What is it does anyone know?? PLZ help me

The work needed to stretch the spring a distance of 5 cm is 1100 N·cm.

To determine the work needed to stretch the spring a distance of 5 cm, we need to calculate the area under the force vs. distance graph within that range. Looking at the graph, we can see that the force initially increases linearly as the distance increases and then levels off.

To calculate the work, we need to find the area of the triangle formed by the initial linear part of the graph and the rectangle representing the constant force. The height of the triangle is the force at 5 cm, which appears to be around 200 N. The base of the triangle is 5 cm. The area of the triangle is given by 0.5 * base * height, which is 0.5 * 5 cm * 200 N = 500 N·cm .The rectangle representing the constant force has a height of 200 N and a base of 3 cm (since it starts at 2 cm and ends at 5 cm). The area of the rectangle is base * height, which is 3 cm * 200 N = 600 N·cm.

Adding the areas of the triangle and the rectangle, we get a total work of 500 N·cm + 600 N·cm = 1100 N·cm.

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Your de Broglie wavelength while walking at 1 m/s with a mass of 80 kg is approximately 8.2825 x 10^--37 meters. This is an incredibly small wavelength, which is typical for macroscopic objects like humans

To estimate your de Broglie wavelength while walking at a speed of 1 m/s, we can use the de Broglie wavelength formula:

λ = h / p

where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the object.

The momentum of an object can be calculated as:

p = m * v

where m is the mass of the object and v is its velocity.

Plugging in the given values, we get:

p = 80 kg * 1 m/s = 80 kg m/s

Using Planck's constant h = 6.626 x 10^-34 m^2 kg/s, we can now calculate the de Broglie wavelength:

λ = h / p = 6.626 x 10^-34 m^2 kg/s / 80 kg m/s ≈ 8.3 x 10^-37 meters

However, it's important to note that the de Broglie wavelength is only significant for objects with very small masses or very high velocities, such as electrons or other subatomic particles.

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Answer:

The graph demonstrates two major trends. First, employment numbers began to drop sharply in 2008. These numbers fell until 2010, when they reached a trough. Employment began to rise again after this point and continued to rise into 2013.

Explanation:

Answer:

\(\huge\boxed{\sf Circumstellar\ discs }\)

Explanation:

Astronomers discovered matter made up of gas, dust or asteroids etc. around nearby stars. This matter is called circumstellar disc (due to it being ring-shaped). These circumstellar discs are materials out of which planets form.

\(\rule[225]{225}{2}\)

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By conservation of momentum, we will find that the mass is 4.97 kg.

So in the original system, we have two objects, the bouquet of flowers of mass M that is not moving and Yvonne, which has a mass of 63 kg and a speed of 10.8 m/s.

Then the total momentum of this system is:

P = (63kg)*(10.8m/s) + M*(0 m/s) = 680.4 kg*m/s.

Remember the conservation of momentum, thus, the final momentum must be equal to the above one.

In the final situation, Yvonne and the bouquet move together with a speed of 10.01 m/s,

Then the final momentum is:

P' = (63kg + M)*(10.01 m/s)

And that must be equal to the initial momentum, then we have the equation:

(63kg + M)*(10.01 m/s) = 680.4 kg*m/s.

630.63kg*m/s + M*10.01 m/s = 680.4 kg*m/s.

M*10.01 m/s = 680.4 kg*m/s - 630.63kg*m/s = 49.77 kg*m/s

M = (49.77 kg*m/s)/(10.01 m/s) = 4.97 kg

So the mass of the bouquet is 4.97 kg

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A radioactive isotope with a mass of 10 g has a 15 Bq activity. Consequently, it is possible to anticipate that 20g of the same isotope will have an activity of 30 Bq. Hence, the statement is true.

The becquerel (Bq), the SI unit of radioactivity, is named after Henri Becquerel. As defined by the unit of radioactivity:

A quantity of radioactive material that is active when one decay occurs every second.

One becquerel refers to ne radioactive decay per second which is numerically equal to 2.703 x 10⁻¹¹ Ci.

So, If  10g of a radioactive isotope has an activity of 15 Bq, 20g of the same isotope will have an activity of 30 Bq. The statement is true.

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Answer:

The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which coefficients would be placed in front of water and oxygen?

2 in front of water and 1 in front of oxygen

Kiley wants to balance the equation H2 + N2 → NH3. Which would be her first step in balancing an equation?

NOT adding the coefficient 3 in front of H2 and the coefficient 2 in front of NH3

Which is the balanced equation for V2O5 + CaS → CaO + V2S5?

V2O5 + 5CaS → 5CaO + V2S5

One half of a balanced chemical equation is shown.

3Mg(OH)2 + 2H3PO4

Which lists the numbers of each atom the other half of the equation would contain?

3 Mg, 2 P, 14 O, 12 H

Substances A and B react in a ratio of A + B → AB. Colin was combining substance A and substance B to form substance AB. He added 10 g of substance A and 45 g of substance B.

How much of substance AB would Colin make?

55g

Lana balanced an equation so that the result was 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr. Which most likely represents the starting equation?

NOT 2C4H3Br + 5O2 → 4CO2 + 2H2O + 2HBr

NOT C4H3Br + O2 → CO2 + H2O + HBr

Which best describes the law of conservation of mass?

The mass of the reactants and products is equal and is not dependent on the physical state of the substances.

Put 2 in front of water and 1 in front of oxygen.

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Answer:F. all are equally likely

Explanation:

F. because there is no way to tell if or what the persentage of tails to heads could be.

If the wire is tipped at an angle of 15.0° with the horizontal, we need to determine the force it will now feel. The length of the wire that will be in the magnetic field is relevant to finding the answer.

When a wire carrying current is placed in a magnetic field, it experiences a force perpendicular to both the direction of current flow and the magnetic field. In this case, when the wire is tipped at an angle of 15.0° with the horizontal, only a component of the wire's length will be in the magnetic field.

To find the force, we need to consider the effective length of the wire in the magnetic field. This can be calculated by multiplying the actual length of the wire by the cosine of the angle between the wire and the magnetic field. Once we have the effective length, we can use the formula for the force on a current-carrying wire in a magnetic field, which is given by the equation F = BIL, where B is the magnetic field strength, I is the current, and L is the length of the wire in the field. By substituting the effective length into the equation, we can determine the force the wire will experience.

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Answer:

A

Explanation:

maximum gravitional potential energy is at point R

Answer:

Explanation:

ave speed is always (frequency) x (wavelength)

Speed  =  (450 /sec) x (0.52 m)

=      234 m/sec .

The type of platform lift that starts with holding the bar off the ground with straight arms, hinging forward until the bar lowers slightly past the knees, and then driving the hips forward until standing at the original position is called a Deadlift.

The Deadlift is a compound exercise that primarily works the muscles in the legs, hips, and back, including the glutes, hamstrings, quadriceps, and erector spinae.

It is a staple exercise in many strength and powerlifting training programs, as well as a functional movement that can improve overall fitness, athletic performance, and daily life activities.

Construction lifts are strong tools. Encompassing boom scissor lifts, lifts, forklifts and other types of aerial and man lifts, construction lifts mention to any lifting machinery that quality an aerial platform supported by a vehicle-mounted extension.

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Answer:

C. 12.4 m/s

Explanation:

Acceleration (a) = -6.2 m/s²

Final velocity (v) = 0 m/s (Comes to rest)

Time taken (t) = 2.0 s

By using equation of motion, we get:

\( \bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow 0 = u + ( - 6.2)(2.0) \\ \\ \rm \longrightarrow 0 = u - 12.4 \\ \\ \rm \longrightarrow u - 12.4 = 0 \\ \\ \rm \longrightarrow u + 12.4 - 12.4 = 0 + 12.4 \\ \\ \rm \longrightarrow u = 12.4 \: m {s}^{ - 1} \)

\( \therefore \) Initial velocity (u) = 12.4 m/s

To draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute values of (a) the shear and (b) the bending moment,

Step 1: Determine the reactions at the supports.
Calculate the total load on the beam and the reactions at each support. You may use the equilibrium equations (ΣFx = 0, ΣFy = 0, and ΣM = 0) to find the reactions.

Step 2: Draw the shear force diagram (SFD).
Starting from one end of the beam, plot the shear force along the length of the beam. Whenever there is a concentrated load or a distributed load, the shear force will change at that point. Use the load and reaction values to calculate the shear force for each segment of the beam.

Step 3: Draw the bending moment diagram (BMD).
Starting from one end of the beam, plot the bending moment along the length of the beam. Calculate the bending moment for each segment of the beam using the shear force and the distance from the end of the beam. The bending moment diagram will be a curve or a series of straight lines.

Step 4: Determine the maximum absolute values of (a) the shear and (b) the bending moment.
Find the maximum absolute value of the shear force from the shear force diagram. Similarly, find the maximum absolute value of the bending moment from the bending moment diagram. These values represent the maximum stresses experienced by the beam due to the given loading conditions.

In summary, to draw the shear and bending-moment diagrams for a beam and determine the maximum absolute values of (a) the shear and (b) the bending moment, you need to calculate the reactions at the supports, draw the shear force and bending moment diagrams, and find the maximum absolute values of the shear force and bending moment from the diagrams.

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Answer:

\(57.24^{\circ}\)

24.21 m/s

Explanation:

x = Displacement in x direction = 55 m

y = Displacement in x direction = 0

\(y_0\) = Height of the ball when the ball is kicked = 1 m

t = Time taken = 4.2 s

\(a_y=g\) = Acceleration due to gravity = \(-9.81\ \text{m/s}^2\)

u = Initial velocity of ball

Displacement in x direction is given by

\(x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}\)

Displacement in y direction is given by

\(y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}\)

\(\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}\)

The ball should be kicked at an angle of \(57.24^{\circ}\)

\(u=\dfrac{13.1}{\cos\theta}=\dfrac{13.1}{\cos57.24^{\circ}}\\\Rightarrow u=24.21\ \text{m/s}\)

The initial speed of the ball is 24.21 m/s.

The number of unknowns in the question are two therefore two equations

are required to find a solution.

Part A: The angle at which the ball should be kicked is approximately 57.56°

Part B: The initial speed with which the ball should be kicked is approximately 24.41 m/s.

Reasons:

The given parameters are;

Time the ball should be in the air = 4.2 s

Point the ball should land, d = 5.5 m from where it was kicked

Height at which the ball leaves the ground = 1.0 m

Part A;

The angle at which the ball should be kicked.

Solution;

The horizontal distance, d = uₓ·t

Therefore;

\(u_x = \dfrac{d}{t} = \dfrac{5.5}{4.2} = \dfrac{55}{42}\)

\(u_x = u \cdot cos(\theta)= \dfrac{55}{4.2}\)

Where;

u = The initial velocity

\(Vertical \ motion , \ h =\mathbf{h_0 + u_y\cdot t - \dfrac{1}{2} \cdot g \cdot t^2}\)

Which gives;

\(0 =1 + u_y\times 4.2 - \dfrac{1}{2} \times 9.81 \times 4.2^2 = 4.2\cdot u_y - 86.5242\)

\(u_y =u \cdot sin(\theta) = \dfrac{86.5242}{4.2} = 20.601\)

\(\dfrac{u_y}{u_x} =\dfrac{20.601}{\left(\dfrac{55}{4.2} \right)} = \mathbf{ \dfrac{u \cdot sin(\theta)}{u \cdot cos(\theta)}} = tan\left( \theta)\)

\(\mathrm{The \ angle \ at \ which \ the \ ball \ should \ kicked } \ \theta = arctan \left(\dfrac{20.601}{\left(\dfrac{55}{4.2} \right)} \right) \approx \underline {57.56 ^{\circ}}\)

Part B

\(u \cdot cos(\theta)= \dfrac{55}{4.2}\)

\(The \ initial \ speed, \ u= \dfrac{55}{4.2 \times cos(57.56^{\circ })} \approx 24.41\)

The initial speed of the ball, u ≈ 24.41 m/s

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The values of t for which the ball's height is 25 feet are t = 5/4 and t = 1.

The values of t for which the ball's height is 25 feet, we can set the height equation equal to 25 and solve for t:

5 + 37t - \(16t^2\) = 25

Rearranging the equation:

\(16t^2\) - 37t + 20 = 0

Solve this quadratic equation, we can factor it:

(4t - 5)(4t - 4) = 0

Setting each factor equal to zero:

4t - 5 = 0 or 4t - 4 = 0

Solving each equation for t:

4t = 5 or 4t = 4

t = 5/4 or t = 1

the values of t when the ball's height is 25 feet, we set the height equation equal to 25 and solved for t. This led to a quadratic equation, which we factored and solved for t.

The solutions were t = 5/4 and t = 1. These represent the values of time when the ball reaches a height of 25 feet. At t = 5/4 seconds, the ball is at its highest point before descending, while at t = 1 second, the ball is on its upward trajectory and hasn't reached its peak height yet.

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Answer:

The troposphere and stratosphere are the only layers where you would find clouds.

Answer: Based on the information provided, the only layers where you would find clouds are the troposphere and the stratosphere.

The troposphere is the layer of the atmosphere that extends from the Earth's surface to about 8-15 kilometers (5-9 miles) above the Earth. This is the layer where most of the Earth's weather occurs, and it is where you will find clouds, such as cumulus clouds, stratus clouds, and cirrus clouds.

The stratosphere is the layer of the atmosphere that extends from about 15-50 kilometers (9-31 miles) above the Earth. This layer is characterized by a rise in temperature with altitude, and it is where you will find clouds called stratospheric clouds, such as nacreous clouds and polar stratospheric clouds.

The mesosphere is the layer of the atmosphere that extends from about 50-85 kilometers (31-53 miles) above the Earth. This layer is characterized by a decrease in temperature with altitude, and it is generally too cold and dry to support the formation of clouds.

The thermosphere is the layer of the atmosphere that extends from about 85 kilometers (53 miles) to the edge of space. This layer is characterized by a rise in temperature with altitude, and it is generally too thin and diffuse to support the formation of clouds.

I hope this helps! Let me know if you have any other questions.

your on here to lol we must be hella failing g

Given,

V1 = 6 ml

T1 = 250 k

V2 = 9 ml

To find T2

According to Charles's Law,

V1/T2 = V2/T2

or, T2 = T1 V2/ V1

          = 250×9/6

          = 375 K

Because it is very unreactive, helium is used to provide an inert shielding ecosystem for making fiber optics and semiconductors, and for arc welding. Helium is also used to stumble on leaks, together with automobile air-conditioning structures, and because it diffuses quickly it's far used to inflate automobile airbags after impact.

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Answer:

Answer in Explanation

Explanation:

Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. This is the main reason why we require height in the P.E formula and calculations.

The derivation of this formula is as follows:

Work = Force * Displacement

For gravitational potential energy:

Work = P.E

Force = Weight = mg

Displacement = Vertical Displacement = Height = h

Therefore,

P.E = mgh

All for the sattement A, B, and E is True and satement C, D, and F is False where you learned about astronomy from space.

A .The Compton, Chandra and Spitzer telescopes are all examples of space telescopes.

-This statement is True.

B. Most types of electromagnetic waves are not visible to ground-based telescopes.

-This statement is True.

C. Enough infrared energy gets to Earth's surface that infrared telescopes do not need to be put in orbit.

-This statement is False.

D. Cosmic rays are the absolute highest-energy type of electromagnetic wave.

-This statement is False.

E. The Hubble Space Telescope makes observations at all wavelengths.

-This statement is True.

F. Earth's atmosphere has no effect on visible light, so there is no need to place an optical telescope in space.

-This statement is False.

Astronomy is the scientific study of celestial objects and phenomena, including stars, planets, galaxies, and the universe as a whole. Astronomers use various tools and techniques, such as telescopes, spectroscopy, and astrophysics, to observe and understand the properties and behavior of these celestial objects.

Through the study of astronomy, scientists have discovered many fascinating things about the universe, such as the existence of black holes, the age and size of the universe, and the formation and evolution of stars and galaxies. Astronomy has also played a significant role in advancing our understanding of fundamental physics, including the laws of gravity and the behavior of matter and energy at extreme conditions.

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The correct option is (B) h(t) = -16t2 + 50t + 3

An equation like this has the generic form h(t) = at2 + v0t + h0,

where; a is the gravitational constant,

v0 is the initial velocity, and

h0 is the initial height.

Given that the gravitational constant is -16.

The equation h(t) = -16t2 + 50t + 3 is given by the starting velocity of 50 and the initial height of 3.

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Answer:

both hit qaulity

Explanation:

Newton's Third Law says that every interaction involves two forces - in this case (1) hammer pushes nail, and (2) nail pushes hammer. When a hammer hits a nail, Newton's Third Law says that the force that the hammer exerts on the nail is exactly the same size as the force the nail exerts on the hammer.

On the moon you weigh 6 times lighter so for this problem you multiply 140n by 6.

Your weight on the Moon is 16.5% what you would experience on Earth. In other words, if you weighed 100 kg on Earth, you would weigh a mere 16.5 kg on the Moon. For you imperial folks, imagine you tipped the scales at 200 pounds.

It’s because of the lower gravity on the Moon. Objects on the surface of the Moon experience only 16.5% of the gravity they would experience on Earth.

The mass of the Moon is only 1.2% the mass of the Earth, so you might expect it to have only 1.2% of the gravity. But it’s only 27% of the size of the Earth, so when you’re standing on the surface of the Moon, you’re much closer to its center of gravity.

Therefore, On the moon you weigh 6 times lighter so for this problem you multiply 140n by 6.

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The mass will remain same of 3.15g.

Mass is used in physics to express inertia, a fundamental function of all be counted.

Essentially, it's far a mass of depend's resistance to converting its direction or velocity in reaction to the application of a pressure. The exchange that an implemented pressure produces is smaller the extra mass a body has.

The kilogram, the unit of mass within the international gadget of units, corresponds to 6.62607015 1034 joule seconds the usage of Planck's constant (SI). One joule is produced with the aid of multiplying one kilogram via one square meter in keeping with second.

The kilogram is decided by using exact measurements of Planck's regular given that the second and the meter have formerly been defined in phrases of different physical constants.

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Answer:

it's the turning effect of force; the product of force and perpendicular distance from line of action of the force to the pivot . Depends on two factors: size of force applied and perpendicular distance from pivot to line of action of the force

The concept law of supply shows that price and quantity supplied move in the same direction.

The law of supply relates price changes for a product to the quantity supplied. In differentiation from the law of demand, the law of supply relationship is coordinated, not converse. The higher the cost, the higher the quantity provided. Lower costs mean decreased supply, all else held equal. Higher prices donate suppliers an incentive to supply more of the item or product, expecting their costs aren't expanding as much. Lower costs result in a fetched crush that checks supply. As a result, supply inclines are upwardly inclined from cleared out to right. As with demand, supply limitations may constrain the cost flexibility of supply for a product, whereas supply shocks may cause an unbalanced cost to alter for a basic commodity.

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The change in the momentum was found to be 16,000 kgm/s

Change in momentum ?

Situations involving impulse and momentum are all around us, whether we notice it or not. Before we begin evaluating these daily instances, we'll go over the fundamental ideas of impulse and momentum.

In physics, an impulse (Ft) is a force (F) operating over a certain time (t) that results in a change in momentum (p) of an object. The impulse-momentum theorem is the equation of impulse with the change in momentum.

the change in the momentum was Δp=mxΔv

Δp=2000x8

Δp=16,000 kgm/s

therefore the change in momentum was found to be 16,000kgm/s

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