What is causing the boat to move toward the shore?
A
The moving water applies a force on the boat, which results in the boat pushing
back on the water, propelling the boat forward.
B
The moving wind applies a force on the boat, which results in the boat pushing
back on the wind, propelling the boat forward.
с
The boy applies a force on the oars, which results in the water pushing back on
the oars, propelling the boat forward.
D
The boy applies a force on the oars, which results in gravity pushing back on the
oars, propelling the boat forward.

Answer:

Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.

Explanation:

To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.

Given:

m1 = 8 kg

x1 = 3 m

x_cm = unknown (to be found)

m2 = 10 kg

y_cm = 4.5 m

Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.

To solve for x2, we can rearrange the formula as follows:

x2 = (x_cm * (m1 + m2) - m1 * x1) / m2

Substituting the given values:

x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg

Simplifying:

x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg

Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:

4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)

Simplifying:

4.5 m = (24 kg + 10 kg * x2) / 18 kg

Multiplying both sides by 18 kg:

81 kg*m = 24 kg + 10 kg * x2

Subtracting 24 kg from both sides:

10 kg * x2 = 81 kg*m - 24 kg

Dividing both sides by 10 kg:

x2 = (81 kg*m - 24 kg) / 10 kg

Simplifying:

x2 = 8.1 m - 2.4 m

x2 = 5.7 m

(brainlest?) ples:(

Answer:

the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.

Explanation:

To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.

The center of mass of a system is given by the equation:

x_cm = (m1x1 + m2x2) / (m1 + m2),

where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.

Given:

m1 = 8 kg,

x1 = 3 m,

m2 = 10 kg,

y_cm = 4.5 m.

To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:

y_cm = (m1y1 + m2y2) / (m1 + m2),

where y1 and y2 are the positions along the y-axis.

Rearranging the equation and substituting the given values:

4.5 = (83 + 10y2) / (8 + 10).

Simplifying the equation:

4.5 = (24 + 10*y2) / 18.

Multiplying both sides by 18:

81 = 24 + 10*y2.

Rearranging the equation:

10*y2 = 81 - 24,

10*y2 = 57.

Dividing both sides by 10:

y2 = 5.7.

Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.

To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2).

Substituting the given values:

x_cm = (83 + 10x2) / (8 + 10).

Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:

0 = (83 + 10x2) / (8 + 10).

Rearranging the equation:

83 + 10x2 = 0.

24 + 10*x2 = 0.

10*x2 = -24.

Dividing both sides by 10:

x2 = -2.4.

Here are five potential-kinetic energy conversions that could be shown in the construction of a device: Pendulum, Roller Coaster, Wind-up Toy, Elastic Slingshot, Windmill.

Pendulum: A pendulum consists of a weight attached to a string or rod, suspended from a fixed point. When the weight is lifted to a certain height, it possesses gravitational potential energy.

As the weight is released, it swings back and forth, converting the potential energy into kinetic energy. At the highest point of each swing, the weight briefly comes to a stop and has maximum potential energy, which is then converted back to kinetic energy as it swings downward.

Roller Coaster: In a roller coaster, potential-kinetic energy conversions occur throughout the ride. When the coaster is pulled up to the top of the first hill, it gains gravitational potential energy.

As the coaster descends, the potential energy is converted into kinetic energy, resulting in a thrilling and high-speed ride. Subsequent hills and loops continue to convert potential energy into kinetic energy and vice versa as the coaster moves along the track.

Wind-up Toy: Wind-up toys typically have a spring mechanism inside. When the toy is wound up, potential energy is stored in the wound-up spring. As the spring unwinds, it transfers its potential energy into kinetic energy, causing the toy to move or perform actions. The kinetic energy gradually decreases as the spring fully unwinds.

Elastic Slingshot: With an elastic slingshot, potential-kinetic energy conversions are evident when the slingshot is stretched. As the user pulls back on the elastic band, potential energy is stored.

Windmill: Windmills harness the kinetic energy of the wind and convert it into other forms of energy. As the wind blows, it imparts kinetic energy to the blades of the windmill. The rotating blades then transfer this kinetic energy into mechanical energy, which can be used for various purposes such as grinding grains or generating electricity.

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Grace Scholars Program is a scholarship program that selects the best and brightest students for the program.

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For a longitudinal wave, the compression region, is the one represented by the densely packed particles with high pressure and the region with loosely packed particles is called rarefaction. Hence, the first part is C he dense region and the second one with some space between the dots is labeled as R.

Longitudinal waves are a type of mechanical waves passing through a medium. Unlike electromagnetic waves, they cannot be passed through vacuum.

In a longitudinal wave, the oscillation of particles is along the direction of wave propagation. The wave is composed of high pressure regions and low pressure regions called compressions and rarefactions respectively.

The regions where, particles are densely packed and shows the thick dote are labelled as compressions and the regions where, some space between particles are labeled as rarefactions.

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Answer:

The greater weight increases the terminal velocity by acting as an extra force against gravity and air resistance.

Hellow!

For this use the next formula:

Vf = Vo + gt

Initial velocity is zero, so the formula simplificate:

Vf = gt

Data:

Vf = Final velocity = ?

g = Gravity = 9.8 m/s²

t = Time = 12 s

Replacing according our data:

Vf = 9.8 m/s² * 12 s

Vf = 117.8 m/s

The final velocity will be 117.8 meters per second.

The speed of the 5.30-kg object when it has fallen 2.50 m is 3.03 m/s.

In the scenario, the two objects (m1 = 5.30 kg and m2 = 3.50 kg) are connected by a light string passing over a light, frictionless pulley. When the 5.30-kg object is released from rest at a point

h = 4.00 m

above the table, we have to find the tension, acceleration, and speed of the 5.30-kg object. The following are the answers to parts a, b, and c of the given question.a. What is the tension in the string?The tension in the string is equal to the weight of the objects. Therefore, we can calculate the tension in the string as:

$$T = (m_1 + m_2)g$$

Substitute the values to get:Tension

T = (5.30 kg + 3.50 kg) × 9.8 m/s² = 82.6 NAns

The tension in the string is 82.6 N.b. What is the acceleration of the 5.30-kg object?We can calculate the acceleration using the formula:

$$a = \frac{m_1 - m_2}{m_1 + m_2}g$$

Substitute the values to get:Acceleration

a = (5.30 kg - 3.50 kg) / (5.30 kg + 3.50 kg) × 9.8 m/s² = 1.84 m/s²Ans:

The acceleration of the 5.30-kg object is 1.84 m/s².c.

What is the speed of the 5.30-kg object when it has fallen 2.50 m?We can use the formula to calculate the final velocity of an object with constant acceleration. The formula is

v² = u² + 2aswhere:v = final velocityu = initial velocity = 0as = acceleration × distance = 1.84 m/s² × 2.50 m = 4.6 m/s²Substitute the values to get:

v² = 0² + 2 × 1.84 m/s² × 2.50 mv² = 9.2 m²/s²v = √9.2 m²/s² = 3.03 m/sAns:

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Answer:

The atomic number of the element in this notation is 15.

Explanation:

The X notation for an atom provides information about its atomic number and mass number. The atomic number is the number of protons in the nucleus of an atom, which determines its chemical properties and identity as an element. The mass number is the sum of the number of protons and neutrons in the nucleus.

In the notation "32 on 15 X", the number on the top (32) represents the mass number of the atom, which is the total number of protons and neutrons in the nucleus. The number on the bottom (15) represents the atomic number of the atom, which is the number of protons in the nucleus.

Therefore, the atomic number of the element in this notation is 15.

This is an experiment that explains the Bernoulli principle. According to this principle, at the bottom of the spoon, there will be a reduction in pressure.

So the pressure at the bottom of the spoon reduces because the velocity of the water is larger and the height of the column of water is smaller at this part.

So the spoon is attracted to the water at the point of lowest pressure because objects gravitate to low-pressure points.

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The battery would have to supply 7.00 μC of charge to the capacitor.

The energy stored in the capacitor in this case is 7.98 μJ.

The battery would have to supply 4.05 mC of charge to store 1.30 J of energy in the capacitor.

The potential difference across the capacitor would be 20.2 V.

The charge supplied by the battery can be calculated using the formula Q = CV,

where Q is the charge, C is the capacitance, and V is the potential difference.

Thus,

Q = (5.00 μF)(1.40 V)

Q = 7.00 μC.

The energy stored in a capacitor can be calculated using the formula:

E = 1/2 CV^2,

where E is the energy, C is the capacitance, and V is the potential difference.

Thus,

E = 1/2 (5.00 μF)(1.40 V)^2

E = 7.98 μJ.

The charge required to store a certain amount of energy in a capacitor can be calculated using the formula:

Q = √(2CE),

where Q is the charge, C is the capacitance, and E is the energy.

Thus,

Q = √(2(5.00 μF)(1.30 J))

Q  = 4.05 mC.

The potential difference across a capacitor can be calculated using the formula V = √(2E/C), where V is the potential difference, C is the capacitance, and E is the energy.

Thus, V = √(2(1.30 J)/(5.00 μF))

V = 20.2 V

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The football left the kicker's foot with an initial vertical speed of 19.62 m/s.

The initial vertical velocity of a football kicked straight up into the air can be calculated using the kinematic equation:
v_f = v_i + a*t
where v_f is the final velocity (0 m/s in this case, as the football stops at its highest point before falling), v_i is the initial velocity (which we want to find), a is the acceleration due to gravity (-9.81 m/s^2, negative because it acts downward), and t is the time taken to reach the highest point.
First, we need to determine the time taken to reach the highest point, which is half of the total time (4 seconds):
t = 4 seconds / 2 = 2 seconds
Now we can substitute the values into the equation:
0 m/s = v_i - 9.81 m/s^2 * 2 seconds
Rearrange the equation to solve for v_i:
v_i = 9.81 m/s^2 * 2 seconds = 19.62 m/s
Therefore, the football left the kicker's foot with an initial vertical speed of 19.62 m/s.

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All substance listed in the chart is made up of the atoms.

Atom is smallest entity of a substance. Body is made up of atoms. it is basic building block of a body. An atom consist of electrons, protons and neutrons as sub atomic particle. whole mass of the atom is concentrated at the center of the atom which we call it as nucleus, nucleus consist of proton and neutron. Electron revolve around the nucleus at determined(fixed) orbit. Total number of protons in the atom decides the atomic number and the elements in the periodic table. The electrons which are completely filled orbitals are called as core shell electrons and which are not filled completely are called as valence electron. valence electrons are responsible for physical and chemical properties of the element. Elements which are on same column in periodic table have same number of valence electrons . Hence they have same properties.

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Explanation:

Rate  X  Time  =   Distance

Distance  / Time  = Rate

225 km / 3.35 hr = 67.2 km/hr

Answer:

The decay constant, or "lambda" (λ), is the rate at which a radioactive isotope decays. It is usually measured in units of inverse time, such as seconds. In this case, the decay constant can be calculated as follows:

16:42

λ = (ln(2)/3.57 x 106) x (5.78 x 1017) = 0.

Explanation:

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

\(v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2\)

Net force, F = ma

\(F=1600\times 1.56\\\\F=2496\ N\)

So, the net force is 2496 N.

The amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N

HOW TO CALCULATE NET FORCE:

Where:

Since a = 1.563m/s²

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The acceleration can be calculated using the formula:

Acceleration = (Final velocity - Initial velocity) / Time taken

Given that the bus starts from rest, the initial velocity is 0 m/s.

Acceleration = (20 m/s - 0 m/s) / 21 sec = 20/21 m/s².

The distance travelled at maximum speed can be calculated by subtracting the distances covered during acceleration and deceleration from the total distance.

Distance during acceleration = (1/2) * acceleration * time² = (1/2) * (20/21 m/s²) * (21 sec)² = 210 m.

Distance during deceleration is the same as distance during acceleration.

Distance travelled at maximum speed = Total distance - 2 * distance during acceleration = 270 m - 2 * 210 m = -150 m.

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The vector product of A=2.00i+3.00j+1.00k and B=1.00i-3.00j-2.00k is C=9.00i+4.00j-9.00k.

To find the vector product (also known as the cross product) of two vectors, A and B, we can use the following formula:

C = A × B

Where C is the resultant vector, A and B are the given vectors, and × denotes the cross product.

Given A = 2.00i + 3.00j + 1.00k and B = 1.00i - 3.00j - 2.00k, we can substitute these values into the formula to find the vector product:

C = (2.00i + 3.00j + 1.00k) × (1.00i - 3.00j - 2.00k)

Now, let's expand the cross product using the properties of vector products:

C = (2.00i × 1.00i) + (2.00i × -3.00j) + (2.00i × -2.00k) +

   (3.00j × 1.00i) + (3.00j × -3.00j) + (3.00j × -2.00k) +

   (1.00k × 1.00i) + (1.00k × -3.00j) + (1.00k × -2.00k)

Now, let's calculate each of these cross products:

C = (2.00 × 1.00) \(i^2\) + (2.00 × -3.00) i × j + (2.00 × -2.00) i × k +

   (3.00 × 1.00) j × i + (3.00 × -3.00) \(j^2\) + (3.00 × -2.00) j × k +

   (1.00 × 1.00) k × i + (1.00 × -3.00) k × j + (1.00 × -2.00) \(k^2\)

Since i × j = k, j × k = i, and k × i = j, we can simplify the expression further:

C = 2.00k - 6.00i + 4.00i - 9.00j + k - 3.00j - 2.00j - 2.00k

Combining like terms, we get:

C = (2.00i + 4.00i) + (-6.00i - 9.00j - 3.00j) + (2.00k + k - 2.00k)

Simplifying further:

C = 6.00i - 12.00j + k

Therefore, the vector product of A and B is C = 6.00i - 12.00j + k, which can be written as C = 9.00i + 4.00j - 9.00k in terms of i, j, and k.

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The vector product of A and B is -3i - 5j - 9k.

The vector product, also known as the cross product, of two vectors A and B is denoted as A x B. It is a vector that is perpendicular to both A and B. To calculate the vector product, you can use the formula A x B = (Ay * Bz - Az * By)i + (Az * Bx - Ax * Bz)j + (Ax * By - Ay * Bx)k.

In this case, we have A = 2.00i + 3.00j + 1.00k and B = 1.00i - 3.00j - 2.00k. Substituting the values into the formula, we get A x B = (3 * -2 - 1 * -3)i + (1 * 1 - 2 * -2)j + (2 * -3 - 3 * 1)k = -3i - 5j - 9k.

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Answer:

Meeee

Explanation:

Answer:

a) The average force that acts on the man is \(2.508\times 10^{8}\) newtons.

b) The average force that acts on the man is 1023.673 newtons.

c) The force of the ground on the man is 1612.093 newtons upwards.

Explanation:

a) After a careful reading of the statement we construct the following model by applying Impact Theorem, that is:

\(m\cdot \vec v_{A} + \vec F \cdot \Delta t = m\cdot \vec v_{B}\) (Eq. 1)

Where:

\(m\) - Mass of the man, measured in kilograms.

\(\vec v_{A}\) - Initial velocity of the man, measured in meters per second.

\(\vec v_{B}\) - Final velocity of the man, measured in meters per second.

\(\Delta t\) - Impact time, measured in seconds.

\(\vec F\) - Average net force, measured in newtons.

Now we proceed to clear average net force within expression:

\(\vec F \cdot \Delta t = m\cdot (\vec v_{B}-\vec v_{A})\)

\(\vec F = \frac{m}{\Delta t}\cdot (\vec v_{B}-\vec v_{A})\) (Eq. 2)

If we know that \(m = 60\,kg\), \(\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right]\), \(\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right]\) and \(\Delta t = 1\times 10^{-6}\,s\), we obtain the following vector:

\(\vec F = \frac{60\,kg}{1\times 10^{-6}\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right]\)

\(\vec F = 2.508\times 10^{8}\,\hat{j}\,\,\,[N]\)

The average force that acts on the man is \(2.508\times 10^{8}\) newtons.

(b) If we know that  \(m = 60\,kg\), \(\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right]\), \(\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right]\) and \(\Delta t = 0.245\,s\), we obtain the following vector:

\(\vec F = \frac{60\,kg}{0.245\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right]\)

\(\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right]\)

The average force that acts on the man is 1023.673 newtons.

(c) From Second Newton's Law we find the following equation of equilibrium:

\(\vec F = \vec N -\vec W\) (Eq. 3)

Where:

\(\vec F\) - Average force that acts on the man, measured in newtons.

\(\vec N\) - Force of the ground on the man, measured in newtons.

\(\vec W\) - Weight of the man, measured in newtons.

By applying the concept of weight, we expand the previous equation:

\(\vec F = \vec N -m\cdot \vec g\) (Eq. 3b)

Where \(\vec g\) is the gravitational acceleration, measured in meters per square second.

And then we clear the force of the ground on the man:

\(\vec N = \vec F +m\cdot \vec g\) (Eq. 4)

If we get that \(\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right]\),  \(m = 60\,kg\) and \(\vec g = 9.807\,\hat{j}\,\,\,\left[\frac{m}{s^{2}} \right]\), the average force is:

\(\vec N = 1023.673\,\hat{j}\,\,\,[N]+(60\,kg)\cdot (9.807\,\hat{j})\,\,\,\left[\frac{m}{s^{2}} \right]\)

\(\vec N = 1612.093\,\hat{j}\,\,\,\left[N\right]\)

The force of the ground on the man is 1612.093 newtons upwards.

ANSWER:

2.7*10^7 N

STEP-BY-STEP EXPLANATION:

The pressure is given by the following equation:

We can solve for F which is the force, but we must know the area (A), which we can calculate with the given data, like this:

We substitute and calculate the force as follows:

Therefore, the force is equal to 2.7*10^7 N

Answer:

I don't know

Explanation:

Please give me Brainliest answer

Answer:

PRESSURE ???

Answer:

The acceleration of the train is 1.581 m/s² inward.

Explanation:

Given;

initial velocity of the train, u = 86.0 km/h = 23.889 m/s

final velocity of the train, v = 56.0 km/h = 15.556 m/s

change in time, Δt = 18 s

The total acceleration of particles moving along a curved path is given as vector sum of the tangential acceleration and radial acceleration

\(a = \sqrt{a_t^2 + a_r^2}\)

where;

\(a_t\) is the tangential acceleration

\(a_r\) is radial acceleration

\(a_t = \frac{v-u}{t} \\a_t = \frac{15.556-23.889}{18} \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)\)

\(a_r = \frac{v^2}{r} \\\\a_r = \frac{15.556^2}{160} \\\\a_r = 1.512 \ m/s^2\)

\(a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{(-0.463)^2+(1.512)^2} \\\\a = \sqrt{2.5005} \\\\a = 1.581 \ m/s^2\)

Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.

Answer:Magnetic fields from two sources add up as vectors at each point, so the strength of the field is not necessarily the sum of the strengths1. Magnetic fields are vectors, which means they have direction as well as size. Therefore, the sum of two magnetic fields is not simply the sum of their magnitudes2.

Explanation:

Answer:

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i am not

Explanation:

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Answer:

the relative density of the gravel in the field is 0.7365 or 73.65%

Explanation:

Given that;

wet density of gravel \(r_{b}\) = 2.3 Mg/m³

field water content w = 16%

density of solids in lab \(r_{s}\) = 2.70 Mg/m³

Maximum void ratio \(e_{max}\) = 0.59

Minimum void ratio \(e_{mini}\) = 0.28

first we determine the dry density of gravel  \(r_{d}\) =  \(r_{b}\) / 1+ w

 \(r_{d}\) = 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³

we know that;  \(r_{w}\) = 1000 kg/m³ = 1 g/cm³

Specific Gravity of soil G = \(r_{s}\)  / \(r_{w}\)  = 2.70 Mg/m³ / 1 = 2.70 Mg/m³  

eo = ((G×\(r_{w}\))/\(r_{d}\)) - 1

eo = (( 2.70 × 1) / 1.9827 ) - 1

eo = 1.3617 - 1

Co = 0.3617

so Relative density \(I_{D}\) will be;

\(I_{D}\) = \(e_{max}\)  - eo  / \(e_{max}\) - \(e_{mini}\)

we substitute

\(I_{D}\) = 0.59  - 0.3617 / 0.59 - 0.28

\(I_{D}\) = 0.2283 / 0.31

\(I_{D}\) = 0.7365 or 73.65%

Therefore;  the relative density of the gravel in the field is 0.7365 or 73.65%

The linear distance and the tangential velocity of the end of the stick are 1.63 m and 10.9 m/s respectively.

Rotational motion is the motion of a body around a circular path, in a fixed orbit. The dynamics for rotational motion are entirely analogous to linear or translational dynamics. The equations for rotating objects are similar to the motion equations for linear motion.

Given, the length of the hockey stick, r = 1.1 m

The hockey stick completes one swing in time, t = 0.15 sec

The angle θ = 85° = 1.483 rad

The average angular velocity of the hockey stick,

ω = θ/t = 1.483rad/0.15s = 9.9 rad/s

The linear distance moved by the end of the hockey stick:

L = rθ = 1.1× 1.483 = 1.63 m

The tangential velocity of the end of the hockey stick:

v = rω = 1.1 × 9.9 = 10.9 m/s

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The velocity of the car 1 can be seen from the calculation as 20 m/s West

A coordinate system or point of view used to observe motion is known as a frame of reference. It can be used as a guide when describing an object's position, speed, and acceleration. Different frames of reference may result in various motion observations.

The relative velocity is the velocity of an object or observer as observed from a particular frame of reference.

We can see that the velocity of the car 1 is;

45 m/s - 25 m/s

= 20 m/s West

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