A motorcycle rider loses control while making a sharp turn on a wet road through the mountains. Dude skids off the road at 30.7 m/s and lands in the valley, 191 m below the road. How long does it take him to hit the valley floor

The acceleration of a 50 g object pushed with a force of 0.5 N is 10 m/s².

To find the acceleration of the object, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:

F = m * a

Given:

Force (F) = 0.5 N

Mass (m) = 50 g = 0.05 kg

Substituting the given values into the equation, we have:

0.5 N = 0.05 kg * a

To find the acceleration (a), we rearrange the equation:

a = F / m

a = 0.5 N / 0.05 kg

a = 10 N/kg

Since acceleration is measured in meters per second squared (m/s²), we convert the unit of N/kg to m/s²:

1 N/kg = 1 m/s²

Therefore, the acceleration of the 50 g object pushed with a force of 0.5 N is 10 m/s².

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Answer:

a    \(D = 3.9 *10^{-12} \ m\)

b    \(\tau_{max} = 1.24 *10^{-25} \ N\cdot m\)

c   \(W = 2.48 *10^{-25} J\)

Explanation:

From the question we are told that

   The magnitude of electric dipole moment is  \(\sigma = 6.2 *10^{-30} \ C \cdot m\)

     The electric field is \(E = 2*10^{4} \ N/C\)

The distance between the positive and negative charge center is mathematically evaluated as

     \(D = \frac{\sigma }{10 e}\)

Where  e is the charge on one electron which has a constant value of  \(e = 1.60 *10^{-19} \ C\)

  Substituting values

     \(D = \frac{6.20 *10^{-30}}{10 * (1.60 *10^{-19})}\)

      \(D = 3.9 *10^{-12} \ m\)

The maximum torque is mathematically represented as

       \(\tau_{max} = \sigma * E * sin (\theta)\)

Here  \(\theta = 90^o\)

This because at maximum the molecule is perpendicular to the field

    substituting values

       \(\tau_{max} = 6.2 *10^{-30} * 2*10^{4} sin ( 90)\)

       \(\tau_{max} = 1.24 *10^{-25} \ N\cdot m\)

The workdone is mathematically represented as

      \(W = V_{(180)} - V_{0}\)

where   \(V_{(180)}\) is the potential energy at 180° which is mathematically evaluated as

     \(V_{(180) } = - \sigma * E cos (180)\)

Where the negative signifies that it is acting against the  field

   substituting values

     \(V_{(180) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (180)\)

      \(V_{(180) } = 1.24*10^{-25} J\)

and

     \(V_{(0)}\) is the potential energy at 0° which is mathematically evaluated as

            \(V_{(0) } = - \sigma * E cos (0)\)

   substituting values

     \(V_{(0) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (0)\)

      \(V_{(0) } =- 1.24*10^{-25} J\)

So \(W = 1.24 *10^{-25} - [-1.24 *10^{-25}]\)

    \(W = 2.48 *10^{-25} J\)

Answer:

Explanation:

Mmmmmmm.

Answer:

50

Explanation:

500/10 = 50

The subatomic particle that is not a part of the nucleus which stays bound to the atom is the electron and is to maintain relative stability.

This is defined as a sub atomic particle which is negatively charged and revolve around the nucleus. In an atom the number of electron is usually equal to the number of the proton which is positively charged and present in the nucleus.

The electron not being in the nucleus is a a result of it wanting to maintain stability because its presence will produce a wavelength which will break the nucleus apart.

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The velocity of propagation is v = 3.836 m/s ±  5%

Data;

The velocity of the flow is calculated as

\(v_h = \sqrt{2gh}\)

h = height of the fluid from the orifice

h = 0.75 ± 10%m

\(v = \sqrt{2*9.81*0.75}\\g = 9.81 m/s^2\\v = 3.836 m/s\)

The uncertainty or error in calculating velocity

\(v = \sqrt{2gh}\)

The uncertainty = 1/2 * error in 'h'

\(\frac{1}{2} * 10 \% = +5 \%\)

The uncertainty = ± 5%

The velocity of propagation is v = 3.836 m/s ±  5%

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Answer:

54.5 kmph

Explanation:

From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle

ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and

f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m

ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle

So,

ΔK = -(f₁d₁ + f₂d₂)

1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)

1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)

v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)

v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)

v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]

substituting the values of the variables into the equation, we have

v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]

v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]

v₀ = √[229.36 (m/s)²

v₀ = 15.14 m/s

v₀ = 15.14 × 3600/1000

v₀ = 54.5 kmph

So, the initial speed of the vehicle is 54.5 kmph

The principal stresses and corresponding principal directions for the stresses in a problem can be found using various methods, one of which is Mohr's circle.

What is Mohr's circle?

Mohr's circle is a graphical representation of a two-dimensional stress state. Given the stress tensor components, the Mohr's circle can be used to visualize and determine the maximum and minimum normal stresses (i.e., the principal stresses) and the orientation of the planes on which they act (i.e., the principal directions).

It is important to note that finding the principal stresses and directions requires a thorough understanding of stress analysis, including stress transformation and the use of Mohr's circle. If you have access to the full problem statement, I would be happy to help you work through the solution.

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Pressure builds under the surface of Earth, causing mountains to form.The correct answer is option D.

Uplift refers to the geological process that elevates the Earth's surface, resulting in the formation of mountains. This process is primarily driven by tectonic forces, including the movement and collision of Earth's lithospheric plates.

When two plates converge, immense pressure builds up beneath the surface, causing the crust to buckle and fold. This deformation leads to the formation of mountains, as rocks are pushed upward and displaced vertically.

As the uplift process continues over millions of years, mountains gradually take shape. Erosion and weathering play significant roles in shaping their features, but it is the initial uplift that initiates the formation of mountains.

As the Earth's surface is elevated, a wide range of landforms can emerge, including rugged peaks, deep valleys, and steep slopes.

Uplift has a profound impact on the Earth's surface and ecosystems. Mountains alter local climates, influencing precipitation patterns and creating variations in temperature and wind patterns.

Therefore, uplift plays a crucial role in shaping the Earth's surface and influencing various geological, biological, and climatic processes.

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Answer:

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Explanation:

The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:

\((m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}\) (1)

Where:

\(m_{1}\), \(m_{2}\), \(m_{3}\) - Masses of the first, second and third fragments, in kilograms.

\(\vec v_{o}\) - Initial velocity of the object, in meters per second.

\(\vec v_{1}\), \(\vec v_{2}\), \(\vec v_{3}\) - Velocities of the first, second and third fragments, in meters per second.

If we know that \(m_{1} = 0.5\,kg\), \(m_{2} = 1.3\,kg\), \(m_{3} = 1.2\,kg\), \(\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right]\), \(\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right]\) and \(\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right]\), the velocity of the third fragment is:

\((-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)\)

\(1.2\cdot \vec v_{3} = (1.4,1.95)\)

\(\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right]\)

The speed of the third fragment is the magnitude of the result found above:

\(v_{3} = 2\,\frac{m}{s}\)

And the direction of the third fragment is:

\(\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)\)

\(\theta_{3} \approx 54.316^{\circ}\)

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

The graph of the slope's acceleration depicts velocity vs time.

reveals the solution.

The distance an object has travelled over time is displayed on a distance-time graph.

A straight-line graph of the time versus distance results is shown. The Y-axis displays the distance. The X-axis displays a time plot. Velocity changes occur more quickly on the graph with the sharpest slope.

It picks up speed the fastest. The acceleration of an item is shown by the slope of a velocity graph. As a result, the value of the slope at a particular time determines the item's acceleration at that specific instant.

A moving item is shown by a sloped line on a distance-time graph. The object's speed is equal to the gradient or slope of the line on a distance-time graph.

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A C and D

C because ofc when electric charge flow electric field is produced

Answer:

A & D

Explanation:

The efficiency of steam engine is  26.66 %.

The efficiency is defined as the work done by the engine divided by the heat supplied.

Work done is the difference between the heat supplied and heat rejected.

So, efficiency η = (Qs -Qr) / Qs

Given is the heat supplied Qs = 225 Joules and heat rejected Qr =285 . then the work done is

W = 225 -225 = - 60 Joules

Efficiency η  = 60/225 x 100%

η = 26.66 %

Thus, the efficiency of the steam engine is  26.66 %.

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When materials vibrate, waves are created that travel through the substance, and this energy is what we hear as sound. Earthquakes are earth vibrations that cause the (potential) energy held within rocks to be released (as a result of their pressure-generating relative positions). Seismic waves are produced by earthquakes.

How do sound waves and earthquakes compare?

The waves lose energy as they move through the air with sound or through the ground with shaking during an earthquake. Therefore, a band can be heard louder close to the stage than farther away, and an earthquake can be felt more strongly close to the fault than farther away.

In actuality, sound in the air cannot match how quickly earthquake waves move. In rock, the compressional or "P" wave of an earthquake moves at the In actuality, sound in the air cannot match how quickly earthquake waves move. The speed of a P wave is typically 10,000 mph. The speed of sound through air is roughly 750 mph.

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There are 90 phones of volume, 10-7 W/m2 of sound intensity, and 0.0314 watts of source power.

A straightforward frequency analysis compares the values of the fields you provide and generates a report listing each value for those fields along with the frequency at which each value occurs.

The rate at which a sound power wave repeats itself, also known as frequency or pitch, is measured in cycles per second. Bullfrog calls and cricket chirps have lower frequencies than drum beats and whistles, respectively.

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Answer:

B. Pre-contemplation

Explanation:

Hope this helps :)

The steps/guidelines for preventing quackery and fraud when shopping are :

Quackery involves the claim of knowledge and provision of a fake and dishonest service to unsuspecting customers or clients and this is very common in medicine.

Fraud is the false representation of a material or service in order to cause loss of money or pain to an unsuspecting client or customer. some steps to prevent fraud are as follows :

Hence we can conclude that The steps/guidelines for preventing quackery and fraud when shopping are as listed above

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The distance covered by a ball in the first second of free fall is approximately 4.9 meters.

When an object falls freely under the influence of gravity, it experiences constant acceleration. In the case of Earth's gravity, the acceleration due to gravity is approximately 9.8 m/s². This means that the velocity of the falling object increases by 9.8 meters per second every second.

To determine the distance covered by the ball in the first second, we can use the equations of motion for uniformly accelerated motion.

The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:

d = ut + (1/2)at²

In this case, the initial velocity is zero (as the ball starts from rest), the acceleration is 9.8 m/s², and we want to find the distance covered in the first second (t = 1 second).

Plugging in the values:

d = 0 * 1 + (1/2) * 9.8 * (1)^2

d = 0 + (1/2) * 9.8

d = 0 + 4.9

d = 4.9 meters

Therefore, the ball covers a distance of approximately 4.9 meters in the first second of free fall.

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Answer:

 ΔV = -2.1 L

Explanation:

To solve this exercise we can use the ideal gas equation for two points

         PV = nRT

          P₁V₁ = P₂ V₂

where point 1 is on the surface and point 2 is at the desired depth,

          V₂ = \(\frac{P_1}{P_2} \ V_1\)

let's calculate

           V₂ = ( \(\frac{1 atm}{2.5 atm}\) ) 3.5 L

            V₂ = 1.4 L

this is the new volume, the change in volume is

          ΔV = V₂ -V₁

          ΔV = 1.4-3.5

          ΔV = -2.1 L

The momentum of the monkey of mass 5 kg is 25 kgm/s.

Momentum is the product of mass and velocity.

To calculate the momentum of the monkey, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the momentum of the monkey is 25 kgm/s.

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Complete question: A 5kg monkey is running with a velocity of 5 m/s to the right. Find the momentum of the monkey

Answer:

0%

Explanation:

There is no purple geneotype with either parents so there will be 0 purple and 100% white

Answer:

One effective procedure for dissolving a cube of sugar in water quickly using equipment available in a standard chemistry lab is:

Boil a specific amount of water in a beaker or flask on a hot plate or Bunsen burner.

Once the water has reached boiling point, remove it from the heat source and add the sugar cube.

Stir the mixture vigorously using a stirring rod or magnetic stirrer until the sugar has completely dissolved.

If necessary, cool the solution by placing the beaker or flask in an ice bath or allowing it to cool naturally to room temperature.

This procedure is effective because the solubility of sugar in water increases with temperature. By heating the water, the energy of the water molecules is increased, which allows them to break apart the sugar molecules more easily. Stirring the mixture further increases the contact between the sugar and the water, which facilitates the dissolution process. The cooling step is necessary to prevent recrystallization of the dissolved sugar as the solution cools down, which would result in a slower dissolution process. Overall, this procedure can dissolve a sugar cube in water quickly and efficiently using equipment readily available in a standard chemistry lab.

Explanation:

Given data

*The given mass of the dummy is m = 75 kg

*The given net force on the dummy is F_n = 825 N

The formula for the dummy's acceleration is given as

Substitute the known values in the above expression as

Hence, the dummy's acceleration is a = 11 m/s^2

When the thistle funnel is moved upwards towards the surface of the liquids, the height (h) decrease.

What happens when the thistle funnel is moved upward?

An important principle outlined by Pascal's law states that any fluid (in this instance, liquids) experiences equal transmission of force in every direction within it. Hence, there is an intimate relationship between the height of a liquid column and its pressure.

Raising a thistle funnel reduces this height causing loss of weight from above this point thus decreasing its pressure.

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See complete question in the attached image.

Answer:

projectile motion

total time of flight= 2V° Sintheta / g

answer is 3second

Hope This would help you!

Explanation:

it's 5 + 5 + 5 + 12 + 5 x 7 x 7 7+ 8 5+6+ +8 +9 + 9 5

Answer:

The horizontal distance between two adjacent crests or troughs is known as the wavelength.

Answer: Wavelength

Explanation:

From crest to crest, it is one full wavelength

Answer:

The correct difference between chronic and acute stress is:

Acute stress is short-term, while chronic stress endures over time.

Explanation:

Acute stress refers to the immediate and temporary response of the body to a specific stressful event or situation. It is often characterized by a rapid increase in heart rate, heightened alertness, and the release of stress hormones like adrenaline. Acute stress is a normal and natural response to perceived threats or challenges, and once the stressor is removed or resolved, the body returns to its normal state.

On the other hand, chronic stress is long-term and persists over an extended period. It is typically caused by ongoing or recurring stressors, such as work pressures, financial difficulties, relationship problems, or chronic health conditions. Chronic stress can have a cumulative and prolonged impact on physical and mental well-being. It may lead to a range of health issues, including cardiovascular problems, weakened immune system, digestive disorders, anxiety, depression, and burnout.

Chronic stress is considered detrimental to overall health, while acute stress, when experienced in moderation, can actually be beneficial as it can enhance performance and help individuals deal with immediate challenges. It is important to manage chronic stress effectively through stress-reducing techniques, self-care practices, and seeking support when needed to prevent its negative consequences on health and well-being.

The angle between vector A and B is 90 degrees.

Let’s consider the vectors A and B with equal magnitudes of 5.

Thus, we can assume that the angle between A and B is θ.

Now, if the sum of A and B is vector 6j, then we can write the vector equation below:

\(A + B = 6j\)

Let's first express vector A and B in terms of their components.

Let’s assume that vector A has components Ax and Ay, while vector B has components Bx and By.

Thus, the vector equation above can be rewritten in terms of components below:

\(Ax + Bx = 0\)       ------------(1)

\(Ay + By = 6\)       ------------(2)

We know that the magnitudes of vectors A and B are equal to 5, thus we can write the following equations:

\(x^2+y^2 = 5^2\)       ------------(3)

\(x^2+y^2 = 25\)      ------------(4)

From equation (4), we can write:

\(y = \sqrt{(25-x^2)}\)      ------------(5)

By substituting equation (5) into equation (2), we have:

\(Ay + By = 6Ay + \sqrt{(25-x^2)} = 6Ay\)

\(= 6 - \sqrt{(25-x^2)}\)    ------------(6)

We can now substitute equation (6) into equation (3):

\(x^2 + (6-\sqrt{(25-x^2)})^2 = 25\)

Solving for x:

\(2x^2 + 12\sqrt{(25-x^2)} = 0x^2 + 6\sqrt{(25-x^2)}\)

\(= 0x^4 + 36x^2 - 225\)

\(= 0x^2\)

\(= 3, -15\)

We choose \(x^2 = 3\) since it is positive and obtain:

\(y = 4By\)

substituting x and y into the original equations \(A = Ax + Ay\) and

\(B = Bx + By\), we obtain:

\(A = (-\sqrt{3}, 6-\sqrt{3})\)

\(B = (\sqrt{3}, -1+\sqrt{3})\)

Thus, the dot product of vectors A and B can be expressed as:

\(A \cdot B = |A||B| cos(\theta)\)

Now, we know that

\(|A| = |B| = 5\)

thus we can write:

\(A \cdot B = 25 cos(\theta)\)

Using the formula above, we can obtain the angle θ between vectors A and B as follows:

\(cos(\theta) = \frac{(A \cdot B)}{|A||B|}\)

\(cos(\theta) = \frac{[(\sqrt{3}(-\sqrt{3}) + (6-\sqrt{3})(-1+\sqrt{3})] }{(5)(5)}\)

\(cos(\theta) = \frac{ [-3 + 3]}{25}\)

\(cos(\theta) = 0\)

\(\theta= cos^{-1}(0)\)

\(\theta =90^{\circ}\)

Therefore, the angle between vector A and B is 90 degrees.

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