a 7.5 water treatment plant operates at its maximum capacity for one week. how many cubic feet of water were processed

Answers

Answer 1

The 7.5 water treatment plant processed approximately 6,997,333 cubic feet of water during its one week of operation.

To determine the cubic feet of water processed by a 7.5 water treatment plant operating at its maximum capacity for one week, we need to use the following formula:

Cubic feet of water = flow rate (gallons per minute) × time (minutes) ÷ 7.48

First, we need to convert the capacity of the plant to gallons per minute. Since there are 60 minutes in an hour and 24 hours in a day, the plant operates for a total of:

7 days × 24 hours per day × 60 minutes per hour = 10,080 minutes

So, the flow rate of the plant is:

7.5 million gallons per day ÷ 24 hours per day ÷ 60 minutes per hour = 5,208.3 gallons per minute

Using the formula, we get:

Cubic feet of water = 5,208.3 gallons per minute × 10,080 minutes ÷ 7.48 = 6,997,333 cubic feet

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Related Questions

the heating element of a hair dryer dissipates 1500 w when connected to a 160 v outlet. part a what is its resistance? express your answer with the appropriate units.

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The heating element of a hair dryer dissipates 1500 w when connected to a 160 v outlet. The resistance of the heating element of a hair dryer is 17.07 ohms.


We can use Ohm's law to find the resistance of the heating element of a hair dryer. Ohm's law states that the resistance of a conductor is equal to the voltage across it divided by the current flowing through it.
In this case, we know that the power dissipated by the heating element is 1500 W and the voltage across it is 160 V. We can use the formula for power, which is power = voltage x current, to find the current flowing through the heating element.
1500 W = 160 V x current
Solving for current, we get:
Current = 9.375 A
Now we can use Ohm's law to find the resistance:
Resistance = Voltage / Current
Resistance = 160 V / 9.375 A
Resistance = 17.07 ohms (rounded to two decimal places)  

Therefore, the resistance of the heating element of a hair dryer is 17.07 ohms, when connected to a 160 V outlet.

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An object moving in a straight line at a constant speed (a=0) is in

Answers

Answer:

uniform motion

Explanation:

Uniform motion is defined as the motion of an object in which the object travels in a straight line and its velocity remains constant along that line as it covers equal distances in equal intervals of time.

at what speed do a bicycle and its rider, with a combined mass of 90 kg , have the same momentum as a 1500 kg car traveling at 6.0 m/s ? express your answer to two significant figures and include the appropriate units.

Answers

100 m/s speed do a bicycle and its rider, with a combined mass of 90 kg , have the same momentum as a 1500 kg car traveling at 6.0 m/s

To find the speed at which the bicycle and its rider have the same momentum as the car, we can use the momentum formula:
momentum = mass × speed
First, let's find the momentum of the car:
momentum car = (1500 kg) × (6.0 m/s) = 9000 kg m/s
Now we want the bicycle and its rider to have the same momentum:
momentum bicycle = momentum car = 9000 kg m/s
We can now use the mass of the bicycle and its rider (90 kg) to find the speed at which they have the same momentum:
speed bicycle = momentum bicycle / mass bicycle
speed bicycle = 9000 kg m/s

90 kg = 100 m/s
Therefore, the bicycle and its rider need to travel at a speed of 100 m/s to have the same momentum as the car.

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With a 1/16" ball penetrator and a penetration depth of 0.082 mm, this makes

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It appears that a penetration depth of 0.082mm would result in a superficial Rockwell hardness value of approximately 18, using a 1/16" ball penetrator and the corresponding test load.

However, as you mentioned, there are various superficial Rockwell scales that use different combinations of penetrators and test loads.

It's important to use the appropriate scale for the material being tested and to follow standardized testing procedures to ensure accurate and reproducible results.

The Rockwell hardness test requires a specific testing procedure, including the use of a calibrated hardness tester, a specific type of penetrator, and standardized testing conditions.

The hardness values obtained from this test are dependent on the material being tested, and cannot be determined solely based on penetration depth.

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correct form of question would be

With a diamond or ball penetrator and a penetration depth of 0.082mm this makes 100 – 0.082/0.001 = 18 superficial Rockwell.

Due to the different combinations of penetrators and test loads, there is a great number of superficial Rockwell scales, whichare labelled with different letters. The respective letter is also preceded by a number which indicates the total load used in the test (see Table 2)

Penetrator- F=441,3N / F=294,2N / F=147,1N /

Diamond Cone = 45 N / 30 N / 15 N

Ball 1/16"1,5875mm= 45 T / 30 T / 15 T

Ball 1/8"* = 45 W / 30 W / 15 W

Ball 1/4"* = 45 X / 30 X / 15 X

Ball 1/2"* = 45 Y / 30 Y / 15 Y

a circular loop of wire lies flat on a level table top. a bar magnet is held stationary above the circular loop with its north pole point downward. as viewed from above, in what direction does the induced current flow in the loop of wire? a circular loop of wire lies flat on a level table top. a bar magnet is held stationary above the circular loop with its north pole point downward. as viewed from above, in what direction does the induced current flow in the loop of wire? an induced current flows clockwise in the loop of wire. an induced current flows counterclockwise in the loop of wire. no current is induced in the loop of wire. the direction of the induced current cannot be determined from the given information.

Answers

The induced current flows clockwise in the loop of wire. When a bar magnet is held stationary above a circular loop of wire with its north pole pointing downward, as viewed from above, the induced current in the loop of wire will flow counter clockwise.

This is due to Lenz's law, which states that the direction of the induced current flows clockwise in the loop of wire. current will be such that it opposes the change in magnetic flux that is producing it. In this case, the counterclockwise current creates a magnetic field opposing the downward magnetic field of the north pole of the bar magnet. the direction of the magnetic field will be down in the plane at the center. The magnetic field produced by a current-carrying wire loop will be in a single direction at the center. The direction of magnetic field at the center of a current-carrying circular loop is perpendicular to the plane of the loop.

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In a Camber airfoil, the Center of Pressure (CP)a.) moves to the rear of the wing at low AOAb.) moves backward as AOA increases c.) moves forward as AOA increases d.) both a and c

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In a cambered airfoil, the Center of Pressure (CP) c.) moves forward as Angle of Attack (AOA) increases.

In a Camber airfoil, the centre of pressure (CP) is the point on the airfoil where the total aerodynamic force can be considered to act. The position of the CP changes with the angle of attack (AOA) of the airfoil.

At low AOA, the CP is located towards the front of the airfoil. As the AOA increases, the CP moves towards the rear of the airfoil. However, as the AOA continues to increase, the CP eventually moves towards the front of the airfoil again.

This means that the CP moves forward as AOA increases in a Camber airfoil. Therefore, the correct answer is c) moves forward as AOA increases.

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10-year-old Sarah stands on a skateboard. Her older brother Jack starts pushing her backward and she starts speeding up. The force of Jack on Sarah isA. greater than the force of Sarah on Jack.B. equal to the force of Sarah on Jack.C. less than the force of Sarah on Jack.

Answers

When Jack pushes Sarah on the skateboard, the force he exerts on her is equal and opposite to the force Sarah exerts on Jack. This is known as Newton's Third Law of Motion.

However, the acceleration of Sarah and the skateboard depends on the net force acting on the system. In this case, the net force acting on the system is the force of Jack on Sarah minus the force of friction between the skateboard and the ground. If the force of Jack on Sarah is greater than the force of friction, then the net force is in the backward direction and Sarah speeds up. Therefore, the answer to the question is A. The force of Jack on Sarah is greater than the force of Sarah on Jack, but this does not mean that Jack is stronger than Sarah. It simply means that he is exerting a greater force on her in this particular situation.When Jack pushes Sarah on the skateboard, the force he exerts on her is equal and opposite to the force Sarah exerts on Jack. This is known as Newton's Third Law of Motion.

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a large block of ice is being pushed on a frozen pond by layla and nadia. layla pushes the block to the right with a force of 40 n and nadia pushes the block to the left with a force of 70 n. what is the net force on the block of ice?

Answers

30N is the net force on the block of ice for a large block of ice is being pushed on a frozen pond by layla and nadia

The net force on the block of ice is the result of combining the forces pushing in opposite directions. Layla pushes to the right with a force of 40 N, while Nadia pushes to the left with a force of 70 N. To find the net force, we need to subtract the smaller force from the larger force, since they are in opposite directions.

When forces are in balance, there is no net force, hence there is no net force.

There is either no movement or steady movement when something is in balance.

Equal in size and directed in the opposite direction, balanced forces are. When forces are evenly distributed, motion remains unchanged.
So, the net force on the block of ice is:
70 N (Nadia's force) - 40 N (Layla's force) = 30 N
Therefore, the net force on the block of ice is 30 N to the left, since Nadia's force is greater than Layla's force.

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Question 20
Which one of the following is the greatest genetic concern in terms of exposure to ionizing radiation?
a. Mutations that accumulate in the gonads
b. Blood-borne infections
c. Increased frequency of dominant gene mutations
d. Chromosomal damage in adolescents

Answers

The greatest genetic concern in terms of exposure to ionizing radiation is mutations that accumulate in the gonads. Option a is correct.

Ionizing radiation has the potential to cause mutations in DNA, which can lead to genetic changes in offspring. Mutations that accumulate in the gonads, which are the cells that produce sperm and eggs, have the potential to be passed down through generations. This can lead to an increased risk of genetic disorders, such as cancer or birth defects, in future offspring.

While other forms of genetic damage can also occur with exposure to ionizing radiation, such as chromosomal damage, mutations in the gonads have the greatest potential impact on future generations. Therefore, it is important to limit exposure to ionizing radiation, particularly during reproductive years. Option a is correct.

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Ch19: For which two out of the following 4 processes entropy of the system increase (ΔS>0)?I. Condensing water vaporII. Heating hydrogen gas from 60° C to 80° CIII. Forming sucrose crystals from a supersaturated solutionIV. Subliming dry ice

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The two processes for which the entropy of the system increases (ΔS>0) are I. Condensing water vapor and IV. Subliming dry ice.

In both these processes, the system undergoes a change from a less ordered state to a more ordered state, which leads to an increase in entropy. In contrast, in process II. Heating hydrogen gas from 60° C to 80° C, the system becomes more disordered as the molecules move faster and the distribution of energy becomes more random, leading to a decrease in entropy. Similarly, in process III. Forming sucrose crystals from a supersaturated solution, the system becomes more ordered as the molecules come together in a specific arrangement, leading to a decrease in entropy.

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Galileo discovered that when air resistance can be neglected, all objects fall with the same _______.

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Galileo discovered that when air resistance can be neglected, all objects fall with the same acceleration, which is approximately 9.81 meters per second squared (m/s^2) near the surface of the Earth.

Galileo's discovery of the universality of free fall was a significant contribution to the development of physics and mechanics. Prior to his experiments, it was commonly believed that heavier objects fell faster than lighter objects. However, Galileo demonstrated through his experiments that this was not the case, and that all objects fall with the same acceleration in the absence of air resistance.

Galileo's experiments involved rolling balls of different masses down inclined planes and measuring their motion. By carefully controlling the angle of the incline and the distance traveled by the balls, Galileo was able to show that the acceleration of the balls was independent of their mass. He also observed that the acceleration due to gravity was constant, and that it did not depend on the velocity or direction of motion.

Galileo's discovery of the universality of free fall laid the foundation for the development of classical mechanics, which is the branch of physics that deals with the motion of objects under the influence of external forces. It also played a crucial role in the development of the theory of gravitation by Isaac Newton, who used Galileo's work as a starting point to develop his laws of motion and the law of universal gravitation. Today, the principle of the universality of free fall is a fundamental concept in physics and is used in a wide range of applications, including in the design of spacecraft and in the study of the structure and evolution of the universe.

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a filter is 18 feet wide and 20 feet long. The maximum filtration rate allowed for this unit is 6.0 gpm/ft2. what is the highest flow rate that this filter can process

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The highest flow rate that this filter can process is 2,160 gpm.

The maximum filtration rate allowed for a filter depends on the filter's size and the permissible flow rate per unit area of the filter. In this case, the filter has an area of 18 feet x 20 feet = 360 square feet.

The surface area of the filter is:

18 feet x 20 feet = 360 square feet

To determine the maximum flow rate, we need to multiply the surface area by the maximum filtration rate allowed:

360 square feet x 6.0 gpm/ft2 = 2,160 gallons per minute (gpm)

Therefore, the highest flow rate that this filter can process is 2,160 gpm.

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Equipment rated 100 ampere or less must have the conductor sized no smaller than the 60 degree column of Table 310-15(B)(16). Equipment rated at more than 100 ampere must have the conductors sized no smaller than for the 75 degree column of Table 310-15(B)(16)(True/False)

Answers

True. According to the National Electric Code (NEC), equipment rated at 100 amperes or less must have conductors sized no smaller than the 60-degree column of Table 310-15(B)(16).

This is because smaller conductors can overheat and cause damage to the equipment or even create a fire hazard. On the other hand, equipment rated at more than 100 amperes requires conductors sized no smaller than the 75-degree column of Table 310-15(B)(16). This is because larger equipment requires more power and larger conductors can handle the increased current without overheating.

It is important to note that these sizing requirements are minimum standards and it is always recommended to consult a licensed electrician to ensure the proper sizing and installation of conductors for your specific equipment. Failure to properly size conductors can result in equipment damage, personal injury, or even death.

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A puck is sliding on the ice with 25 J of kinetic energy. After 3 seconds, the puck comes to a stop and has no more kinetic energy. Find the work done on the puck.

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The work done on the puck is -25 J, after  3 seconds, the puck comes to a stop and has no more kinetic energy.

What is work done?

Work done is described  as the amount of force needed to move an object a certain distance.

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.

The change in kinetic energy is:

ΔK = Kf - Ki = 0 - 25 = -25 J

Note that the  negative sign indicates that the kinetic energy of the puck decreased.  

W = ΔK = -25 J

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a straight length of wire carries a current of 50 a in a region where a uniform magnetic field has a magnitude of 0.100 t. the field is directed at an angle of 30 degrees away from the wire. there is a force on the wire measured to be 10n. how long is the wire?

Answers

The equation F = BILsinθ, where F is the force on the wire, B is the magnitude of the magnetic field, I am the current in the wire, L is the length of the wire, and θ is the angle between the magnetic field and the wire. Plugging in the given values, we get10 = 50 Lsin30Simplifying this equation, we get. L = 4 meters Therefore, the length of the wire is 4 meters.


The solve this problem, we will use the formula for the magnetic force on a current-carrying wire.F = I * L * B * sin(θ)
where F is the force, I am the current, L is the length of the wire, B is the magnitude of the magnetic field, and θ is the angle between the magnetic field and the direction of the current. We are given the following information = 10 I = 50 A
B = 0.100 T θ = 30 degrees First, we need to convert the angle to radians θ = 30 degrees × π radians / 180 degrees = π/6 radians Now, we can plug the given values into the formula and solve for L10 N = 50 A * L * 0.100 T * sin(π/6) Divide both sides by 50 A * 0.100 T * sinπ/6 L = 10 N / 50 A * 0.100 T * sinπ/6 Calculate the length L ≈ 3.464 m So, the length of the wire is approximately 3.464 meters.

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Two slits spaced 0. 0720 mm apart are 0. 800 m from a screen. Coherent light of wavelength λ passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is 3. 00 mm. The intensity at the peak of the central maximum is 0. 0700 W/m2. What is the intensity at point on the screen that is 2. 00 mm from the center of the central maximum? What is the intensity at point on the screen that is 1. 50 mm from the center of the central maximum?

Answers

The intensity at a point on the screen 2.00 mm from the center of the central maximum is approximately 0.034 W/m². The intensity at a point on the screen 1.50 mm from the center of the central maximum is approximately 0.024 W/m².

I = Imax cos² (πd sin θ / λ),

where Imax is the intensity at the center of the interference pattern, d is the distance between the two slits, θ is the angle between the line connecting the point on the screen to the center of the interference pattern and the line perpendicular to the screen, and λ is the wavelength of the light.

To find the angle θ, we can use the small angle approximation:

sin θ ≈ θ ≈ y/L,

where y is the distance from the center of the interference pattern to the point on the screen, and L is the distance between the slits and the screen.

We are given d = 0.0720 mm, λ = unknown, L = 0.800 m, Imax = 0.0700 W/m², and the distance from the center of the central maximum to the first minimum y = 3.00 mm.

Using the given distance y, we can find the value of sin θ:

y/L = sin θ,

3.00 mm / 0.800 m = sin θ,

sin θ = 0.00375.

Now we can solve for the wavelength λ:

Imax cos² (πd sin θ / λ) = I,

0.0700 W/m² cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I,

cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I / 0.0700 W/m²,

π(0.0720 × 10⁻³ m)(0.00375) / λ = ± cos⁻¹ (√(I / 0.0700 W/m²)),

λ = π(0.0720 × 10⁻³ m)(0.00375) / cos⁻¹√(I / 0.0700 W/m²)),

λ = 5.70 × 10⁻⁷ m (for the positive root).

Now we can find the intensities at the given distances from the center of the central maximum.

For y = 2.00 mm:

sin θ = y/L = 2.00 mm / 0.800 m = 0.00250,

I = Imax cos² (πd sin θ / λ)

I = 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.00250) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².

So the intensity at a point on the screen 2.00 mm from the center of the central maximum would be approximately 0.034 W/m².

For y = 1.50 mm:

sin θ = y/L = 1.50 mm / 0.800 m = 0.001875,

I = Imax cos² (πd sin θ / λ)

I= 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.001875) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².

I ≈ 0.024 W/m².

So the intensity at a point on the screen 1.50 mm from the center of the central maximum would be approximately 0.024 W/m².

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What kind of expansion do ideal gases undergo?

Answers

Answer:

Isothermal Expansion

Explanation:

This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings

Question 11
Perhaps our single most significant source of radiation exposure is:
a. Radon
b. Cosmic radiation
c. Alpha particles
d. Gamma rays

Answers

Perhaps our single most significant source of radiation exposure is, (B).  Cosmic radiation is correct option.

High-energy particles that come from the universe and have the ability to enter the atmosphere of the Earth are referred to as cosmic radiation. Protons, gamma rays, and other radiation types fall within this category. For people who frequently fly, such as airline crew members and frequent travelers, as well as astronauts who spend a lot of time in space, cosmic radiation is a substantial source of radiation exposure. People who work in environments with greater elevations, such as mountain climbers and pilots, may potentially be exposed to cosmic radiation.

However, based on variables like height, latitude, and solar activity, the exposure to cosmic radiation can differ. Cosmic radiation is regarded as one of the most major sources of radiation exposure, along with radon, alpha particles, and gamma rays.

Therefore, the correct option is (b).

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(Table 352-30) 1 inch rigid nonmetallic conduit must be supported every _____ feet.

Answers

According to Table 352.30 of the National Electrical Code (NEC), 1 inch rigid nonmetallic conduit must be supported at intervals not exceeding 10 feet.

The National Electrical Code (NEC) is a standard that provides guidelines for the safe installation and use of electrical wiring and equipment in the United States. The NEC is updated every three years to incorporate new technology, safety advancements, and other changes in the electrical industry.

Table 352.30 of the NEC specifies the maximum spacing between supports for rigid nonmetallic conduit. The spacing requirements are based on the diameter of the conduit, the weight of the conduit and the contents it carries, and the temperature of the surrounding environment.

In the case of 1 inch rigid nonmetallic conduit, Table 352.30 specifies that the conduit must be supported at intervals not exceeding 10 feet. This means that there must be a support bracket or hanger installed at least every 10 feet along the length of the conduit to prevent it from sagging or breaking under its own weight.

Proper support of conduit is important for ensuring that electrical systems are safe and reliable. Unsupported conduit can become damaged, causing electrical faults, shorts, or even fires. By following the NEC guidelines for conduit support, electricians and contractors can ensure that electrical systems are installed and maintained safely and effectively.

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can someone do my physics test please? 100P

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your answer is c bro but if not then it's b

because you just choose c or b either one should work if you right it down right

(362-28) All cut ends of electrical nonmetallic tubing shall be trimmed inside and ____ to remove rough edges.

Answers

All cut ends of electrical nonmetallic tubing shall be trimmed inside and deburred to remove rough edges.

When working with electrical nonmetallic tubing, it is important to ensure that all cut ends are properly trimmed and deburred.

Deburring is the process of removing any rough edges or burrs that may be present on the cut end of the tubing. This is important because rough edges can damage wires or cables that are being pulled through the tubing, or can cause injury to the person handling the tubing. To deburr the tubing, a deburring tool or file can be used to smooth out the edges of the cut. Once the edges are smooth, the tubing can be safely used for electrical installations.

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Question 63
Which one of the following is probably least susceptible to microwave induced injury?
a. Eyes
b. Urinary bladder
c. Gastrointestinal test
d. liver

Answers

The urinary bladder is probably the least susceptible to microwave-induced injury among the given options. Microwave energy has a higher chance of affecting tissues with higher water content, such as the eyes, gastrointestinal tract, and liver.

The urinary bladder, on the other hand, has less water content and is less likely to be affected by microwave radiation.The eyes, gastrointestinal tract, and liver are organs that contain tissues with higher water content and are therefore more susceptible to microwave-induced injury, as microwaves can be absorbed by water molecules and generate heat. However, the urinary bladder is a muscular organ that stores urine and does not contain as much water content compared to other organs, making it less likely to be as susceptible to microwave-induced injury. Nonetheless, it's important to note that microwave radiation should be used with caution and in accordance with safety guidelines to minimize potential risks to all organs and tissues in the body.

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at steady state, a 1 m thick wall has a temperature difference (between the left and right surfaces) of capital delta t equals 5 k. if the wall's thermal conductivity is k equals 10 space fraction numerator w over denominator m k end fraction, what is the heat flux across this wall? (assume 1d conduction heat transfer.)

Answers

The magnitude of the heat flux across the wall is [tex]50 W/m^2[/tex] at steady state, a 1 m thick wall has a temperature difference.

To calculate the heat flux across the 1 m thick wall at steady state, we can use Fourier's Law of Heat Conduction. The formula is: q = -k * (dT/dx)
where q is the heat flux ([tex]W/m^2[/tex]), k is the thermal conductivity (10 W/m·K), dT is the temperature difference (5 K), and dx is the thickness of the wall (1 m).
Now, plug in the given values:
q = -10 * (5 K / 1 m)
q = [tex]-50 W/m^2[/tex]
Since we're considering 1D conduction heat transfer and the heat flux is negative, it means the heat is transferred from the higher temperature side to the lower temperature side.

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Place the following in sequence: A) Hubble makes his discoveries; B) Cosmic background radiation is first detected; C) Lemaitre proposes his theory

Answers

The correct sequence is:

C) Lemaitre proposes his theory --> A) Hubble makes his discoveries --> B) Cosmic background radiation is first detected.

Lemaitre proposed his theory of the expanding universe, which later became known as the Big Bang theory, in the 1920s. Hubble's observations in the 1920s and 1930s provided evidence for the expansion of the universe and the relationship between distance and recession velocity for galaxies.

The cosmic microwave background radiation, which is the afterglow of the Big Bang, was first detected in 1964 by Arno Penzias and Robert Wilson.

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The average distance from Earth to the sun is 9.3 × 107 miles. How many kilometers isthis?A) 1.5 × 108 km D) 1.7 × 10-8 kmB) 1.5 × 105 km E) 1.5 × 1011 kmC) 5.6 × 107 km

Answers

The  distance from Earth to the sun is approximately 1.5 x 10^8 kilometers.

To convert miles to kilometers, we can use the conversion factor 1 mile = 1.609344 kilometers.

So, to find the distance from Earth to the sun in kilometers, we can multiply the given distance in miles by the conversion factor:

d (km) = 9.3 x 10^7 miles x 1.609344 km/mile
d (km) = 1.496 x 10^8 km

Therefore, the distance from Earth to the sun is approximately 1.5 x 10^8 kilometers.

The closest answer choice is A) 1.5 x 10^8 km, which is the correct answer.

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a mass m is attached to an ideal massless spring. when this system is set in motion, it has a period t . what is the period if the mass is doubled to 2 m ?

Answers

The period of a mass-spring system is given by T = 2π√(m/k), where m is the mass of the object attached to the spring and k is the spring constant. Since the spring is ideal and massless, k remains constant when the mass is changed.

Using the equation T = 2π√(m/k), we can find the period when the mass is doubled. Let's call the new period T2 and the original period T1.

T1 = 2π√(m/k)
T2 = 2π√(2m/k)

To find the relationship between T1 and T2, we can take the ratio of the two equations:

T2/T1 = √(2m/k)/√(m/k)
T2/T1 = √(2)

Therefore, when the mass is doubled, the period of the system increases by a factor of √(2).

The period of the mass-spring system will increase by a factor of √(2) when the mass is doubled.

We can conclude that increasing the mass of an ideal massless spring system will increase its period. In this case, doubling the mass will increase the period by a factor of √(2).

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a man pushes a 15 kg block to the west with an acceleration of 0.1 m/s/s. using newton's second law of motion, what is the total force used?

Answers

The total force used by the man to push the block to the west is 1.5 N (Newtons).

Hi! I'd be happy to help you with your question. To find the total force used by a man pushing a 15 kg block to the west with an acceleration of 0.1 m/s², we can use Newton's second law of motion.



Newton's second law states that Force (F) equals mass (m) multiplied by acceleration (a), or F = m × a.

Step 1: Identify the mass (m) and acceleration (a).
Mass (m) = 15 kg
Acceleration (a) = 0.1 m/s²

Step 2: Apply Newton's second law of motion formula.
F = m × a

Step 3: Substitute the values and calculate the force.
F = 15 kg × 0.1 m/s²

F = 1.5 N

So, the total force used by the man to push the block to the west is 1.5 N (Newtons).

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The man applied 1.5 N (Newtons) of force in total to move the block in the west.

Hi! I'd be delighted to answer your query. Newton's second equation of motion can be used to calculate the total force applied by a man pushing a 15 kg block with an acceleration of 0.1 m/s2 to the west.

According to Newton's second law, force (F) is equal to mass (m) times acceleration (a), or F = m a.

Determine the mass (m) and acceleration (a) in step 1.

Weight (m) = 15 kilogramme

0.1 m/s2 is the acceleration (a).

Step 2: Use the calculus for Newton's second law of motion.

F = m × a

Step 3: Calculate the force by substituting the values.

F = 15 kg × 0.1 m/s²

F = 1.5 N

The man utilised 1.5 N (Newtons) of force in total to push the block in a westward direction.

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A company is developing a system which can heat up and melt ice on roads in the winter. This system is called 'energy storage'.
During the summer, the black surface of the road will heat up in the sunshine.
This energy will be stored in a large amount of soil deep under the road surface. Pipes will run through the soil. In winter, cold water entering the pipes will be warmed and brought to the surface to melt ice.
The system could work well because the road surface is black.
Suggest why.​

Answers

Answer:

Explanation:

The color of a surface can affect how much solar energy it absorbs or reflects. Black surfaces absorb more solar radiation than lighter-colored surfaces, which reflects more of the sunlight. This is because black surfaces have a lower albedo, which is a measure of how much light a surface reflects.

When a black surface, such as a road, is exposed to sunlight, it absorbs more of the sun's energy, which is then converted into heat. This heat is then transferred to the soil underneath the road surface, where it can be stored for later use. During winter, the pipes running through the soil can be used to extract this stored heat and warm the cold water, which can then be used to melt ice on the road surface.

Therefore, the black surface of the road is beneficial for this energy storage system because it allows for more efficient absorption of solar energy, which can then be used to heat the soil and melt ice during the winter months.

44. What is the magnitude of the centripetal acceleration of a point on the rim of the grindstone?
A) zero m/s2
B) 0.5 m/s2
C) 1.0 m/s2
D) 2.0 m/s2
E) 4.0 m/s2

Answers

The centripetal acceleration of a point on the rim of a grindstone is determined by the formula a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius of the circle. In this case, we assume that the grindstone is rotating at a constant speed, which means that the velocity of any point on the rim is constant.

Therefore, the magnitude of the centripetal acceleration depends only on the radius of the circle.Since the question does not provide any information about the radius of the grindstone, we cannot determine the magnitude of the centripetal acceleration. However, we can conclude that options A and B are incorrect because the centripetal acceleration cannot be zero if the grindstone is rotating, and it cannot be less than 0.5 m/s^2 because that is the minimum acceleration required to keep an object moving in a circle.
Therefore, the correct answer must be either C, D, or E, depending on the radius of the grindstone. If the radius is relatively small, the acceleration will be closer to 4.0 m/s^2 (option E), while if the radius is relatively large, the acceleration will be closer to 1.0 m/s^2 (option C). The centripetal acceleration of a point on the rim of a grindstone is determined by the formula a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius of the circle. In this case, we assume that the grindstone is rotating at a constant speed, which means that the velocity of any point on the rim is constant.
In summary, the magnitude of the centripetal acceleration of a point on the rim of a grindstone depends on the radius of the circle and is given by the formula a = v^2/r. We cannot determine the exact answer without knowing the radius of the grindstone, but we can eliminate options A and B as incorrect.

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In an oscillating LC circuit, the total stored energy is U and the maximum charge on the capacitor is Q. When the charge on the capacitor is Q/2, the energy stored in the inductor is closest to:

Answers

In an oscillating LC circuit, the total stored energy is divided between the capacitor and the inductor. When the charge on the capacitor is Q/2, the energy stored in the capacitor is also half of its maximum value.

Therefore, the energy stored in the inductor is also closest to half of its maximum value, which is U/2. This is because the energy oscillates back and forth between the capacitor and the inductor, with the charge on the capacitor and the current in the inductor both reaching their maximum values at opposite times during each cycle. So, when the charge on the capacitor is at its midpoint, the energy stored in the inductor is also at its midpoint.

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