How many significant figures? 53,598 m/sec

Answer:

x = 10

Explanation:

One of the properties of a rhombus is that the diagonals bisect each other at 90°

This means AC and BD are perpendicular to each other.

Let us call this intersection point as O

Then AOC forms a triangle with angles 3x - 13, 8x-7 and 90°

Since the sum of 3 angles of a triangle = 180°

(3x - 13) + (8x-7) + 90 = 180

(3x - 13) + (8x-7) = 180 - 90

(3x - 13) + (8x-7)  = 90

3x - 13 + 8x - 7 = 90

3x + 8x - 13 - 7 = 90

11x - 20 = 90

11x = 90 + 20 = 110

x = 10

Answer: 25.04m/s

Explanation:

v^2 = v^2o + 2a(x-xo)

v^2 = velocity 

v^2o = initial velocity 

a = acceleration 

x = final position/distance 

xo = initial position/distance

In this case, the initial velocity is 0 since the ball wasn't moving before it was dropped. The final position is 32 as the motion ended after the ball traveled 32m. The initial position is 0. The acceleration is 9.8m/s (free fall). Plug these numbers into the formula: 

v^2 = 0 + 2(9.8)(32)

v^2 = 25.04396135 

Round to get 25.04m/s

Answer:

The modern periodic table of elements contains

Eighteen main groups.

The problem concerning an interesting effect observed with specific non-Newtonian fluids investigating the impact of shear stress on the flow behavior of non-Newtonian fluids.

In this experiment, a non-Newtonian fluid is placed in a cylindrical container and its flow behavior is observed when a shear stress is applied to its surface which is created by a rotating paddle wheel which is placed at the bottom of the container. The paddle wheel is rotated at different speeds to increase the shear stress applied to the non-Newtonian fluid.

The results of this experiment demonstrate that when the shear stress is increased, the viscosity of the non-Newtonian fluid decreases. This decrease in viscosity causes the non-Newtonian fluid to flow more easily and with less resistance. This effect can be seen when the paddle wheel is rotated at higher speeds; the non-Newtonian fluid flows more quickly and with less effort.

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The mass of the meter stick is 543.6g

What is mass?

Mass is the measure of an object's resistance to acceleration. It is a fundamental property of matter and is usually expressed in kilograms (kg). Mass is related to weight, which is the measure of an object's gravitational force.

Mass of meter stick = Mass of 52.0 g mass × (Distance moved by fulcrum) / (Distance of 52.0 g mass from fulcrum)

Mass of meter stick = (52.0 g × (39.2 cm - 49.7 cm)) / (12.5 cm - 49.7 cm)

Mass of meter stick = (52.0 g × (-10.5 cm)) / (-37.2 cm)

Mass of meter stick = 543.6 g

hence, The mass of the meter stick is 543.6g

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A hydrogen atom is in its ground state (nᵢ = 1) when a photon impinges upon it. The atom absorbs the photon, which has precisely the energy required to raise the atom to the nf = 3 state. (a) The photon's energy that was absorbed is approximately 1.51 eV (negative sign indicates absorption).(b)option B and C are correct.

To determine the photon's energy and the energies of photons that might be emitted when the hydrogen atom returns to the ground state, we can use the energy level formula for hydrogen atoms:

E = -13.6 eV / n^2

where E is the energy of the electron in the atom, and n is the principal quantum number.

(a) To find the energy of the photon that was absorbed by the hydrogen atom to raise it from the ground state (nᵢ = 1) to the nf = 3 state, we need to calculate the energy difference between the two states:

ΔE = Ef - Ei = (-13.6 eV / 3^2) - (-13.6 eV / 1^2)

Calculating the value of ΔE:

ΔE = -13.6 eV / 9 + 13.6 eV

= -1.51 eV

Therefore, the photon's energy that was absorbed is approximately 1.51 eV (negative sign indicates absorption).

(b) When the hydrogen atom returns to the ground state, it can emit photons with energies corresponding to the energy differences between the excited states and the ground state. We need to calculate these energy differences and check which values are present among the given options.

ΔE1 = (-13.6 eV / 1^2) - (-13.6 eV / 3^2) = 10.20 eV

ΔE2 = (-13.6 eV / 1^2) - (-13.6 eV / 4^2) = 10.20 eV

ΔE3 = (-13.6 eV / 1^2) - (-13.6 eV / 5^2) = 12.10 eV

ΔE4 = (-13.6 eV / 1^2) - (-13.6 eV / 6^2) = 12.10 eV

ΔE5 = (-13.6 eV / 1^2) - (-13.6 eV / 7^2) = 13.55 eV

ΔE6 = (-13.6 eV / 1^2) - (-13.6 eV / 8^2) = 13.55 eV

ΔE7 = (-13.6 eV / 1^2) - (-13.6 eV / 9^2) = 13.55 eV

Comparing the calculated energy differences with the given options:

(A) 1.89 eV: This energy difference does not match any of the calculated values.

(B) 12.1 eV: This energy difference matches ΔE3 and ΔE4.

(C) 10.2 eV: This energy difference matches ΔE1 and ΔE2.

(D) 13.6 eV: This energy difference does not match any of the calculated values.

Therefore option B and C are correct.

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Answer:

personality disorder

Explanation:

Certain family dynamics, such as abusive parenting, can increase the susceptibility of someone developing a(n) personality disorder.

A disorder is diagnosed when these personality traits are significantly “deviated” from others and are very inflexible, impairing a person's adaptation to everyday situations and causing distress and discomfort in their personal relationships and other areas of life.

Personality disorders are generally pervasive and persistent patterns of thinking, perceiving, reacting, and relating that cause significant distress or functional impairment.

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The passband gain for the given first-order circuit is 5, and the cutoff frequency is 15,000 rad/s.

To find the passband gain and cutoff frequency of the voltage transfer function T(s) = 5s/(s + 15,000), follow these steps:

1. Passband gain: In a first-order circuit, the passband gain is simply the coefficient of 's' in the numerator, which is 5 in this case.
2. Cutoff frequency: The cutoff frequency is the value of 's' when the denominator is equal to the numerator. In this case, set 5s = s + 15,000, which gives s = 15,000 rad/s as the cutoff frequency.

To plot the Bode gain response using MATLAB, use the following code:

```MATLAB
numerator = [5 0];
denominator = [1 15000];
sys = tff(numerator, denominator);
bode(sys);
```


This code creates a transfer function object 'sys' with the given numerator and denominator coefficients, then uses the 'bode()' function to plot the Bode gain response.

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electromagnet

When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.

Answer:the sample of solid has less energy than the sample of gas

Explanation:

Apex

Answer:

Illegal drugs are dangerous because once someone starts doing drugs they go insane for more and their demand for my drugs can be so dangerous that they would even sometimes harm the people close to them and, In some cases people start to hate them selves and some even killed themselves.

A bar magnet that holds a chain of nails illustrates "a. magnetic induction."

Magnetic induction occurs when a magnetic field induces magnetism in a nearby material. In this case, the magnetic field produced by the bar magnet induces magnetism in the nails, causing them to become temporarily magnetized and stick to the magnet. This is due to the alignment of the magnetic domains in the nails with the magnetic field of the bar magnet, creating an attraction force between them. This phenomenon is also known as ferromagnetism, which is the ability of certain materials to become magnetized in the presence of a magnetic field.

A magnetic field is the area surrounding a magnet where its magnetic force is exerted. In this scenario, the bar magnet produces a magnetic field that affects the nails. Magnetic induction refers to the process where a magnet's magnetic field induces magnetism in nearby ferromagnetic materials, such as the nails. As a result, the nails become temporarily magnetized and attract each other, forming a chain.

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In order to find the horizontal distance from the release point at which the raft strikes the water, we need to apply the kinematic equations of motion. First, let's find the time it takes for the raft to hit the water. We can use the equation for displacement: s =\(ut + 1/2 at^2\).

Since the raft is only moving in the horizontal direction, the acceleration is zero and we can use the following equation to find the time it takes for the raft to hit the water: s =\(u t t\) = \(s/u\) . Since the initial horizontal velocity of the raft is 120 m/s,  

We can use the equation for displacement again: s = \(u t + 1/2 at^2\).In this case, the acceleration is the acceleration due to gravity, g = 9.8 m/s^2. The initial vertical velocity is zero since the raft is released from rest in the vertical direction. Thus, we have: s = \(0 + 1/2 * 9.8 * (10/3)^2s\) = 163.333 m Therefore, the horizontal distance from the release point at which the raft strikes the water is\(120 * 10/3\) = 400 m.

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Answer:

The unique physical properties, including a high heat of vaporization, strong surface tension, high specific heat, and nearly universal solvent properties of water are also due to hydrogen bonding.

To draw a free-body diagram, identify all the forces acting on an object and draw them including their magnitude and direction. Use a point to represent the object.

a)

While the ball is flying through the air, the only force acting on the ball is its weight (due to gravitational attraction to the Earth).

b)

There are three forces acting on the cart: its weight, the pull to the right and the rolling friction to the left:

c)

There are three forces acting on the cart: The gravitational pull downwards, the push upwards of the air that compensates the weight, and the air resistance to the right (since the cart is traveling toward the left).

The force exerted by the object having a mass of 5 kg is 10.79 N

The mass of the object, m = 5 kg

The velocity of the object from rest, v = 47.5 m/s

The time taken, t = 22 s

The formula to find the force exerted on the object is

              F = ma

where m is the mass of the object

           a is the acceleration of the object

But we know that

          a = v/t

where v is the velocity of the object

           t is the time taken

So, let us substitute the equation of the acceleration in the force equation, we get

          F = mv/t

Now, let us substitute the known values in the above equation,

        F = 5 x 47.5 / 22

           = 237.5 / 22

           = 10.79

Therefore, the force exerted by the object is 10.79 N

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At the lowest point of her swing, she can withstand a maximum speed of 10.78 m/s.

Given that,

Mass of the carpenter = 61.7 kg

Length of the rope = 11.5 m

Capable force = 1229 N

Centripetal force acting on the body,

F = mv²/r = (61.7× v²)/11.5 = 5.37 v²

Gravitational force acting on her is

F = m × g = 61.7 × 9.81 = 605.28 N

By summing up gravitational and centripetal forces to get the total available force,

5.37 v² + 605.28 = 1229

5.37 v² = 623.72

v² = 116.15

v = 10.78 m/s

Hence, the maximum speed at the low point of her swing is 10.78 m/s.

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Answer:

2 m/s2

Explanation:

the answer is 2 m/s2

Answer:

A 1.4Km

Explanation:

angle of inclination= \(8^0\).

distance d = \(10.0 Km\).

                                    \(d_{y}=d sin(\theta)\\d_{y}=10 sin(8^0)\\d_{y}=1.4 Km\)

Answer:

A 1.4

Explanation:

A DVD that is initially at rest, begins to turn with a constant angular acceleration of 5.11 rad/s².

At 2.4 s, what is the angular displacement of the disc in degrees? When the DVD begins to turn with a constant angular acceleration of 5.11 rad/s², we can use the formula;θ = ω₀t + 1/2 αt²Where;θ = angular displacementω₀ = initial angular velocity of 0 rad/sα = angular acceleration of 5.11 rad/s²t = 2.4 s

Using the formula above;θ = 0 + 1/2 (5.11 rad/s²) (2.4 s)²θ = 0 + 1/2 (5.11 rad/s²) (5.76 s²)θ = 14.7 rad Therefore, the angular displacement of the disc in degrees is;θ (in degrees) = (θ in radians) × (180°/π)θ (in degrees) = (14.7 rad) × (180°/π)θ (in degrees) = 842.6°Therefore, the angular displacement of the disc in degrees is 842.6°.

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The unusually bright centers found in some galaxies are called "active galactic nuclei" (AGN).

These centers are characterized by their extreme brightness, which is due to the intense emission of energy across the electromagnetic spectrum, from radio waves to X-rays. The source of this energy is believed to be the accretion of mass onto supermassive black holes at the heart of these galaxies. When the matter surrounding the black hole is drawn towards it, it forms an accretion disk, which heats up due to friction and emits radiation. The emitted radiation can be so powerful that it outshines the entire galaxy, making the AGN easily observable from great distances. Examples of galaxies with active galactic nuclei include quasars, blazars, and Seyfert galaxies.

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Answer:

1.5 m/s²

Explanation:

Applying,

a = (v-u)/t ...................... Equation 1

Where a = acceleration of the particle, v =  final velocity of the particle, u = Initial velocity of the particle, t = time.

From the question,

Given: u = 0 m/s (From rest), v = 15 m/s, t = 10 seconds

Substitute these values into equation 1

a = (15-0)/10

a = 15/10

a = 1.5 m/s²

0.03M is  the concentration of sodium hydroxide.

Here from calculation concentration of NaOH -

100×0.019/ 20 = .095 M

For acetic acid-

Volume acid used, V1= 25 ml

Concentration of acetic acid N1= ?

Volume of NaOH used, V2 = 8.99 ml

Concentration of NaOH used, N2= .095 M

From N1×V1= N2×V2

N1 = N2×V2 / V1

N1 = .095 × 8.99/25

N1 = 0.03M

concentration is the quantity of a material in a certain area. The ratio of the solute in a solution to the solvent or whole solution is another way to define concentration. In order to indicate concentration, mass per unit volume is typically used. The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume. Although concentration is often used to describe chemical solutions, it may be computed for any combination.

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A commercial motor vehicle’s (CMV’s) engine controls are used for a variety of purposes such as managing fuel consumption, controlling speed, and maximizing engine performance.

Engine controls are an essential component of CMVs as they determine how the engine operates, which ultimately impacts the vehicle's overall performance and safety. Engine controls are primarily used to regulate fuel injection and the air/fuel mixture to ensure efficient combustion and optimum engine performance. They also help to manage the transmission system, providing the driver with smooth gear transitions and optimal acceleration.

Engine controls also manage the exhaust gas recirculation (EGR) system, which helps to reduce emissions and improve fuel economy. Other important features of CMV engine controls include cruise control and speed limiting systems. Cruise control allows the driver to set a specific speed, which the vehicle then maintains without any additional input. Speed limiting systems are designed to restrict the maximum speed of the vehicle, providing additional safety for drivers and other road users.Finally, engine controls include safety features such as automatic shutdown systems, which automatically switch off the engine if the vehicle is idling for an extended period or if the driver forgets to turn off the ignition. This prevents accidents and helps to save fuel.

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The engine controls of a commercial motor vehicle (CMV) regulate and manage the performance of the engine. Throttle control, ignition control, and fuel system control are examples of engine controls in a CMV.

The engine controls of a commercial motor vehicle (CMV) are used to regulate and manage the performance of the engine. These controls include various components and systems that enable the driver to control the speed, power, and operation of the vehicle's engine.

Examples of engine controls in a CMV may include:

These engine controls work together to optimize engine performance, fuel efficiency, and emissions in a commercial motor vehicle.

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