the gravitational force is considered a very weak force. yet, it is strong enough to hold earth in orbit around the sun. explain this apparent disparity.

Answer:

232 N

Explanation:

By Hooke's law, the force applied to a spring is proportional to the stretch of the spring, so

F = kx

Where F is the force, k is the spring constant and x is how much it is stretched.

So, replacing k by 58N/cm and x by 4 cm, we get

F = (58 N/cm)(4 cm)

F = 232 N

Therefore, the force applied is 232 N

Answer:

The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.

Explanation:

SHOCKING

Explanation:

Answer:

D i promises.

Explanation:

light waves can go 2 direction

An organized way to collect and record scientific observations is with a data table (option C).

An experiment is a test under controlled conditions made to either demonstrate a known truth, examine the validity of a hypothesis, or determine the efficacy of something previously untried.

The scientific method involves the collection of data and organizing them in a way that they can be easily interpreted.

However, the best way to collect and record scientific observations is with the aid of a data table.

A data table is any display of information in tabular form, with rows and/or columns named.

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True. A Type Sa galaxy is a spiral galaxy with a large bulge and tightly wound arms.

Galaxies are categorized based on their morphology, and the Hubble classification system is commonly used to classify galaxies into different types. According to the Hubble sequence, spiral galaxies are classified as Type Sa, Type Sb, and Type Sc, depending on the characteristics of their bulges and arms.

In the case of a Type Sa galaxy, it is characterized by a large, prominent bulge at its center and tightly wound arms. The bulge is a concentrated, spherical or ellipsoidal region containing older stars, while the arms extend from the bulge and contain younger stars, gas, and dust.

The tightly wound arms in a Type Sa galaxy indicate a more tightly wound spiral structure compared to other types. This results in a more compact appearance and less pronounced separation between the arms.

A Type Sa galaxy is indeed a spiral galaxy with a large bulge and tightly wound arms. Its classification within the Hubble sequence indicates specific characteristics related to the structure and morphology of the galaxy. Understanding these classifications helps astronomers categorize and study the diverse range of galaxies in the universe.

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When it takes 1 second for the echo sounder to send and receive one sound pulse, the approximate depth of the water beneath the boat is 750 meters.

To determine the approximate depth of the water beneath the boat when it takes 1 second for the echo sounder to send and receive one sound pulse, we can use the formula:

Depth = (Speed of Sound × Time) / 2

Given that the speed of sound in water is roughly 1500 meters/second and the time taken is 1 second, we can calculate the depth:

Depth = (1500 m/s × 1 s) / 2

Depth = 750 meters

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Given data

*The given mass of the uniform disk is m = 2.01 kg

*The given radius of the disk is r = 0.100 m

*The given frequency is

The formula for the angular momentum is given as

Substitute the known values in the above expression as

empty space or vaccum the Electro magnetic wave travels the fastest.

The second step of the combustion process in an incinerator requires a high temperature of at least 1500 to 1800 degrees F. So, the correct answer is option b.

The correct answer is d. 1800 to 1900 degrees F. The second step of the combustion process in an incinerator requires a high temperature to ensure the complete combustion of the waste materials. This temperature range is necessary to break down any remaining organic matter and convert it into ash and gases.
Combustion, or combustion, is a high-temperature exothermic redox reaction between a fuel and an oxidizer (usually atmospheric oxygen) that produces oxidized, mostly gaseous products in a mixture called smoke. Combustion does not always lead to a fire because the flame is only seen when the burning material has evaporated, but when this happens, the flame is indicative of a reaction. The energy that must be overcome to initiate combustion, and the heat produced by the flame can provide enough energy for the reaction to take place.

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The life expectancy of the given stars are:a) 0.2-solar mass, 0.01-solar luminosity red dwarf: 10 trillion yearsb) 3-solar mass, 30-solar luminosity star: 10 million yearsc) 10-solar mass, 1000-solar luminosity star: 10 million years.

The life expectancy of a star is determined by its mass and luminosity. The more massive and luminous the star is, the shorter its life expectancy is. Hence, using this information, we can estimate the life expectancy of the following stars:a) 0.2-solar mass, 0.01-solar luminosity red dwarfRed dwarfs are known to have the longest life expectancies among all types of stars. They can live for trillions of years.

Hence, a 0.2-solar mass, 0.01-solar luminosity red dwarf is expected to have a much longer life expectancy than the Sun. It could live for up to 10 trillion years or more.b) 3-solar mass, 30-solar luminosity starA 3-solar mass, 30-solar luminosity star is much more massive and luminous than the Sun. As a result, it will have a much shorter life expectancy than the Sun.

Based on its mass and luminosity, it is estimated to have a lifetime of around 10 million years.c) 10-solar mass, 1000-solar luminosity starA 10-solar mass, 1000-solar luminosity star is extremely massive and luminous. It will burn through its fuel much faster than the Sun, resulting in a much shorter life expectancy. Based on its mass and luminosity, it is estimated to have a lifetime of only around 10 million years as well.

Therefore, the life expectancy of the given stars are:a) 0.2-solar mass, 0.01-solar luminosity red dwarf: 10 trillion yearsb) 3-solar mass, 30-solar luminosity star: 10 million yearsc) 10-solar mass, 1000-solar luminosity star: 10 million years.

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The weight of the original mass on the spring was approximately 6.75 Newtons (N). The period of oscillation of a spring/mass system is determined by the mass and the spring constant.

Let's assume the original mass on the spring is represented by M and the corresponding weight is W.

Given that the original period is 5 seconds and the modified period is 3 seconds, we can set up the following equation using the formula for the period of an oscillating spring/mass system:

T = \(2\pi \sqrt{M/k}\), Where T is the period, M is the mass, and k is the spring constant.

For the original system with a period of 5 seconds, we have:

5 = \(2\pi \sqrt{M/k}\) ...(1)

When 12 N are removed from the spring, the modified system has a period of 3 seconds. This implies that the spring constant has changed, but the mass remains the same. Let's assume the new spring constant is k'.

3 = \(2\pi \sqrt{M/k'}\) ...(2)

Dividing equation (1) by equation (2), we can eliminate the mass M:

5/3 = \(\sqrt{k'/k}\)

Squaring both sides of the equation gives:

25/9 = k'/k.

Rearranging the equation gives:

k' = (25/9)k.

Since the spring constant is directly proportional to the weight of the mass, we can conclude that the weight of the original mass W is also reduced by a factor of (25/9).

Let's assume the weight of the original mass on the spring is W0. Thus, the weight of the modified mass is (W0 - 12 N).

Using the proportionality, we have: (W0 - 12 N) = (25/9)W0.

Simplifying the equation, we find: (9/9)W0 - (12 N) = (25/9)W0, (-16/9)W0 = 12 N.

Multiplying both sides by (-9/16) gives: W0 = (-9/16)(12 N), W0 = -6.75 N

Therefore, the weight of the original mass on the spring was approximately 6.75 Newtons (N).

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Answer: A method of classifying stars by their temperatures and compositions.

Explanation:

The Morgan-Keenan Spectral Classification system is a method of classifying stars by their temperatures and compositions.

The primary factor used to categories stars is their temperature.

The Morgan-Keenan luminosity class (MK or MKK) was developed as a result of the Harvard spectral classification classification scheme does not completely describe the star as it cannot distinguish between stars with the same temperature but different luminosities.

Since then, class I stars have been separated into Ia-O, Ia, and Ib, and classes VI (sub-dwarf), and D (white dwarf) have been added, replacing the original roman numerals between I (supergiant star) and V (main sequence).

The original Harvard classification of the star is supplemented by the MK luminosity class in order to fully define it. For instance, the entire classification of the Sun, which is a main sequence G2 star, is G2V.

The MK luminosity class is added to the Harvard classification of the star in order to fully define it.

A method of classifying stars by their temperatures and compositions is applied in Morgan-Keenan Spectral Classification system.

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The fact that the blood type matches that of the defendant is no proof that the defendant is guilty as charged.

There are about four blood types that people can have, these blood types are; A, B, AB and O. Everybody must posses one of these blood types. A person's blood type is determined through laboratory tests.

The  prosecuting attorney was wrong in concluding that the defendant is guilty as charged only based on the blood type evidence. different people could have the same blood type. Only DNA analysis can conclusively prove that the defendant is actually guilty as charged.

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Answer:

2 g / cm^3

Explanation:

Density = mass / volume

              = 4 g / 2 cm^3   = 2 g / cm^3

First, compare two balls from the heavier group to Group A. There must be two weighings; lower bound because there are three options per weighing.

a) Follow these steps to determine which ball is heavier in just two weighings:

1. nine balls should be divided into three groups of three balls each—Group A, Group B, and Group C.

2. Using the balanced scale, weigh Group A in comparison to Group B:

3. Using just two weighings, you can determine which ball is heavier by following these steps.

4.  Compare those two balls to one another:

Using just two weighings, you can determine which ball is heavier by following these steps.

(b) The decision tree lower bound demonstrates that fewer than two weighings are required to identify the defective ball. This is due to the fact that there are nine balls, and each one can only tell you how much each group weighs. We can only reduce the number of possibilities by a factor of three with each weighing.

We would need to compare all possible ball combinations, which would require more than two weighings, in order to distinguish between all nine balls in a single weigh. Therefore, the minimum required to determine which of the nine balls with identical appearances is the heavier is two weighings.

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Answer: Final speed,Vf= -4.5m/s

It is the same speed but in opposite direction,coming down against gravity.

Explanation:

Calculating distance,d= (Vf^2-Vo^2)/2g

d=(0-4.5^2)/(2×9.8)

d= -20.25/19.6

d= -0.053m

Vf^2=Vo^2 + 2gd

Vf= Sqrt(0 +2×9.8(×-0.053))

Vf=V=-4.5m/s

Answer:

I think it's D!!

cuz protons are in the atoms

Answer: Just took it!

Explanation:

Mechanical energy flows out of the magnets and into the electric charges in the loop of wire. Therefore,  the correct option is option A among all the given options.    

A current is the movement of particles (electrons) across wires and components. It is the charge flow rate. If an electric charge passes through a conductor, scientists say the conductor has an electric current. Electrons form a charge flow in circuits that use metallic conductors.

Electromagnetic induction in a moving loop of wire cause an electric generator to produce an electric current by mechanical energy flows out of the magnets and into the electric charges in the loop of wire cause.

Therefore,  the correct option is option A among all the given options.    

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The ball's velocity is 40/91 m/s if its angular velocity around the elbow joint is 20.0 rad/s and it is 1.25 m away from the elbow joint.

In essence, the angular velocity is a vector quantity and represents the speed at which an item rotates. The angular velocity of an object is determined by its angular displacement over a specified amount of time.

In contrast to orbital angular velocity, spin angular velocity describes how quickly a rigid body rotates with regard to its center of rotation and is independent of the choice of origin. An angle per unit of time is the usual unit of measurement for angular velocity.

w = angular velocity of the ball about the elbow joint = 30.3 rad/s

r = radius of circular turn = distance of the ball from the elbow joint = 1.35m

v = linear velocity of the ball

Linear velocity of the ball is given as

v = r w

Inserting the values given above in the equation

v = (1.35) (30.3)

v = 40.91 m/s

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The large size of insects and amphibians during the Pennsylvanian period has been suggested to be due to higher atmospheric oxygen levels.

During this period, which lasted from about 323 to 298 million years ago, oxygen levels in the atmosphere are estimated to have been as high as 35%, compared to the current level of about 21%.

This phenomenon is known as the oxygen hypothesis, and it is supported by fossil evidence showing larger body sizes of insects and amphibians during this period.

During the Pennsylvanian Period (approximately 323 to 298 million years ago), the Earth's atmosphere contained significantly higher levels of oxygen compared to the present day.

Insects and amphibians, which rely on passive diffusion for respiration, can benefit from higher oxygen levels as it allows for more efficient oxygen uptake, supporting larger body sizes. The increased oxygen availability likely facilitated the growth of these organisms to larger proportions.

Furthermore, the Pennsylvanian Period was characterized by the absence of large land-dwelling vertebrate predators. Without substantial predators, insects and amphibians faced fewer constraints on their size, allowing them to evolve and grow larger over time.

This lack of predation pressure provided an opportunity for these organisms to exploit ecological niches and evolve to larger sizes.

Together, the combination of higher oxygen levels and the absence of large land-dwelling vertebrate predators likely contributed to the impressive size of insects and amphibians during the Pennsylvanian Period.

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The spectrum observed from the Sun (or any star) would exhibit an absorption spectrum. This is because the outer gaseous atmosphere of the star absorbs specific wavelengths of light, resulting in dark absorption lines in the spectrum.

In the cooler, lower-density outer atmosphere, where white light from the star travels, some atoms or molecules in the atmosphere absorb photons with particular energy. In the spectrum, these absorptions show up as black lines at specific wavelengths. The specific set of absorption lines that each element or molecule generates results in a distinctive pattern that can be used to identify the elements that are present in the star's atmosphere.

The absorption spectrum offers insightful data on the chemical make-up and physical characteristics of the star. Astronomers can ascertain the elements present, their abundances, and other characteristics like the temperature, pressure, and velocity of the star's atmosphere by examining the absorption lines.

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Answer:

Ekaryote cells

an organism consisting of a cell or cells in which the genetic material is DNA in the form of chromosomes contained within a distinct nucleus. Eukaryotes include all living organisms other than the eubacteria and archaea.

prokaryote cells

microscopic single-celled organism which has neither a distinct nucleus with a membrane nor other specialized organelles, including the bacteria and cyanobacteria

Answer:

\(11.3\ \text{g/cm}^3\)

\(7.92\ \text{g/cm}^3\)

\(238.1\ \text{cm}^3\)

Explanation:

8) Volume of lead

\(V=4.5\times 5.2\times 6\\\Rightarrow V=140.4\ \text{cm}^3\)

m = Mass of block = 1587 g

Density is given by

\(\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{1587}{140.4}\\\Rightarrow \rho=11.3\ \text{g/cm}^3\)

Density of lead is \(11.3\ \text{g/cm}^3\)

9) Volume of lead = Volume of water displaced = (49.1-45.5) = 3.6 mL = \(3.6\ \text{cm}^3\)

m = Mass of iron = 28.5 g

Density is given by

\(\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{28.5}{3.6}\\\Rightarrow \rho=7.92\ \text{g/cm}^3\)

The density of iron is \(7.92\ \text{g/cm}^3\)

10) m = Mass of silver = 2500 g

\(\rho\) = Density of silver = \(10.5\ \text{g/cm}^3\)

Volume is given by

\(V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{2500}{10.5}\\\Rightarrow V=238.1\ \text{cm}^3\)

The volume of silver is \(238.1\ \text{cm}^3\)

Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

Answer:

A resistor of 160 ohms should be used to replace the current resistor

Explanation:

Applying,

Ohm's Law

V = IR................... Equation 1

Where V = Voltage of the battery, I = current, R = Resistance

From the question,

Given: I = 0.2 A, R = 120 ohms.

Substitute these values into equation 1

V = 0.2×120

V = 24 V

IF a current of 0.15 A is required,

make R the subject of equation 1

R = V/I............... Equation 2

Given: V = 24 V, I = 0.15 A

Substitute these values into equation 2

R = 24/0.15

R = 160 ohms.

Hence a resistor of 160 ohms should be used to replace the current resistor

Answer:

Socialist

Explanation:

Answer: socialist

Explanation: A socialist form of government is where the goods and services are equally shared, and the political power is distributed among the people.

Answer:

a. 26.7 cm. b. 11.4 cm.

Explanation:

a. We know h'/h = d'/d where h' = image height = + 2 cm (since it is a real image), h = object height = + 1.5 cm, d' = image distance from mirror and d = object distance from mirror = 20 cm

So, from h'/h = d'/d

d = h'd/h

= 2 cm × 20 cm/1.5 cm

= 40/1.5 cm

=  26.67 cm

≅ 26.7 cm

The position of the image is 26.7 cm from the mirror

b. Using the mirror formula

1/d + 1/d' = 1/f where d = object distance from mirror = + 20 cm, d' = image distance from mirror = + 26.7 cm (its positive since its a real image) and f = focal length of mirror.

So, 1/d + 1/d' = 1/f

⇒ f = dd'/(d + d')

= 20 cm × 26.7 cm/(20 cm + 26.7 cm)

= 534/46.7

= 11.43 cm

≅ 11.4 cm

The focal length of the mirror is 11.4 cm

The calculated data aligns with the general rule that a normal soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil.

For the given soil sample, the following properties can be calculated: 1. Bulk density = 0.75 g/cm3, 2. Particle density = 0.45 g/cm3, 3. Percentage of total porosity = 40%, 4. Percentage of water at field capacity (mass basis) = 25%, 5.

Percentage of water at field capacity (volume basis) = 20%, 6. Total volume of pores = 220 cm3, 7. Total volume of water at field capacity = 100 cm3, 8. Percentage of pore space filled with water at field capacity = 45%, 9.

Total volume of air spaces at field capacity = 120 cm3, 10. Percentage of pore space filled with air at field capacity = 55%. The calculated data agrees with the general rule that a soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil.

Bulk density is calculated by dividing the mass of dry soil by its volume, which gives a value of 0.75 g/cm3.

Particle density is calculated by dividing the mass of solid particles by their volume, resulting in a value of 0.45 g/cm3.

Percentage of total porosity is obtained by subtracting the particle density from the bulk density, dividing the result by the bulk density, and multiplying by 100, resulting in 40%.

Percentage of water in the soil at field capacity (mass basis) is calculated by dividing the mass of water by the mass of dry soil, which gives 25%.

Percentage of water in the soil at field capacity (volume basis) is obtained by dividing the volume of water by the total volume of soil, resulting in 20%.

Total volume of pores is calculated by subtracting the volume of solid particles from the total volume of soil, resulting in 220 cm3.

Total volume of water in the pores at field capacity is given as 100 cm3.

Percentage of the total pore space filled with water at field capacity is calculated by dividing the volume of water by the total volume of pores and multiplying by 100, resulting in 45%.

Total volume of air spaces at field capacity is obtained by subtracting the volume of water from the total volume of pores, resulting in 120 cm3.

Percentage of the total pore space filled with air at field capacity is calculated by dividing the volume of air by the total volume of pores and multiplying by 100, resulting in 55%.

The calculated data aligns with the general rule that a normal soil near field capacity contains approximately 50% water and 50% air in the total pore space of the soil, as the percentages obtained for water and air are close to this expected distribution.

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Answer:

WDYM?

Explanation:

Answer: Get therapy (not like actually lol but like you know...)

Explanation: Listen to calming music or your favorite music, read, draw, color, literally anything that makes you calm

Answer:

The  value is    \(v = 47 \  m/s\)

Explanation:

From the question we are told that

   The initial  speed of the roller coaster is \(u =  13 \  m/s\)

    The  length of the hill is  \(l   = 400 \  m\)

    The  acceleration of the  roller coaster is \(a=4.0 \ m/s^2\)

Generally the acceleration is mathematically represented as

      \(a =  \frac{ v - u}{ t_f -  t_i }\)

Here  \(t_i\) is the initial time which is equal to zero

         \(v_f\) is the final velocity which is mathematically represented as

          \(v_f  =  \frac{d}{ t_f}\)

So  

     \(a =  \frac{ \frac{d}{d_f}  - u }{ t_f - t_i}\)

     \(4 = \frac{\frac{400}{ t_f}  - 13}{t_f - 0}\)

      \(4 =  \frac{400 - 13t_f}{ t_f} *  \frac{1}{t_f}\)

     \(4t_f ^2  +13f  + 400 =\)

Solving this using quadratic formula we obtain

    \(t_f =  8.5 \ s\)

     \(t_f =  -11.8 \ s\)

Generally  time cannot be negative so

       \(t_f =  8.5 \ s\)

Generally the  final velocity is mathematically represented as

         \(v = \frac{400}{8.5}\)

         \(v = 47 \  m/s\)

Answer:58 m/s

Explanation:

That’s what it says