calculate the acceleration of a 250000- kgkg jumbo jet just before takeoff when the thrust on the aircraft is 80000 nn .

Answer:

f(x) = mx + b

Explanation:

The image is lοcated apprοximately 7.69 x 10² mm tο the right οf the lens.

The height οf the image is apprοximately -4.99 x 10¹ cm.

The ray diagram is attached.

An image may be defined as that pοint, where the light rays cοming frοm an οbject meet οr appears tο meet after reflectiοn οr refractiοn. In this definitiοn, the wοrd 'οbject' may be defined as anything frοm which light rays are cοming.

Based οn the given infοrmatiοn, we can use the lens equatiοn tο determine the lοcatiοn and height οf the image fοrmed by the lens.

Given:

Height οf the οbject (h₀) = 2.49 cm = 24.9 mm

Object distance (d₀) = -38.3 mm (since it is tο the left οf the lens)

Fοcal length οf the lens (f) = 36.5 mm

Using the lens equatiοn:

1/f = 1/d₀ + 1/dᵢ

Plugging in the values:

1/36.5 = 1/(-38.3) + 1/dᵢ

Simplifying and sοlving fοr dᵢ:

dᵢ = 1 / (1/36.5 - 1/38.3)

= 1 / (0.0274 - 0.0261)

= 1 / 0.0013

= 769.23 mm

≈ 7.69 x 10² mm

Therefοre, the image is lοcated apprοximately 7.69 x 10² mm tο the right οf the lens.

Tο find the height οf the image (hᵢ), we can use the magnificatiοn equatiοn:

m = -hᵢ / h₀ = dᵢ / d₀

Plugging in the values:

m = (7.69 x 10² mm) / (-38.3 mm)

Simplifying:

m ≈ -20.07

Therefοre, the height οf the image is apprοximately -20.07 times the height οf the οbject.

Height οf the image (hᵢ) = -20.07 x 24.9 mm

≈ -498.84 mm

≈ -49.88 cm

≈ -4.99 x 10¹ cm

Hence, the height οf the image is apprοximately -4.99 x 10¹ cm.

As fοr the ray diagram, since the οbject is placed tο the left οf the lens, a cοnverging lens is assumed. A cοnverging lens fοrms a real, inverted image οn the οppοsite side οf the lens when the οbject is placed at a distance greater than the fοcal length. The image wοuld be lοcated further tο the right and have a reduced height cοmpared tο the οbject. The ray diagram wοuld shοw the incident rays frοm the οbject cοnverging thrοugh the lens tο fοrm the image.

The diagram is given below.

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The motion of the two boxes will depend on the net horizontal force acting on them and the frictional force opposing the motion.

To determine which graph describes the motion of the two boxes, consider the net horizontal force acting on each box and the frictional force. In scenario 1, the student pushes with a force F0 at an angle θ from the horizontal. The horizontal component of the force is F0*cos(θ). In scenario 2, the student pulls with the same force F0 at the same angle θ, resulting in the same horizontal component, F0*cos(θ).

The frictional force acting against the motion will be the same for both scenarios. Therefore, since the net horizontal forces and frictional forces are the same in both scenarios, the motion of the two boxes as they are moved across the room will be described by the same graph.

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When capacitors are connected in parallel, their equivalent capacitance is simply the sum of their individual capacitances. Therefore, the equivalent capacitance of the 5.0 μF capacitor, 10 μF capacitor, and 19 μF capacitor connected in parallel is: Ceq = 5.0 μF + 10 μF + 19 μF Ceq = 34 μF. Therefore, the equivalent capacitance of the three capacitors connected in parallel is 34 μF.

The equivalent capacitance (C_eq) of capacitors connected in parallel can be found by adding their individual capacitances. In this case, you have a 5.0 μF capacitor (C1), a 10 μF capacitor (C2), and a 19 μF capacitor (C3). The formula for the equivalent capacitance is:

C_eq = C1 + C2 + C3

C_eq = 5.0 μF + 10 μF + 19 μF

C_eq = 34 μF

So, the equivalent capacitance of these capacitors connected in parallel is 34 μF.

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Answer: repeatable

Explanation: Accurate repeatable is needed in a valid experiment. I think

Answer:

Greatest position of the ball are z

The correct option is D. the volume of the ventricle when it is most full.

The volume of the ventricle when it is most full is the most correct description of end-diastolic volume.

It can be defined as the amount of blood in the ventricle immediately before a cardiac contraction or systole occurs. End-diastolic volume is the volume of blood in the ventricles at the end of diastole, after filling with blood from the atria, before the ventricles contract to begin systole.

ventricles are hollow chambers or cavities found in the heart and brain. In the heart, there are two ventricles responsible for pumping blood, while in the brain, there are four interconnected ventricles that produce and circulate cerebrospinal fluid.

Therefore, the correct option is D. the volume of the ventricle when it is most full.

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The distance, s, travelled by the car is 66 m.

The distance travelled by the toy car is calculated using the formula below:

Where:

Distance, s = 5.0 * 6.0 +  2.0 * 6.0²/2

Distance,  s = 66 m

In conclusion, the distance travelled is calculated from the initial velocity, the time taken, as well as the acceleration.

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The final speeds of the spheres are 3.47 m/s and 3.08 m/s.

We can use conservation of momentum to solve this problem since there are no external forces acting on the system.

The initial momentum of the system is:

p_initial = m₁ * v₁ + m₂ * v₂

where m₁ and m₂ are the masses of the spheres, and v₁ and v₂ are their initial velocities (4.00 m/s and 0 m/s, respectively).

After the collision, the momentum of the system is:

p_final = m₁ * v1' + m₂ * v₂'

where v₁' and v₂' are the final velocities of the spheres. We also know that the angle between the first sphere's final path and its initial path is 60 degrees, which means that the angle between the two spheres after the collision is 150 degrees (90 + 60).

Using conservation of momentum, we can set the initial and final momenta equal to each other:

m₁ * v₁ + m₂ * v₂ = m₁ * v₁' + m₂ * v₂'

We can also break down the final velocities into their x and y components using trigonometry. Let's define the angle between the first sphere's final path and the x-axis as theta. Now we can use conservation of momentum to solve for the final velocities:

m₁ * v₁ + m₂ * v₂ = m₁ * v₁' * cos(theta) + m₂ * v₂' * cos(150 degrees)

0 = m₁ * v₁' * sin(theta) + m₂ * v₂' * sin(150 degrees)

Solving the first equation for v₂', we get:

v₂' = (m₁ * v₁ + m₂ * v₂ - m₁ * v₁' * cos(theta)) / (m₂ * cos(150 degrees))

Substituting this expression into the second equation and solving for v₁', we get:

v₁' = (m₂ * sin(150 degrees) * v₁ + m₂ * sin(150 degrees) * v₂ + m₁ * sin(theta) * v₁' - m₁ * sin(theta) * m₂ * v₁ * cos(theta) / cos(150 degrees)) / (m₁ * sin(theta))

Plugging in the given values and solving, we get:

v₁' = 3.47 m/s

v₂' = 3.08 m/s

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Answer:

The geological features that is not created by the movement of the Earth's plates is Canyons.

His final velocity at the end of the 2.6 seconds will be  25.48 m/s

given

time = 2.6 seconds

initial velocity = u = 0

acceleration = acceleration due to gravity = 9.8 \(m/s^{2}\)  (since , it is a free fall )

final velocity = ?

using kinematics equation

v = u + at

v = u - g*t

v = - 9.8 * 2.6 = - 25.48 m/s

His final velocity at the end of the 2.6 seconds will be  25.48 m/s in downward direction

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Answer:

45

Explanation:

because you divide volume and mass

Therefore for cart to stay here on track when it arrives at point P, the net force applied to it must be greater or equal than for the heaviness of the cart.

The force is an impact that has the power to alter an object's motion. A mass-containing object's velocity can be altered by a force. An obvious way to describe force is as a pushed or a pull. A vector quantity is called a force since it has both strength and the direction.

The strongest force in existence is thought to be the strong force. The gravitation, weak, and electromagnetism forces come after it in lowest to highest.

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Answer:

The truck must travel at a speed of 98.6677 km/hr to stay beneath the airplane

Explanation:

The given parameters are;

The velocity of the airplane, v = 105 km/hr

The direction of the airplane, θ = 25 degrees to the ground = 25°

Therefore, we have that the horizontal component of the velocity of the plane, vₓ is given as follows;

vₓ = v × cos(θ) = 105 km/hr × cos(25) ≈ 98.6677 km/hr

The vertical component of the velocity of the plane, \(v_y\), is given as follows;

\(v_y\) = v × sin(θ)  = 105 km/hr × sin(25) ≈ 44.37 km/hr

The truck only has to travel at a speed equal to the horizontal velocity of the airplane to stay beneath the airplane

Therefore, the travelling speed the truck must be vₓ which is 98.6677 km/hr to stay beneath the airplane.

The graph that corresponds to 0.1 s in one complete cycle is graph D.

option D is the correct answer.

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years.

Also, the period of a wave is the amount of time it takes for a wave to complete one wave cycle or wavelength.

From the given parameter, the coil rotates 10 times in one second. The period of the coil is calculated as;

Period = 1 s / 10

Period = 0.1 s

From the graphs, the only option that has one complete cycle in one second is option D.

Check option D, half cycle is 0.05 s and one full cycle is 0.1 s.

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To calculate the height of the cliff and the time it takes for the rock to reach the ground when thrown straight down, we can use the equations of motion.

(a) Height of the cliff:

When the rock is thrown straight up, it reaches its highest point before falling back down. The time it takes for the rock to reach its highest point is equal to the time it takes for the rock to fall back down to the ground.

Using the equation:

s = ut + (1/2)at^2

Where:

s is the distance traveled (height of the cliff),

u is the initial velocity (8.19 m/s),

t is the time (2.32 s),

a is the acceleration due to gravity (-9.8 m/s^2, taking downward direction as negative).

Rearranging the equation:

s = ut + (1/2)at^2

s = (8.19)(2.32) + (1/2)(-9.8)(2.32)^2

s = 19.004 - 25.798

s = -6.794 m

Since the height of a cliff cannot be negative, we take the absolute value of the result:

Height of the cliff = |s| = 6.794 m

So, the height of the cliff is approximately 6.794 meters.

(b) Time to reach the ground when thrown straight down:

When the rock is thrown straight down with the same speed, the initial velocity (u) is still 8.19 m/s, but the acceleration due to gravity (a) remains -9.8 m/s^2.

Using the equation:

s = ut + (1/2)at^2

Where:

s is the distance traveled (height of the cliff, which is now negative),

u is the initial velocity (8.19 m/s),

t is the time we want to find,

a is the acceleration due to gravity (-9.8 m/s^2, taking downward direction as negative).

Substituting the known values:

-6.794 = (8.19)t + (1/2)(-9.8)t^2

Rearranging the equation:

-6.794 = 8.19t - 4.9t^2

Rearranging further:

4.9t^2 - 8.19t - 6.794 = 0

Solving this quadratic equation, we find two possible values for t: 0.828 seconds and 1.303 seconds. Since we are considering the time it takes to reach the ground, the valid solution is t = 0.828 seconds.

Therefore, when the rock is thrown straight down, it takes approximately 0.828 seconds to reach the ground.

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Answer:he's not applying force or motion

Explanation:

p=f & m

The Fad,x, the x-component of the force exerted by the infinite wire on segment AD of the loop is 2.39 X 10⁻⁸ N. The Fbc,x, the x-component of the force exerted by the infinite wire on segment BC of the loop is 9.56 X 10⁻⁹ N. The Fnet,y, the y-component of the net force exerted by the infinite wire on the loop is zero.

1. The Fad,x, the x-component of the force exerted by the infinite wire on segment AD of the loop =

= Fad,x = u₀i₁i₂H / (2πL)

= Fad, x = [ (4π X 10⁻⁷)(0.672)(0.256) X (0.25) ] /2π X 0.36

= Fad,x= 2.39 X 10⁻⁸ N

2. The Fbc,x, the x-component of the force exerted by the infinite wire on segment BC of the loop =

= Fbc,x = u₀i₁i₂H / (2π) X (L+W)

= Fbc,x = (4π X 10⁻⁷)(0.672)(0.256)((0.25)/2π X (0.36+0.54)

= Fbc,x = 9.56 X 10⁻⁹ N

3. The Fnet,y, the y-component of the net force exerted by the infinite wire on the loop =

= Fnet,y = 0, since force acting on perpendicular wire is zero.

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In the given scenario, different types of heat transfer are occurring.

Radiation: Chris getting sunburned during the day is an example of heat transfer through radiation. The sun emits electromagnetic waves, including infrared radiation, which can directly heat the skin and cause sunburn.

Convection: When Chris walks by the fire started by his father, he feels the heat. This heat transfer is due to convection. The fire heats the air surrounding it, causing the hot air to rise and create convection currents. As Chris walks through the rising hot air, he feels the heat being transferred to his body.

Conduction: When Chris accidentally touches a hot pan while his mother is cooking breakfast, he gets burned. This is an example of heat transfer through conduction. The hot pan transfers its heat directly to Chris's hand upon contact. Heat travels from the higher temperature object (the pan) to the lower temperature object (Chris's hand) through direct molecular collisions.It is important to note that all three types of heat transfer can occur simultaneously in various situations. In this scenario, radiation from the sun causes sunburn, convection from the fire creates heat in the air, and conduction from the hot pan leads to a burn on Chris's hand. Despite the minor incidents, Chris still had a fun camping trip.

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Answer:

your ball exherts more pressure on your hand because your hand is smaller then the glove. The pressure gets distributed a lot more when you catch it with your glove.

Explanation:

PLEASE DO NOT ASK SUCH QUESTIONS AGAIN.

Just kidding :)

Answer:

v = 31.89 m/s

Explanation:

p = m × v

1,850 = 58 × v

58v = 1,850

v = 1,850/58

v = 31.89 m/s

Forces on object must be concurrent. Object must not deform under the forces. Object must not change shape. Object must not be large. Object must not rotate.

Force is a physical quantity that represents the interaction between two objects. Force can cause an object to accelerate or change its velocity. It can also cause a change in an object's shape or orientation. There are several types of force, such as gravitational force, electromagnetic force, and nuclear force.

An object can be treated as a particle under the following conditions:

Forces on object must be concurrent. Object must not deform under the forces. Object must not change shape. Object must not be large. Object must not rotate.

It's important to note that the concept of treating an object as a particle is a simplifying assumption used in some branches of physics, such as mechanics, and is only valid under certain conditions.

When an object is treated as a particle, it is assumed to have no internal structure, and that all forces acting on it act at a single point. The object is also assumed to be small enough that its deformation or shape change under forces can be ignored. Additionally, it is also assumed that the object does not rotate.

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The correct statement regarding the potential energy of a system is: A. The potential energy of a system can convert into kinetic energy.

Potential energy is the energy stored within a system due to its position or configuration. It represents the potential for that system to do work. When potential energy is released, it can be converted into kinetic energy, which is the energy of motion. This conversion occurs as the system moves and changes position or configuration.

Option B is incorrect because the potential energy of a system can be either positive or negative, depending on the reference point chosen. It represents the energy difference between the current state of the system and a reference state.

Option C is also incorrect because the potential energy of a body typically depends on its position or height, not its speed. Speed is related to kinetic energy, not potential energy.

Therefore, the correct statement is option A: The potential energy of a system can convert into kinetic energy.

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Therefore, the magnitude of vector C is 25.

Given:A = 10x - 2yB = 5x + 4yC=2A + BNow we have to calculate the magnitude of vector C.Let's calculate each part of the vector C first;2A = 2(10x-2y) = 20x - 4yB = 5x + 4yC = 2A + B= (20x-4y)+(5x+4y)=25xNow we can calculate the magnitude of vector C by using the formula;|C| = √(Cx²+Cy²+Cz²)Here, we only have two dimensions, so the formula becomes;|C| = √(Cx²+Cy²)|C| = √(25²) = 25. Therefore, the magnitude of vector C is 25.

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Answer:

6.266829864 g

Explanation: