Which statement is true about divorce in the United States? `



A. The divorce rate is lower for second marriages than first marriages.



B. Divorce usually leads to a higher income and higher standard of living.



C. Similarities between spouses guarantee that they will not divorce.



D. About half of first marriages end in divorce.

Explanation:

The primary to secondary voltage ratio is given as 4:1 .

That is

                 [tex]\frac{V_p}{V_s}=\frac{4}{1}=4[/tex]

We need to find what is the secondary voltage if the primary voltage is 460 volts.

That is

                    [tex]V_p=460V[/tex]

Substituting in ratio equation

                    [tex]\frac{V_p}{V_s}=4\\\\\frac{460}{V_s}=4\\\\V_s=\frac{460}{4}\\\\V_s=115V[/tex]

The secondary voltage is 115 V.                  

Final answer:

To find the secondary voltage in a transformer with a primary to secondary voltage ratio of 4:1 and a primary voltage of 460 volts, divide the primary voltage by 4, resulting in a secondary voltage of 115 volts.

Explanation:

To solve for the secondary voltage in a transformer when given a voltage ratio of 4:1 and a primary voltage of 460 volts, we can use the transformer equation.

This equation states the ratio of the secondary voltage (Vs) to the primary voltage (Vp) is equal to the ratio of the number of turns in the secondary coil (Ns) to the number of turns in the primary coil (Np).

In this question, the primary to secondary voltage ratio is 4:1. If the primary voltage (Vp) is 460 volts, we can calculate the secondary voltage (Vs) by dividing the primary voltage by the ratio, which in this case is 4. Therefore, Vs = 460 / 4 = 115 volts.

Answer:

A.

Explanation:

A PERT chart is a project management tool that provides a graphical representation of a project's timeline. The Program Evaluation Review Technique (PERT) breaks down the individual tasks of a project for analysis. But PERT charts are considered preferable to Gantt charts because they identify task dependencies, but they're often more difficult to interpret.

Answer:

lithosphere and asthenosphere

Explanation:

Lithosphere is the part of the earth surface which is rigid and hard. It consists of whole crust and the upper mantle.It comprises of number of plates called plate tectonic.

The asthenosphere is the Earth's upper mantle's extremely viscous, mechanically fragile and ductile area. This sits below the lithosphere, some 80 to 200 km below the ground at depths. The boundary between lithosphere and asthenosphere is commonly referred to as LAB.

Answer: sensory, motor, associative

Explanation:

Answer:

Normal Magnetic Polarity

Explanation:

Normal Magnetic polarity.

The polarity can be either normal or reversed Natural polarity is where the north magnetic points (roughly) towards the north pole. ... The reversed polarity is in the opposite direction and the northern end of the magnetic field is near the current southern pole.

Answer:

[tex]\frac{R}{1} = \frac{44}{9}\ohm[/tex]

Explanation:

Let us imagine that there are three wire of length equal length having equal resistances each of 44/3 Ω

Now connect these wires in parallel to so that their equivalent resistance is R.

then

[tex]\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]

[tex]\frac{1}{R} = \frac{3}{44}+\frac{3}{44}+\frac{3}{44}[/tex]

[tex]\frac{1}{R} = \frac{9}{44}[/tex]

⇒[tex]\frac{R}{1} = \frac{44}{9}\ohm[/tex]

Answer:

4.89 Ω

Explanation:

we know that resistance is directly proportional to length. hence as the wire is cut in three pieces, the resistance of each piece becomes one-third of the original resistance of the wire.

[tex]R[/tex] = Resistance of wire = 44 Ω

[tex]r[/tex] = resistance of each piece

Resistance of each piece  is given as

[tex]r = \frac{R}{3}\\r = \frac{44}{3}[/tex]

The three pieces are connected in parallel,

[tex]R_{p}[/tex] = Resistance of parallel combination of three pieces

Resistance of parallel combination is given as

[tex]\frac{1}{R_{p}}= \frac{1}{r} +  \frac{1}{r} +  \frac{1}{r} \\\frac{1}{R_{p}}= \frac{3}{r}\\R_{p}= \frac{r}{3}\\R_{p} = \frac{\frac{44}{3} }{3}\\R_{p} = \frac{44}{9} \\R_{p} = 4.89 ohm[/tex]

Applying Newton's laws of motion, the water droplets will continue in the direction they were moving at the moment of detachment, thus will travel in straight-line paths. This assumes that the effect of gravity is negligible.

The subject of this question lies in the vicinity of Physics, specifically, the principles of Newton's laws of motion and projectile motion. To answer the question, when the dog shakes and the droplets leave the fur, the motion of these droplets will be in straight-line paths (Option B).

Let's consider why this happens. The movement of the water droplets is set in motion by the dog's shake, and upon detachment from the fur, the droplets will continue in the direction they were moving at the moment of detachment, following Newton's first law of motion. As we're assuming negligible gravity during this short time, the droplets will not fall towards the earth, hence move in straight-line paths rather than parabolic arcs.

It's important to remember the core tenets of Newton's laws of motion. An object at rest will stay at rest, and an object in motion will continue in a straight line unless acted upon by an external force.

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Answer:

We can explain it in term of center of mass and torque. When the sprinters start their race then the ground exerts the force towards the center of mass and thats how the torque can be prevented from acting on the sprinter.In this way the crouched position allows them to accelerate quickly without tipping over backward.

Explanation:

Answer:

3 Hz

Explanation:

A beat is the pattern of interference among two waves with frequencies which are different.

The difference of the frequencies of the waves gives us the beat frequency.

Here, one frequency is [tex]f_1=440\ Hz[/tex], the other frequency is [tex]f_2=443\ Hz[/tex]

So, the beat frequency

[tex]\Delta f=|f_1-f_2|\\\Rightarrow \Delta f=|440-443|\\\Rightarrow \Delta f=3\ Hz[/tex]

The beat frequency that is produced is 3 Hz

The beat frequency is the difference between two interfering frequencies. In this case, it's the absolute difference between the violin string's frequency (443 Hz) and the electronic tuner's frequency (440 Hz), which results in a beat frequency of 3 Hz.

In the context of music and acoustics, a beat frequency is the pulsating sound that results when two slightly different frequencies interfere with each other. It can be calculated by taking the absolute difference between the two frequencies. So when a violinist is tuning his violin and compares a string's frequency (443 Hz) with that of an electronic tuner (440 Hz), the beat frequency produced would be 443 Hz - 440 Hz = 3 Hz.

This means that the violinist would hear an oscillating sound 3 times per second due to the interference of the two frequencies. Just like a piano tuner, the violinist would then adjust the tuning of the string until this beat frequency is minimized, signifying that the string's frequency matches that of the tuner.

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Answer:

B. 10 cubic inches

Explanation:

Boyle's Law: States that the volume of a given mass of gas is inversely proportional to its pressure provided that temperature is constant.

Mathematically, Boyle's law can be expressed as

P₁V₁ =P₂V₂......................... Equation 1

making V₂ the subject of formula in the equation above,

V₂ = P₁V₁/P₂ ..................... Equation 2

P₁ =initial pressure, V₁ = initial volume, P₂ = final pressure, V₂ = final Volume.

Given: P₁ = 5 psi, V₁ = 20 cubic inches, P₂ = 10 psi.

Substituting these values into equation 2

V₂ = (5 × 20)/10

V₂  = 10 cubic inches.

Therefore the new volume = 10 cubic inches

The right option is B. 10 cubic inches.

Answer: 1.34A

Explanation:

Since the wire will break if it exceeds 1.90A current, this will be the maximum current needed.

The root mean square current

Irms = I0/√2

Where I0 is the maximum current given

Irms = 1.90/√2

Irms = 1.34A

Answer:

450N

Explanation:

1) F=ma F2=50kg*3m/s^2=150N

2)Ff=F1-F2=600N-150N=450N

A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². Friction force acting on the cart is 450 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question a student is pushing a 50 kg cart, with a given force of [tex]F_1[/tex] = 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s²,

F = ma

[tex]F_2[/tex] = 50 kg*3 m/s²

     =150 N

Frictional force,

[tex]F_f[/tex] = [tex]F_1 - F_2[/tex]

   = 600 N-150 N

   = 450 N

A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². Friction force acting on the cart is 450 N.

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a) The acceleration in absence of friction is [tex]3.5 m/s^2[/tex]

b) The acceleration in presence of friction is [tex]0.6 m/s^2[/tex]

Explanation:

a)

First of all, we notice that the crate accelerates only along the horizontal direction, since the forces in the vertical direction are balanced (in fact, the weight is balanced by the normal reaction + the vertical component of the applied force).

This means that we can consider only the horizontal component of the applied force, which is

[tex]F_x = F cos \theta = (280)(cos 35^{\circ})=229.4 N[/tex]

where

F = 280 N is the applied force

[tex]\theta=35^{\circ}[/tex] is the angle

Now we can find the acceleration of the crate, by using Newton's second law:

[tex]F_x = ma[/tex]

where

[tex]F_x = 229.4 N[/tex] is the net force along the horizontal direction

m = 65 kg is the mass of the crate

a is the acceleration

Solving for a, we find

[tex]a=\frac{F_x}{m}=\frac{229.4}[65}=3.5 m/s^2[/tex]

b)

In this case, we also have the force of friction, so we have to compute the force of friction.

We start by writing the equation of the forces along the vertical direction:

[tex]R+F sin \theta - mg = 0[/tex]

where

R is the normal reaction

[tex]F sin \theta[/tex] is the vertical component of the applied force

[tex]mg[/tex] is the weight

We can re-write this as

[tex]R=mg-Fsin \theta[/tex] (1)

Now we can write the equation of the forces along the horizontal direction

[tex]F cos \theta - \mu_k R = ma[/tex] (2)

where

[tex]F cos \theta[/tex] is the horizontal component of the applied force

[tex]\mu_K R[/tex] is the force of friction, with [tex]\mu_k = 0.4[/tex] is the coefficient of friction

By substituting (1) into (2) and solving for a, we find the new acceleration of the crate:

[tex]F cos \theta - \mu_k (mg-F sin \theta) = ma\\a=\frac{F cos \theta - \mu_k mg +\mu_k F sin \theta}{m}=\frac{(280)(cos 35)-(0.4)(65)(9.8)+(0.4)(280)(sin 35)}{65}=0.6 m/s^2[/tex]

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Answer:

3.53 m/s^2

Explanation:

Use the net x value.

The question relates to Simple Harmonic Motion of a mass-spring system, and asks to calculate the period of the motion. The approach involves the use of Hooke's law and Newton's second law, and understanding that at maximum amplitude, gravitational potential energy converts to elastic potential energy.

The question discusses a concept in physics known as Simple Harmonic Motion, specifically in relation to a mass attached to a spring. The object of unknown mass is hung on the unstretched spring, let go and then allowed to move down under the force of gravity. This object comes to rest for the first time when it has moved to its maximum amplitude, which is given as 2.75 cm. The period of the motion is what we are required to find.

The period of oscillation for a simple harmonic oscillator (in this case, a mass-spring system) can be calculated using the formula T = 2π√(m/k), where m is the mass and k is the spring constant. However, from the given information, we do not have the mass or the spring constant and need an alternative approach.

Let's use Hooke's law (F = kx) and Newton's second law (F = ma). We know that at the maximum amplitude, all the gravitational potential energy is converted to elastic potential energy. We can write mgx = 0.5 kx^2, where g is the acceleration due to gravity and x is the displacement (2.75 cm). From here we can calculate k. Then we substitute this value of k obtained into the formula for T, we could then find the period of the motion of the object.

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Answer:

D

Explanation:

According to research and data, the average cost for a severe injury in a collision is $247,000.

These include traumatic brain injury, severe damage to limbs which could result to loss of the limbs, spinal cord injuries that could result to partial or total paralysis as well as internal damage to organs in the collision.

Answer:

Angle of reflection will be [tex]40^{\circ}[/tex]

Explanation:

We have given angle of incidence from the smooth surface is [tex]40^{\circ}[/tex]

So angle of incidence = [tex]40^{\circ}[/tex]

We have to find the angle of reflection

From l;aw of reflection we know that angle of reflection is equal to the angle of incidence

So angle of reflection will be equal to [tex]40^{\circ}[/tex]

Answer:

40 degrees

Explanation:

got it right

Explanation:

It is known that entropy is the degree of randomness. More is the speed of molecules of a substance more will be its degree of randomness. Hence, more will be the entropy of substance.

When we place a teaspoon of sugar in the bottom of a glass of water, it will dissolve completely over time. This dissolution will cause the sugar molecules to spread out in the water.

And, with time water will evaporate and this will lead to the formation of crystals again. This reappearance of sugar crystals will cause a decrease in entropy as there occurs decrease in randomness.

But this decrease in entropy is balanced by increase in the entropy of water molecules as they are evaporating.

The process of sugar dissolving and reappearing in water can be explained in terms of entropy.

The process described in the question can be explained in terms of entropy, which is a measure of the disorder or randomness in a system. When the teaspoon of sugar is placed in the water, the sugar crystals break down and mix with the water molecules, increasing the entropy of the system. With time, the water molecules continue to move and mix, increasing the disorder even further.

However, if the water is left to evaporate, the water molecules will escape into the air, and the sugar molecules will not be able to dissolve in the evaporating water. As a result, the sugar crystals will eventually reappear.

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The electric potential difference is the electric potential energy per unit charge

Explanation:

First of all, we define the concept of electric potential. The electric potential is a measure of the gradient of the electric field at a certain point of the space. The electric potential at a distance [tex]r[/tex] from a positive charge of magnitude [tex]q[/tex] is given by

[tex]V(r) = \frac{kq}{r}[/tex]

where k is the Coulomb's constant.

Now we can define the electric potential energy and the electric potential difference:

The electric potential energy is also defined as the work done on a charge q moved through a potential difference of [tex]\Delta V[/tex]. Consequently, the potential difference [tex]\Delta V[/tex] represents the work per unit charge done, i.e. the work done when moving a unitary charge through a potential difference [tex]\Delta V[/tex].

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Final answer:

Electric potential energy is related to the energy of a charge's position in a field, whereas electric potential difference, or voltage, relates to the work required to move a charge between two points in a field.

Explanation:

Electric potential energy is a measure of the potential energy stored in a system of charges based on their positions, much like gravitational potential energy in a system of masses. On the other hand, electric potential difference, often referred to as voltage, is the work needed to move a charge between two points in an electric field without producing an acceleration. An analogy that can be helpful in differentiating the two is thinking of electric potential energy as the energy associated with the charge's position, similar to a rock perched on a hill, while electric potential difference relates to the effort required to move a rock up or down that hill.

The most reliable form of identifying potentially effective reinforcers is known as Preference Assessment. It is an adaptive procedure vastly used in the field of reinforcers.

Explanation:

Preference Assessment identifies items that are likely to be effective as reinforcers by identifying a particular learner's preference for them.

Answer:

Gravitational potential energy, kinetic energy and mechanical energy

Explanation:

The body is at a displacement to the gravity and therefore has energy stored in it as a result of its displacement against gravity. When this body is released the gravitational potential energy is converted to translational kinetic energy as well as rotational kinetic energy as it rolls down the the ramp. The sum of the kinetic energies and potential energy of the body amount to the mechanical energy of the ball down the ramp.

Answer:

e. design programming

Explanation:

The planning techniques are responsible for structuring the tasks to be performed within the project, defining the duration and the order of execution of the same, while the programming techniques try to organize the activities so that the logical temporal relationships between them, determining the calendar or the moments of time in which each one must be realized. The programming must be consistent with the objectives pursued and respect existing restrictions (resources, costs, workloads).

The programming therefore consists in setting, in an approximate way, the moments of beginning and termination of each activity. Some activities may have slack and others are critical activities (fixed over time).

STEPS:

Build a time diagram (moments of beginning and slack of activities).

Establish the times of each activity.

Analyze project costs and adjust clearances (minimum cost project).

Final answer:

When equal forces of 3.0 newtons are applied on both ends of a spring with an equilibrium length of 2.0 meters, the resulting length of the spring remains 2.0 meters because the forces cancel each other out.

Explanation:

To find the resulting length of the spring, we should first determine the amount of stretch or compression caused by the forces applying on the spring. Since Alice and Bob are both pulling with a force of 3.0 newtons each in opposite directions, the net force on the spring is zero. Therefore, the spring neither stretches nor compresses from its equilibrium length.

The spring has an equilibrium length of 2.0 meters and remains at this length because the forces applied by Alice and Bob cancel each other out. So, the resulting length of the spring when 3.0 newtons is applied at each end in opposite directions is 2.0 meters, which is answer (B).

The spring stiffness constant of the fisherman's scale, based on Hooke's Law, is 653.3 N/m. If a fish is pulled down further by 2.1 cm, it will oscillate with an amplitude of 5.7 cm and a frequency of approximately 0.816 Hz.

To answer part (a) of your question about what the spring stiffness constant is, we use Hooke's Law. This law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This relationship is formally shown as F = k * x, where F is the force (in this case, the weight of the fish influences this force), k is what we are looking for (the spring stiffness constant), and x is the displacement (how much the spring stretches due to the force).

In this problem, the fish's weight becomes the force, which you can calculate as mass * gravity, or 2.4 kg * 9.8 m/s² = 23.52 N. The scale stretches 3.6 cm, or 0.036 m. Substituting these into Hooke's law gives us 23.52 N = k * 0.036 m, and solving for k provides a value of approximately 653.3 N/m. Thus, the spring stiffness constant is 653.3 N/m.

As for part (b) of the question, determining the amplitude and frequency of the oscillation if the fish is pulled down a further 2.1 cm and released, we know that the amplitude simply the displacement from the equilibrium position. Hence, total displacement from equilibrium is 3.6 cm (initial stretching) + 2.1 cm (additional pull) = 5.7 cm, or 0.057 m. This is the amplitude of the oscillation.

For the frequency of the oscillation, we need to understand the fish-spring system as a simple harmonic oscillator. The frequency (f) of a simple harmonic oscillator system is given by f = 1 / (2π) √(k/m), where k is the spring constant and m is the mass. Substituting our known values gives a frequency of approximately 0.816 Hz. Therefore, the oscillations will have an amplitude of 5.7 cm and a frequency of 0.816 Hz.

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When a star is moving away from an observer at a constant speed, the wavelengths of the absorption lines in its spectrum appear to shift to longer wavelengths. This is known as the redshift.

When a star is moving away from an observer at a constant speed, the wavelengths of the absorption lines in its spectrum appear to shift to longer wavelengths. This phenomenon is known as the redshift. Redshift occurs because the motion of the star causes the wavelengths of light to stretch as the star moves away, just like the stretching of sound waves when a source moves away from an observer.

As the star gets farther and farther away, the amount of redshift increases, meaning the wavelengths of the absorption lines will shift even further towards the longer end of the spectrum. This shift in wavelengths can be measured using spectroscopy and is an important tool in determining the motion and distance of celestial objects. One example of redshift is the expansion of the universe itself, where galaxies are moving away from each other due to the expansion of space. The redshift of the light from distant galaxies provides evidence for this cosmic expansion.

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The wavelengths of absorption lines from a star moving away at a constant speed increase, exhibiting a redshift due to motion.

As a star moves away at a constant speed, the wavelengths of its absorption lines shift toward the red end of the spectrum. This is known as "redshift." It occurs because of the Doppler effect, where the motion of the source (star) affects the observed wavelength of light. When the star moves away, the observed wavelengths of light become longer, causing a redshift.

The Doppler effect explains that when an object is moving away from an observer, the waves it emits, including light, get stretched out, causing a shift to longer wavelengths. This redshift is used in astronomy to measure the recessional velocity of celestial objects, helping us determine the expansion of the universe and the distances between objects.

Redshift also plays a crucial role in the Hubble's law, which is fundamental in cosmology for understanding the universe's expansion.

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Answer:

C: look left then right then left again

Explanation:

At an intersection, there three routes, the left, the right and straight ahead.

For a car stopped at an intersection, once the light turn green, it is advisable that the driver should watch the left and right then left again before accelerating. This is because there might be any car coming from any of the sides unnoticed which might leads to accident especially the first side u looked at (left) must be take much cognisance of hence the need to look left again before accelerating.

Final answer:

When stopped at an intersection and your light turns green, you should A. look right and then left before accelerating.

Explanation:

When stopped at an intersection and your light turns green, you should look left, then right, then left again before accelerating. This practice ensures that you are aware of any vehicles that may not have stopped at their red light or are trying to beat a yellow light, as well as any pedestrians or cyclists that may be crossing.

Remember that while traffic signals are there to regulate flow, driver vigilance is crucial in preventing accidents and ensuring safety. Thus, it is important to check both directions to ensure the intersection is clear before proceeding. Safe driving practices include being cautious and aware of potential risks on the road.

Answer:

The force of impact is same for both objects.

Explanation:

It is given that,

Speed of the car, v = 100 km/h

It is understood that the mass of car is more than that of the bug. When one object strikes another object, they both applies a force on each other. One object applies the same force on another object when they strikes as per Newton's third law of motion.

In this case, when a car strikes a bug, the force of impact is the same for both. Hence, the correct option is (c).

According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, the force of impact when a car strikes a bug is the same for both. However, due to its smaller mass, the impact has a much larger effect on the bug.

The subject of this question is Physics and it relates to the principle of Newton's Third Law which states that 'for every action, there is an equal and opposite reaction'. When the car and the bug collide, they apply forces to each other of equal magnitude and opposite direction. Thus, the force of impact is actually the same for both, as suggested by option C.

However, the consequences of this force are more notable for the bug due to its much smaller mass compared to the car. The high acceleration experienced by the bug (according to Newton's second law 'Force = Mass x Acceleration') causes it to splatter, while the car continues largely unaffected.

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Answer:

TRUE.   The weight component is responsible for acceleration and therefore the increase in vehicle speed,

Explanation:

Let's write Newton's second law for the car down a slope

Y Axis

         N-Wy = 0

         N = Wy

X axis

        Wx -fr= m a

Using trigonometry

        sin θ = Wx / W

        cos θ = Wy / W

         Wx = W sin θ

         Wy = W cos θ

We replace

             W sin θ - fr = m a

The weight component is responsible for acceleration and therefore the increase in vehicle speed,

fr is the force of car brakes

Consequently, of the above TRUTH

Answer:

a.V(t/365)=30,000(0.89)^(t/365).

b. V(5)=$17000

Explanation:

An automobile is depreciating at 11% per year, every year. A $30,000 car depreciating at this rate can be modeled by the equation V(t) = 30,000(0.89)t. What is an equivalent equation for this vehicle at a daily depreciation and what is it worth (rounded to the nearest thousand dollar) 5 years after purchase?

i will like to work back from te question by solving questin 2 first

V(t) = 30,000(0.89)^t

let t=5

V(5) = 30,000(0.89)^5

V(5)=$17000

2.yearly depreciation will be

30,000(0.89)^1

26700=it amounts to tis after a year,

so it depreciates by 3300 for one year

daily depreciation will be 3300/365 ,

since there are 365 days in a year

$9.041

the equation that satisfy this is

V(t/365)=30,000(0.89)^(t/365).

Final answer:

Mathematically, an automobile depreciating at 11% annually can be modeled for daily depreciation, but for the calculation of value after 5 years, we use the original annual model V(t) = 30,000(0.89)^t to find the car is worth approximately $13,000.

Explanation:

The student's question about the depreciation of an automobile can be classified under Mathematics, specifically dealing with exponential decay and depreciation models. To find the daily depreciation rate equivalent to 11% annual depreciation, we can use the formula for continuous compounding, where the annual rate is converted into a daily rate using the expression (1 + annual rate)^(1/365) - 1. However, since the problem does not ask for continuous daily depreciation but rather the equivalent discrete daily rate, we can instead divide the annual rate by 365, assuming each day the car depreciates by an equal fraction of the annual rate. Therefore, the daily depreciation rate would be 11% / 365.

However, because the student is seeking the value of the car after 5 years, we can directly use the provided equation V(t) = 30,000(0.89)^t, where t is the number of years. After 5 years, this gives us V(5) = 30,000(0.89)^5. Calculating this we get V(5) ≈ $13,000 when rounded to the nearest thousand.

Answer:

3.33 x 10⁻⁵ T

Explanation:

given,

current carried by two parallel wire = 25 A

distance between two wire = 13 cm

1) Let p be the distance between two wire

         p = 13 cm = 0.13 m

         a = 10 cm = 0.1 m

         b = 6 cm = 0.06 m

 using cosine law

 [tex]p^2 = a^2 + b^2- 2ab cos\theta[/tex]

 [tex]cos\theta =\dfrac{a^2+b^2}{2ab}[/tex]

 [tex]\theta =cos^{-1}(\dfrac{0.1^2+0.06^2}{2\times 0.1 \times 0.06})[/tex]

       θ = 105.96°

       θ = 106°(approximately)

now, magnetic field on one wire

[tex]B_1 = \dfrac{\mu_oI}{2\pi r}[/tex]

[tex]B_1 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.1)}[/tex]

      B₁ = 5 x 10⁻⁵ T

now, magnetic field on another wire

[tex]B_2= \dfrac{\mu_oI}{2\pi r}[/tex]

[tex]B_2 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.06)}[/tex]

      B₂ = 8.33 x 10⁻⁵ T

resultant field,

= [tex]\sqrt{B_1^2+B_2^2 - 2 B_1B_2 cos \theta}[/tex]

= [tex]\sqrt{(5\times 10^{-5})^2+(8.33\times 10^{-5})^2 - 2\times 5 \times 10^{-5}\times 8.33 \times 10^{-5} cos 106^0}[/tex]

= 3.33 x 10⁻⁵ T

1) The magnitude of the magnetic field vector at the given point is; -3.195 * 10⁻⁵ T

2) The direction of the magnetic field vector at the given point is;  129.71° from the x-axis

We will start with finding the direction of the field for each wire from the tangent to the circle around the wire.

For their magnitudes, we have;

B₁ = (µ₀/4π) * 2I₁/L₁

B₁ = (4π * 10⁻⁷/4π) * 2 * 25/0.1

B₁ = 5 * 10⁻⁵ T

B₂ = (µ₀/4π) * 2I₂/L₂

B₂ = (4π * 10⁻⁷/4π) * 2 * 25/0.06

B₂ = 8.33 * 10⁻⁵ T

If we draw a vector diagram, we will have;

B = B₁(-sin θ₁ i + cos θ₁ j) + B₂(-sin θ₂ i + cos θ₂ j)

B = (5 * 10⁻⁵ )(-sin 26.34 i + cos 26.34 j) + (8.33 * 10⁻⁵)(-sin 47.7 i + cos 47.7 j)

B = -(8.38 * 10⁻⁵) i + (10.09 * 10⁻⁵) j

Direction of the field is;

θ = tan⁻¹((10.09 * 10⁻⁵)/-(8.38 * 10⁻⁵)

θ = 129.71° from the x-axis

Thus, magnitude is;

B_mag = B₁ * cos 129.71°

B_mag = -3.195 * 10⁻⁵ T

Read more about magnetic field at; https://brainly.com/question/7802337

Answer:

Both objects will undergo the same change in velocity

Explanation:

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of object

Any object which is falling has only the acceleration due to gravity.

[tex]ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2[/tex]

The acceleration due to gravity on Earth is 9.81364 m/s²

So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.

Hence, both objects will undergo the same change in velocity.