if the spring pushes the block up the incline, what distance, l in meters, will the block travel before coming to rest? the spring remains attached to both the block and the fixed wall throughout its motion.

Answer:

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Explanation:

idont know

Answer:

P = 2 pi (L / g)^1/2

Let 1 represent earth and 2 the moon

P2^2 / P1 ^2 = L2 g1 / (L1 g2)       dividing equations

L2 = (P2 / P1)^2 * g2 / g1 * L1

L2 = 1 * 1.62 / 9.80 * 1.9 = .314 m

a) The average annual evaporation from the catchment is approximately 180.29 mm/year.

b) The exact value of evapotranspiration from the irrigated area cannot be calculated due to missing information.

a) Average annual evaporation from the catchment in mm/year:

First, we calculate the total annual rainfall that is collected by the catchment area:

800,000,000 m² × 0.2 m = 160,000,000 m³/year

Since this is the only source of water for the catchment, the total amount of water available to the catchment area per year will be 160,000,000 m³/year.

We know that the average annual discharge at the outlet of a catchment is 0.5 m³/s, and since there are 31,536,000 seconds in a year, we can calculate the total volume of water that is discharged per year:

0.5 m³/s × 31,536,000 s = 15,768,000 m³/year

So, the total volume of water that is lost through evaporation per year will be:

160,000,000 m³/year - 15,768,000 m³/year = 144,232,000 m³/year

To convert this into millimeters, we need to divide this value by the area of the catchment in square meters, and then multiply by 1000 (since 1 m = 1000 mm):

144,232,000 m³/year ÷ 800,000,000 m² × 1000 mm/m = 180.29 mm/year

Therefore, the average annual evaporation from the catchment is approximately 180.29 mm/year.

b) Evapotranspiration from the irrigated area in mm/year:

Since we know that the size of the irrigated area is 10 km² = 10,000,000 m², we can calculate the total volume of water that is used for irrigation each year by multiplying this area by the amount of discharge that is lost as a result of the irrigation project:

10,000,000 m² × (0.5 m³/s - 0.175 m³/s) × 31,536,000 s/year = 4,422,480,000 m³/year

To calculate the amount of water that is lost through evapotranspiration from the irrigated area, we need to know the crop coefficient and the reference evapotranspiration (ET0) for the area. However, since this information is not provided in the question, we cannot calculate the exact value of evapotranspiration from the irrigated area.

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Answer:

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Dont know the answer but have a good day :)

Answer:

An ideal "Carnot" engine

E = (Q2 - Q1) / Q2 or 1 - Q1 / Q2

Q1 (heat wasted) would have to be zero for 100% efficiency

Answer:

George Washington

Explanation:

Answer:

well for me

Explanation:

It was George Washington

Answer: find the answer in the explanation as kinetic energy converts to potential energy.

Explanation:

Before the truck driver sees a dog running into the road, The mechanical energy state of the truck will be kinetic energy at maximum.

Immediately he applied the brakes, the mechanical energy of the truck will be combination of kinetic energy and potential energy.

The kinetic energy will gradually decrease as potential energy continue to increase till it reaches maximum potential energy.

The truck will come to a stop at maximum potential energy

Answer:

Hello your question lacks some very vital information hence I will give you a general equation where you can input the missing values and get your answer

answer :  mass of Bananas ( m ) = \(\frac{A^2}{V^2} * K\)

To calculate for weight of Bananas  = m * g      where g = 9.8

Explanation:

Given an amplitude ( A )= 29.8981

Spring constant = x ( unknown )

maximum speed of Bananas( v ) = wA  ( unknown )

question : calculate the weight of the bananas

First  step : calculate the period of oscillation

T = \(\frac{2\pi }{vA}\) --- ( 1 )

next step : express T in terms of mass ( m ) of the bananas

= \(2\pi \sqrt{\frac{m}{k} } = \frac{2\pi }{vA}\)

∴ m/k = A^2 / v^2

hence  mass of Bananas ( m ) = \(\frac{A^2}{V^2} * K\)

To calculate for weight = m * g

Explanation:

We have,

Initial speed of a softball is 3.89 m/s

Deceleration of the ball is -1.44 m/s²

Distance covered by the ball is 4.8 m

It is required to find the time that the ball take when it slides. Let t is time. Using second equation of motion :

\(h=ut+\dfrac{1}{2}at^2\)

Putting all the values we get :

\(4.8=3.89t+\dfrac{1}{2}(-1.44)t^2\\\\4.8=3.89t-0.72t^2\\\\t=1.9\ s \ \text{and}\ t=3.49\ s\)

Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. when astronomers make such a statement, they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find.

SO option B is correct.

An astronomer is described as a scientist in the field of astronomy who focuses their studies on a specific question or field outside the scope of Earth by observing astronomical objects such as stars, planets, moons, comets and galaxies – in either observational or theoretical astronomy.

Astronomers are known to use redshifts to measure how the universe is expanding, and thus to determine the distance to our universe's most distant and therefore oldest objects.

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Answer:

The amount of pressure delivered to them

Explanation:

Answer:

The force that holds the gases in the sun.

The force that causes a ball you throw in the air to come down again.

The force that causes a car to coast downhill even when you aren't stepping on the gas.

The force that causes a glass you drop to fall to the floor.

Explanation:

Hope, these examples may help you

We are given the following information about an inelastic collision.

Mass of 1st object = 4 kg

Mass of 2nd object = 10 kg

Initial velocity of 1st object = 25 m/s

Final velocity of both objects = 14 m/s

Initial velocity of 2nd object = ?

In an inelastic collision, the momentum is conserved but the kinetic energy is not conserved.

Recall that the total momentum is conserved and given by

Let us substitute the given values and solve for initial velocity of the body (u2)

Therefore, the initial velocity of the body of mass 10 kg is 9.6 m/s

\(\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}\)

\(\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}\)

\(\\ \tt\longmapsto Acceleration=\dfrac{25}{30}\)

\(\\ \tt\longmapsto Acceleration=0.8m/s^2\)

Answer:

We can use Coulomb's law to solve this problem:

F = k * q1 * q2 / r^2

where F is the force between the two charges, k is Coulomb's constant (k = 9 x 10^9 N m^2 / C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them.

We know the force F, the distance r, and the magnitude of one of the charges q1. We can rearrange the equation to solve for the magnitude of the other charge q2:

q2 = F * r^2 / (k * q1)

Substituting the values we have:

q2 = (250 N) * (0.15 m)^2 / (9 x 10^9 N m^2 / C^2 * 6.5 x 10^-5 C)

Simplifying:

q2 = 8.65 x 10^5 C

Therefore, the magnitude of the other charge is 8.65 x 10^5 C.

Answer:

Student science investigatory projects provide students practical experience in using the scientific method and help stimulate their interest in scientific inquiry. Given the benefits society can realise from professional scientific inquiry, such goals in themselves have importance.

electronic configuration CO is [Ar]3d 4s

it loses two electrons from 4s orbital to form CO+2ion

The time period of most time drafts ranges from 10 days to 60 days. So option b is correct.

Time drafts are a type of short-term credit used to finance international transactions. The buyer is given a certain amount of time to pay for the goods, usually between 10 and 60 days. This gives the buyer time to sell the goods and generate the cash to pay for them.

The other options are not as common for time drafts. A time draft of 1 year to 5 years would be considered a long-term loan, and a time draft of 2 weeks to 52 weeks would be considered a regular invoice.Therefore option b is correct.

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Fluid ounces (fl oz) and liters (L) are units of volume, with the former being more commonly used in the United States and the latter being more commonly used in most other parts of the world.

To convert from fluid ounces to liters, you can use the following conversion factor:

1 fl oz = 0.0295735 L

Volume is the measure of space occupied by an object or substance. It is expressed in different units depending on the system of measurement used. In the International System of Units (SI), the standard unit of volume is cubic meters (m³). However, in practical situations, other units are commonly used.

In the metric system, liters (L) and milliliters (mL) are used as units of volume. A liter is equal to one cubic decimeter (1 dm³), while a milliliter is one-thousandth of a liter. In the US customary system, fluid ounces (fl oz), cups (c), pints (pt), quarts (qt), and gallons (gal) are used. One fluid ounce is equal to 29.5735 milliliters, while one gallon is equal to 3.78541 liters.

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Answer:

your girl is very very very very very very very pretty

Answer:

very very pretty

Explanation:

Answer:

Explanation:

Given:

P = 12 W

R = 270 Ω

t = 2.5 s

____________

Q - ?

Power:

P = I²·R

Current strength:

I = √ (P / R) = √ (12 / 270) ≈ 0.21 A

Charge:

Q = I·t = 0.21·2.5 = 0,525 C

Answer:

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The speed of the rock is 56.8 m/s.

Wavelength, λ = 2.03 x 10⁻³⁴m

Mass of the rock, m = 115 x 10⁻³kg

So, the kinetic energy,

1/2 mv² = hc/λ

v = √(2hc/mλ)

v = √(2 x 6.63 x 10⁻³⁴/115 x 10⁻³x2.03 x 10⁻³⁴)

v = 56.8 m/s

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It should be noted that kinetic energy simply means a form of energy that an object has due to its motion.

The information is incomplete as the experiment isn't given. Therefore, an overview will be given. Kinetic energy is the energy of motion.

It should be noted that the two vital factors that affect kinetic energy are mass and speed. This is because the motion of an object depends on how fast the object is traveling and the mass that it has.

Therefore, an increase in speed will bring about a large increase in kinetic energy and vice versa

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In this case, a low-pass clear-out is wanted to dispose of the strong 60 Hz noise from the measurements of stress at the tractor suspension. By designing a filtering circuit with appropriate values of resistance (R) and capacitance (C), we can reap a nook frequency of 50 Hz.

This will attenuate the 60 Hz noise while permitting the 1 Hz sign to bypass thru with minimum distortion. The actual attenuations and segment shifts will rely on the clear-out design and the unique frequencies worried.

In the given scenario, the Senior Cornhusker desires to measure the trade-in strain on a tractor suspension through the years. However, there is a sturdy 60 Hz noise from fluorescent light ballasts that are interfering with the measurements. To deal with this problem, we want to decide whether a low-pass or excessive-skip filter out is required and lay out the filtering circuit for this reason.

Determining the Filter Type:

Since the preferred signal of the hobby is predicted to change at a frequency of 1 Hz and the interfering noise is at a frequency of 60 Hz, we want to get rid of the better-frequency noise at the same time as retaining the decreased-frequency signal. Therefore, a low-pass filter out is suitable in this example. It will attenuate frequencies above the corner frequency and allow frequencies underneath it to pass thru.

Designing the Filtering Circuit:

To design the low-pass clear-out, we want to select appropriate values of resistance (R) and capacitance (C). The nook frequency (fc) of the clear-out determines the frequency at which the attenuation starts off evolved. It may be calculated using the components:

fc = 1 / (2πRC)

To reduce the noise at 60 Hz, we are able to choose a nook frequency slightly under this price. Let's anticipate we want a corner frequency of 50 Hz.

Corner Frequency and Attenuation:

Using the favored nook frequency of 50 Hz, we are able to rearrange the components to remedy the fabricated from RC:

RC = 1 / (2πfc) = 1 / (2π * 50)

Now, we will select suitable values for R and C based on sensible concerns. Let's say we pick R = 10 kΩ and clear up for C:

C = 1 / (2π * R * fc) = 1 / (2π * 10,000 * 50)

This will supply us with the capacitance value to obtain the preferred nook frequency.

Attenuations and Phase Shifts:

The low-bypass filter will attenuate frequencies above the corner frequency even as permitting frequencies beneath it to skip via with minimum attenuation. The quantity of attenuation and segment shift relies upon the filter layout and the specific frequencies involved. With the chosen corner frequency, the 60 Hz noise will be attenuated, while the 1 Hz signal of hobby could be largely preserved.

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The correct question is:

"Senior Cornhusker is setting up instrumentation to measure the change in strain on a tractor suspension over time. The strain is expected to change at a frequency of 1 Hz, but there is a strong 60- Hz noise from fluorescent light ballasts that are interfering with his measurements. First, help him to determine whether he needs a low-pass or high-pass filter in this situation. Second, design the filtering circuit by selecting appropriate values of R and C. With the selected R and C, what is the corner frequency? What are the attenuations and phase shifts for the signal and noise, respectively?"

The most common injury related to electrical hazards are Electrical burns.

What are electrical hazards?

An electrical burn is a type of skin burn that develops when your body comes into contact with electricity. It is possible for electricity to pass through your body when it comes into contact with it. When this occurs, the voltage has the potential to harm tissues and organs.

Hence, The most common injury related to electrical hazards are Electrical burns.

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Answer:

Explanation:

F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?

We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that

15 = 5a so

a = 3

Then the F is cut into thirds, so

5 = 5a so

a = 1

The second acceleration is one-third of the first one, where the Force is 3 times greater.