearing, which is about 0 dB or 10^-12 W/m².

Explanation:

The frequency at which most people perceive the least intense sounds is typically around the threshold of human hearing. This threshold is at about 0 dB, which corresponds to an intensity level of approximately 10-12 W/m². However, frequency and intensity are not the only factors that contribute to the perception of sound, as the sensitivity of the human ear varies across different frequencies. The human ear has the maximum sensitivity to frequencies ranging from 2000 to 5000 Hz. Sounds at these frequencies are perceived as louder compared to those at lower or higher frequencies, even if they have the same intensity level. Therefore, the least intense sounds that humans can perceive are at the lower end of this range, around 2000 Hz, at the absolute threshold of hearing of 0 dB.

A 2.00 kg block is placed against a spring on a frictionless 36° incline. the spring, whose spring constant is 19.8n/cm, is compressed 22.8 cm and then released. what is the elastic potential energy of the compressed spring?

Answers

Final answer:

The elastic potential energy of the compressed spring with a spring constant of 19.8 N/cm and compressed 22.8 cm is approximately 51.5 Joules.

Explanation:

The elastic potential energy (EPE) stored in a spring is calculated using the formula EPE = 1/2 k x², where k is the spring constant, and x is the displacement from the spring's equilibrium position. Given that the spring constant is 19.8 N/cm and the spring is compressed 22.8 cm, we must first convert the spring constant to N/m by multiplying by 100, giving us 1980 N/m. Using the formula, the elastic potential energy is thus EPE = 1/2 (1980 N/m) (0.228 m)² = 1/2 (1980) (0.051984) J = 51.504 J. Therefore, the potential energy of the compressed spring is approximately 51.5 Joules.

If a violin string vibrates at 195 hz as its fundamental frequency, what are the frequencies of the first four harmonics? first harmonic hz second harmonic hz third harmonic hz fourth harmonic hz

Answers

The frequency of the nth-harmonic is n times the fundamental frequency f1:
[tex]f_n = n f_1[/tex]
The first harmonic is the fundamental frequency: [tex]f_1 = 195 Hz[/tex], then the following harmonics are
[tex]f_2 = 2 f_1 = 2 \cdot 195 Hz = 390 Hz[/tex]
[tex]f_3 = 3 f_1 = 3 \cdot 195 Hz = 585 Hz[/tex]
[tex]f_4 = 4 f_1 = 4 \cdot 195 Hz = 780 Hz[/tex]

Answer:

A violin is a both ends fixed, and so successive harmonics are simply multiples of the fundamental:

Therefore, f1 = 440Hz (fundamental frequency)

f2 = 2f1 = 2 x 440 = 880 Hz

f3 = 3f1 = 3 x 440 = 1320 Hz

f4 = 4f1 = 4 x 440 = 1760 Hz

Eddie and Val observed the picture of an athlete running in a race. Eddie stated the picture shows potential energy being transformed to kinetic energy. Val stated that it shows chemical energy being transformed to mechanical energy. Which best explains who is correct?

Only Eddie is correct because the picture does not show a transformation that starts with chemical energy.

Only Eddie is correct because the picture does not show a transformation that ends with mechanical energy.

Both Eddie and Val are correct because chemical energy is a type of potential energy and mechanical energy includes kinetic energy.

Both Eddie and Val are correct because chemical energy is a type of kinetic energy and mechanical energy includes potential energy.


ITS NOT THE LAST ONE I GOT THAT ONE WRONG LAST TIME

Answers

Answer:

"Both Eddie and Val are correct because chemical energy is a type of potential energy and mechanical energy includes kinetic energy."

Explanation:

Answer:

C

Explanation:

now give me brainly please

For a sine wave depicting simple harmonic motion, the smaller the amplitude of the wave, the smaller the of the pendulum from the equilibrium position. The shorter the period, the the pendulum’s rod.

Answers

displacement   , shorter 

Answer:

The smaller the amplitude of the wave, the smaller the displacement of the pendulum from the equilibrium position. The shorter the period, the shorter the pendulum’s rod.

To find the right answer, we could look the analysis of a simple pendulum movement. The relation that defines this movement is:

[tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex]

[tex]T: \ period\\l: \ pendulum's \ rod \ length\\g: \ gravity[/tex]

From the equation, we can see the relation between the period and the length of the pendulum's rod. They are directly proportional, this means that if the period increases, the length increases, or vice versa. So, the shorter the period, the shorter the pendulum's rod, because they are directly proportional.

On the other hand, the amplitude of the harmonic motion refers to the displacement of the object from the equilibrium point which is in the center of the periodic movement. This allow us to deduct that if the amplitude is small, the displacement of the object is small, if the amplitude is big, the displacement from the equilibrium point is bid.

What change occurs to the mass of an object when an unbalanced force is applied to it? The mass increases. The mass decreases. The mass remains the same. The mass reduces to zero. The mass changes by an indeterminate amount

Answers

the mass remains the same

Third option is correct. The mass of an object remains same when an unbalanced force is applied to it.

the mass on an object doesn't depend on the applied force. it is constant.When an unbalanced force is applied to an object, it will get accelerated. The velocity of the object will increase or decrease depending on the direction of the force but the mass of the object remains same. That doesn't change with the applied force.

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s glove, what average force did the ball exert on the catcher?

Answers

To solve for force, you need to get the product of mass and acceleration. 
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration. 
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

[tex]a = \frac{vf-vi}{t} [/tex]

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s. 

So we can put it into our formula now:

[tex]a = \frac{0m/s-50m/s}{20ms} [/tex]

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s. 

[tex]20ms[/tex] x [tex] \frac{1s}{1000ms} [/tex] = [tex] \frac{20s}{1000} = 0.02s [/tex]

So your new time is 0.02s. Now we put this time into the formula:


[tex]a = \frac{0m/s-50m/s}{0.02s} [/tex]
[tex]a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2} [/tex]

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop. 

So now we have our acceleration. Now using this, we can get our force. 

[tex]F= ma[/tex]

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is [tex]kg.m/ s^{2} [/tex] but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first. 

[tex]150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg[/tex]

Our new mass is 0.150kg. 

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 [tex]m/ s^{2} [/tex]

Now that all our units match, we can put that into our formula:

[tex]F= ma[/tex]
[tex]F= (0.150kg)(-2,500m/s^{2})[/tex]
[tex]F = -375kg.m/ s^{2} or -375N[/tex]

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign. 



What is the kinetic energy of a 3000 lb car traveling at 60 mph?

Answers

5 × 12 = 60 I got 60mph I multiply and that's how I get 60

Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a 4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?

Answers

i use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop

The force require to stop the second object is greater because it has bigger momentum when compared to the first object.

The given parameters;

mass of the first object, m₁ = 8 kgspeed of the first object, u₁ = 0.2 m/smass of the second object, m₁ = 4 kgspeed of the second object, u₂ = 1 m/s

The force required to stop each object in the given time is determined by applying Newton's second law of motion as shown below;

[tex]F = ma = \frac{mv}{t} \\\\F_1 = \frac{8 \times 0.2}{10} \\\\F_1 = 0.16\ N[/tex]

For the second object;

[tex]F_2 = \frac{4 \times 1}{10} \\\\F_2 = 0.4 \ N[/tex]

Thus, the force require to stop the second object is greater because it has bigger momentum when compared to the first object.

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As the army approached the bridge into the city, the commander told them to break step. He did not want the soldiers to march in cadence. Why did the commander give this order?

Answers

At a definite point, the bridge would begin oscillating to the matching rhythm as that of the marching footsteps. This oscillation would touch a determined peak when the bridge can no longer tolerate its own power and later collapses. So, soldiers are systematic to break their steps while passing a bridge.

Soldiers are ordered to 'route step' across a bridge to prevent forced oscillations and resonance that can lead to structural failure if marching is synchronized.

The reason soldiers are usually ordered to "route step" or walk out of step across a bridge has to do with forced oscillations and resonance. When soldiers march in cadence, their synchronized steps act as a periodic force that can cause the bridge to oscillate at its natural frequency if the cadence matches this frequency. This can amplify the oscillations to dangerous levels, potentially leading to structural failure of the bridge. By breaking step, the soldiers apply random forces to the bridge, which decreases the chances of resonance occurring, thus maintaining the structural integrity of the bridge.

What condition occurs when the eyeball is too long?

Answers

Hello!

Myopia, also known as nearsightedness, occurs when the eyeball is too long. This condition is very common in the U.S and throughout the world. It can be corrected by wearing glasses or contact lenses.

Explanation :

There can be many eye defects in the human eye. Some of them are Myopia, Hypermetropia, Astigmatism etc.

In Myopia the size of the eyeball is too long and the image is formed in front of the retina. A person suffering from this defect is not able to see far objects clearly but can see nearby objects normal.

Concave lenses are used to correct this defect so that the image is focused on the retina.

A strong lightning bolt transfers about 25 c to earth. how many electrons are transferred? the elementary charge is 1.60218 × 10−19
c.

Answers

The total charge transferred is [tex]Q=25 C[/tex]. This total charge consists of N electrons, each one with charge [tex]e=1.6 \cdot 10^{-19}C[/tex]. To find N, the number of electrons transferred, we must simply divide the total charge by the charge of a single electron:
[tex]N= \frac{Q}{e}= \frac{25 C}{1.6 \cdot 10^{-19}C}=1.56 \cdot 10^{20} [/tex]
Final answer:

To determine how many electrons are transferred in a lightning strike, you divide the total charge transferred (in this case, 25 coulombs) by the elementary charge which is the charge per electron. The result approximates to 1.56 × 1020 electrons.

Explanation:

The number of electrons transferred during a lightning strike can be determined by dividing the total charge transferred by the charge of a single electron. The charge of a single electron, also known as the elementary charge, is approximately 1.60218 × 10-19 Coulombs. Therefore, if a lightning bolt transfers about 25 Coulombs to the ground, the number of electrons transferred can be calculated as follows:

Number of electrons = Total charge transferred / Charge per electron

= 25 Coulombs / 1.60218 × 10-19 Coulombs

Therefore, a lightning bolt that transfers about 25 Coulombs to the earth transfers approximately 1.56 × 1020 electrons.

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While scuba diving with some friends, Deshawn notices that the air bubbles exhaled by his friends increase in size as they get closer to the surface of the water.

Why does this occur and which gas law explains it?
hurrry plzz/////

Answers

Hello!

What Deshawn notices is that the volume of the bubbles increase as they get closer to the surface of the water.

This occurs because of the Boyle's Law. That Law states that when pressure decreases at constant temperature (like in water), the volume of a gas increases ([tex]P1*V1=P2*V2[/tex]). 

What is happening is that as the bubbles get closer to the surface of the water, the pressure on them decreases, causing their volume to increase, and that is seen by the increase in the size of the Bubbles that Deshawn noticed.

Have a nice day!

Answer:

Sample Response: As the air bubbles rise to the surface, the pressure on them decreases. This decrease in pressure allows the air to expand and each bubble increases in size. Boyle’s law explains why this occurs.

Explanation:

The flower of the species Rosa verdus can be either green or red. in the species a single gene with two alleles determines flower color. Small g is the recessive allele that results in red flowers in the homozygous condition. Which two genotypes will result in plants with green flowers.

Answers

Since G is a dominate trait the 2 genotypes would be Gg or GG

Answer:

SS and Ss

I did the test!!!!

Which statement best describes the characteristics of white light? A. White light is composed of a spectrum of many different colors. B. White light is composed of radiation with a wide range of amplitudes. C. White light is composed of various wavelengths of light not seen by the naked eye. D. White light is composed of radiation that travels at different speeds producing various colors

Answers

a white light is, by definition, apparently colorless light, for example ordinary daylight. It contains all the wavelengths of the visible spectrum at equal intensity.

so A.

White light is composed of a spectrum of many different colors (Option A).

What is the spectrum?

The spectrum is the variety of colors created when light travels through a glass prism or a drop of water.

White light is a mixture of all the colors of the visible spectrum, which includes red, orange, yellow, green, blue, indigo, and violet. When light passes through a prism, it is separated into its individual colors, creating a rainbow-like effect.

This is because different colors of light have different wavelengths, and a prism bends the light of different wavelengths by different amounts, causing them to spread out and form a spectrum.

So, white light is actually composed of many different colors of light, each with its own unique wavelength.

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Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force on the hammer when it has a tangential speed of 31.95 m/s? The answer is rounded off to the nearest whole number. 190 N 840 N 1,400 N 6,100 N

Answers

In the circular motion of the hammer, the centripetal force is given by
[tex]F=m \frac{v^2}{r} [/tex]
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
[tex]F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N [/tex]

Answer:

6,100 N

Explanation:

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall acceleration. Is a person in this ride experiencing apparent weightlessness, true weightlessness, or neither? Explain.

Answers

An object experiences true weightlessness when the net force of all gravitational forces acting upon the object is zero. In this case, the gravitational force exerted by the earth on the people that are on the park ride while it's free falling never ceases to act on the people. If the person on the ride were in a case of true weightlessness then they would not fall in any direction in the first place. The answer is the apparent weightlessness.

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what is the x position of the center of mass? m 2) what is the y position of the center of mass? m 3) what is the speed of the center of mass? m/s 4) when a fifth mass is placed at the origin, what happens to the horizontal (x) location of the center of mass? it moves to the right. it moves to the left. it does not move. it can not be determined unless you know the mass. 5) when a fifth mass is placed at the center of mass, what happens to the vertical (y) location of the center of mass? it moves up. it moves down. it does not move. it can not be determined unless you know the mass.

Answers

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
[tex]R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i[/tex]
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
[tex]R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i[/tex]
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
[tex]R_x=0.96m[/tex]
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
[tex]R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i[/tex]
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
[tex]R_y=-0.84m[/tex]
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
[tex]v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}[/tex]
[tex]v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}[/tex]
Total velocity is:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
Once we calculate velocities we get:
[tex]v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}[/tex]
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing
Final answer:

The center of mass and speed can be calculated using the formulas provided. Adding a fifth mass at either the origin or the current center of mass will affect the location of the center of mass, but it can only be determined with knowledge of the mass of the new particle.

Explanation:

The center of mass for a system of particles can be calculated using the following formulas: Xcm = ∑mi xi/mi and Ycm = ∑mi yi / ∑ mi, where 'mi' denotes the mass of the ith particle and 'xi' and 'yi' are the x and y coordinates respectively. This principle extends not just to two dimensions but also to three dimensions.

The speed of the center of mass of a group of particles can be calculated by the formula Vcm = ∑mi vi / ∑ mi, where 'vi' is the vector sum of the velocities of the particles, and 'mi' is the mass of the ith particle.

When a fifth mass is placed at the origin, the location of the center of mass can only be determined if the mass is known. Similarly, when a fifth mass is placed at the center of mass, the new location depends on the mass of the fifth particle and cannot be determined otherwise.

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A fan blade rotates with angular velocity given by ωz(t)= γ − β t2, where γ = 4.90 rad/s and β = 0.750 rad/s3 . part a calculate the angular acceleration as a function of time.

Answers

The angular velocity as a function of time is given by
[tex]\omega (t)=\gamma-\beta t^2[/tex]
where [tex]\gamma=4.90 rad/s[/tex] and [tex]\beta=0.750 rad/s^3[/tex]. The angular acceleration as a function of time is equal to the derivative of the angular velocity. If we calculate the derivative of w(t), we find:
[tex]\alpha(t)= \frac{d\omega}{dt} =-2\beta t[/tex]
and this is the angular acceleration of the fan blade.

Answer:

Angular acceleration, α = -2βt

Explanation:

Angular velocity of fan is [tex]\omega_{z(t)}=\gamma -\beta t^2[/tex]

[tex]\gamma=4.90\ rad/s[/tex]

[tex]\beta=0.750\ rad/s^3[/tex]

Angular acceleration is given by :

[tex]\alpha=\dfrac{d\omega}{dt}[/tex]

[tex]\alpha=\dfrac{d(\gamma -\beta t^2)}{dt}[/tex]

[tex]\alpha=-2\beta t[/tex]

Hence, the above equation is the angular acceleration as a function of time.   

If two oscillators connected to speakers create frequencies of 594 hz and 600 hz, what is the beat frequency when you hear both at the same time? hz

Answers

When we have two sound waves with frequencies f1 and f2 that interfere, the beat frequency is equal to the absolute value of the difference of the two frequencies, so in our problem:
[tex]f_{beat}=f_2-f_1 = 600 Hz-594 Hz = 6Hz[/tex]

1. How is electric potential energy similar to gravitational potential energy? How is it different? Where will an electron bound in an atom and have the largest electrical potential energy?

Answers

both ve similar equations 
both are energies of one object w.r.t another 
differences- electric pe is due to electrostatic force and gravitational pe is due to gravitational force 
electric pe is > than gravitational pe since electrostatic force> gravitational force 
electron bound in an atom ll ve largest potential enegy in its ground state. i think hope it helps

A 15 kg block is attached to a very light horizontal spring of spring constant 500 N/m and is resting on a frictionless horizontal table as shown in the figure. Suddenly it is struck by a 3 kg stone traveling horizontally at 8 m/s to the right; the stone rebounds at 2 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Answers

Let's separate the problem in two parts:

Part 1): Collision between the stone and the block.
In the collision, the total momentum of the system stone+block is conserved.
Before the collision, only the stone is moving, so the total momentum is:
[tex]p_i = m_s v_s[/tex]
where [tex]m_s = 3 kg[/tex] is the mass of the stone and [tex]v_s = 8 m/s[/tex] is the speed of the stone, traveling towards the block (to the right).
After the collision, both the stone and the block are in motion, so the total momentum is:
[tex]p_f = m_s v_s'+m_b v_b[/tex]
where [tex]v_s' = -2 m/s[/tex] is the new speed of the stone (with a negative sign, since the stone is now moving in the opposite direction, to the left), [tex]m_b = 15 kg[/tex] is the mass of the block and [tex]v_b [/tex] is the mass of the block just after the collision.
Since the momentum must be conserved,
[tex]p_i = p_f[/tex]
So we can rewrite everything
[tex]m_s v_s = m_s v_s' + m_b v_b[/tex]
to find [tex]v_b[/tex]
[tex]v_b = \frac{m_s v_s - m_s v_s'}{m_b}= \frac{(3kg)(8m/s)-(3kg)(-2m/s)}{15kg}=2 m/s [/tex]
So the block slides with speed 2 m/s to the right after the collision.

Part 2) Block compressing the spring
At this point we can ignore the stone and focus only on the block and the spring. The block starts to move with speed 2 m/s, so its kinetic energy is
[tex]K= \frac{1}{2}m_bv_b^2 [/tex]
As it compresses the spring, the speed of the block decreases and its kinetic energy is converted into elastic potential energy of the spring, which undergoes through a compression [tex]x[/tex] with respect to its rest position. When the block completely stops, the compression of the spring is maximum, [tex]x_{max}[/tex], and the elastic potential energy of the spring is:
[tex]U= \frac{1}{2} k x_{max}^2 [/tex]
where [tex]k=500 N/m[/tex] is the constant of the spring.
For the conservation of energy, we must have
[tex]K=U[/tex]
So we can write
[tex] \frac{1}{2}m_b v_b^2 = \frac{1}{2}kx_{max}^2 [/tex]
and we can solve to find the compression of the spring:
[tex]x_{max}= \sqrt{ \frac{m_b v_b^2}{k} }= \sqrt{ \frac{(15 kg)(2m/s)^2}{500 N/m} }=0.35 m [/tex]

0.35 m is the maximum distance that the block will compress the spring after the collision.

Given :

Mass, [tex]\rm m_b = 15 \;Kg\;and\; m_s = 3\;Kg[/tex]

Spring constant, K = 500 N/m

Solution :

Before the collision the total momentum is,

[tex]\rm p_i = m_s \times v_s[/tex]

Whare [tex]m_s[/tex] is mass of stone and [tex]\rm v_s[/tex] is the velocity of stone beforethe collision.

After the collision stone and block both are moving therefore total momentum is,

[tex]\rm p_f = m_sv_s' + m_bv_b[/tex]

where, [tex]\rm v_s'[/tex] is the velocity of stone after collision, [tex]\rm m_b[/tex] is the mass of block and [tex]\rm v_b[/tex] is the speed of the block after collision.

Now, momentum should be conserve

[tex]\rm p_i = p_f[/tex]

[tex]\rm m_sv_s= m_sv_s'+ m_bv_b[/tex]

[tex]\rm v_b = \dfrac{m_s}{m_b}(v_s - v_s')[/tex]  ------- (1)

[tex]\rm v_s = 8\;m/sec[/tex]      (Given)

[tex]\rm v_s' = 2 \;m/sec[/tex]      (Given)

Now putting the values in equation (1) we get,

[tex]\rm v_b = \dfrac{3}{15}(8-2)[/tex]

[tex]\rm v_b = 2\;m/sec[/tex]

Therefore, block slides with the speed of 2 m/s to the right after the collision.

We know that the block starts to move with speed 2 m/s, so its kinetic energy is

[tex]\rm K = \dfrac{1}{2}m_bv_b^2[/tex]

As the spring is compress, block speed decreases and its kinetic energy is converted into elastic potential energy of the spring and the elastic potential energy of the spring is,

[tex]\rm U = \dfrac{1}{2} Kx^2_m_a_x[/tex]

Now through conservation of energy we have,

[tex]\rm K = U[/tex]

[tex]\rm m_bv_b^2 = Kx^2_m_a_x[/tex]

[tex]\rm x_m_a_x = \sqrt{\dfrac{m_bv_b^2}{K}}[/tex]----- (2)

Putting the values of  [tex]\rm m_b,\;v_b\;and\; K[/tex]  in equation (2),

[tex]\rm x_m_a_x= \sqrt{\dfrac{15\times2^2}{500}}[/tex]

[tex]\rm x_m_a_x=0.35\;m[/tex]

0.35 m is the maximum distance that the block will compress the spring after the collision.

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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 792.7 rad / s while it is being read, and then is allowed to come to rest over 0.234 seconds, what is the magnitude of the average angular acceleration of the disk?

Answers

(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²

Describe the relationship between the force between two magnetic poles and the distance of separation.

The magnetic force is directly proportional to the distance of separation.
The magnetic force is inversely proportional to the distance of separation.
The magnetic force is directly proportional to the square of the distance of separation.
The magnetic force is inversely proportional to the square of the distance of separation.

Answers

The magnetic force is inversely proportional to separation, and this is quite unique as electric field and gravitational fields are both inversely squared with distance. This means magnetic fields don’t lose strength as quickly as other fields.

The relationship between the force between two magnetic poles and the distance of separation is The magnetic force is inversely proportional to the square of the distance between them.

What is Coloumb's law of magnetism?

Coloumb's law of magnetism states the relationship between the magnetic strength of two magnets. The force of attraction or repulsion between two magnets is directly proportional to the strength of magnetic poles and inversely proportional to the square of the distance between them.

Coloumb's law of magnetism describes the electric field between two charged objects. The force of attraction or repulsion is inversely proportional to the absolute permeability of the surrounding medium.

The attractive and repulsive force, F∝ m₁×m₂ / μ₀r², where m₁, and m₂ are the magnetic strength or magnetic poles and r is the distance between two magnets.

The attractive or repulsive force, F = k (m₁×m₂) / μ₀r², where μ₀ is the permeability of the medium and k is the constant of proportionality. The unit is Newton/Tesla. The magnetic force is inversely proportional to the square of the distance of separation.

Hence, the ideal solution is option D.

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Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of 4.5 × 104 m/s.

Answers

Considering the rocket+gases as an isolated system, the variation of momentum of the rocket should be equal to the variation of momentum of the gases, which is
[tex]\Delta p = \Delta m \cdot v[/tex]
since the speed v at which the gas is expelled is constant.
If we divide both terms per the time interval, [tex]\Delta t[/tex], we get
[tex] \frac{\Delta p}{\Delta t}= \frac{\Delta m}{\Delta t} v [/tex]
But [tex] \frac{\Delta p}{\Delta t}[/tex] is equal to the force F exerted on the gas by the rocket (and for Newton's third law, this is equal to the force exerted by the gas on the rocket), while [tex] \frac{\Delta m}{\Delta t} [/tex] is the rate at which the gas is expelled, 1300 kg/s. Therefore, the force exerted on the rocket is
[tex]F= \frac{\Delta m}{\Delta t} v = (1300 kg/s)(4.5 \cdot 10^4 m/s)=5.85 \cdot 10^7 N[/tex]

This question involves the concept of Newton's Second Law of Motion.

The force exerted on the rocket is "5.85 x 10⁷ N".

NEWTON'S SECOND LAW OF MOTION

According to Newton's Second Law of Motion, whenever an unbalanced force is applied to an object, it produces an acceleration in the direction of the force. Mathematically,

[tex]F=ma[/tex]

where,

F = force = ?a = acceleration = [tex]\frac{\Delta v}{t}[/tex]

Therefore,

[tex]F = \frac{m\Delta v}{t}[/tex]

where,

[tex]\frac{m}{t}[/tex] = mass flow rate of gases = 1300 kg/sΔv = change in velocity = 4.5 x 10⁴ m/s

Therefore,

F = (1300 kg/s)(4.5 x 10⁴ m/s)

F = 5.85 x 10⁷ N

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An object in motion will stay in motion unless acted upon by what type of force

Answers

An object will stay in motion unless acted upon by an UNBALANCED force. 
The answer Is unbalanced On USATESTPREP

According to Einstein’s theory, how does an increase in the number of photons affect a beam of light?

The wavelength of the beam of light increases.

The speed of the beam of light increases.

The brightness of the beam of light increases.

The frequency of the light beam increases.

The energy of each photon in the beam of light increases.

Answers

The answer is the third choice, "the brightness of the beam of light increases"

According to Einstein’s theory, an increase in the number of photons (per unit are) affects a beam of light by causing a higher intensity.  Sometimes, the term "brightness" is used when referring to the intensity of a color, although there are instances where this can be a misleading term when we try to describe intensity

Answer:

The brightness of the beam of light increases.

Explanation:

As per Einstein's experimental verification he proves that light is of particle nature and his equation for energy balance is given as

[tex]h\nu = \phi + KE[/tex]

so incident photons on a photosensitive plate will eject the electrons from the plate with some sufficient KE.

So here light is considered as particles which means one photon will eject one electron. So here intensity of light is considered as

Intensity = (number of photons per second) (energy of one photon)

so it is given as

[tex]I = \frac{N}{t}(\frac{hc}{\lambda})[/tex]

now if the intensity of light is increased for same wavelength of light then it means the number of photon must have to be increase.

so correct answer will be

The brightness of the beam of light increases.

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