Hw2b. 8. Position from polar velocity a particle starts at time at the position the velocity of the particle is written in the polar basis associated with its current position, and is: what is the position of at ?

Answer:

Spring constant: approximately [tex]28\; {\rm N\cdot m^{-1}}[/tex].

Unloaded length: approximately [tex]7.0\; {\rm cm}[/tex].

(Assume that the weight of the spring is negligible, and that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)

Explanation:

Divide the tension [tex]F[/tex] on the spring by the displacement [tex]x[/tex] to find the spring constant. [tex]k[/tex]:

[tex]\begin{aligned}k &= \frac{F}{x}\end{aligned}[/tex].

Let [tex]L_{0}[/tex] denote the unloaded length of the spring (in meters.)

When the [tex]0.225\; {\rm kg}[/tex] mass is on the spring, the tension on the spring would be [tex]m\, g = (0.225)\, (9.81) \; {\rm N} \approx 2.207\; {\rm N}[/tex].

It is given that the length of the spring with this [tex]0.225\; {\rm kg}[/tex] mass attached is [tex]L = 0.15\; {\rm m}[/tex]. Subtract the initial length of the spring [tex]L_{0}[/tex] from the current length to find the displacement of the spring: [tex]x = L - L_{0} = 0.15 - L_{0}[/tex].

Divide the tension on the spring by the displacement to find an expression for the spring constant:

[tex]\begin{aligned}k &= \frac{F}{x} \approx \frac{2.207}{0.15 - L_{0}}\end{aligned}[/tex].

Similarly, when the [tex]1.85\; {\rm kg}[/tex] object is on the spring, the tension on the spring would be [tex]m\, g = (1.85)\, (9.81) \; {\rm N} \approx 18.15\; {\rm N}[/tex].

Subtract the initial length [tex]L_{0}[/tex] from the current length [tex]L = 0.725\; {\rm m}[/tex] to find the displacement: [tex]x = L - L_{0} = (0.725 - L_{0})\; {\rm m}[/tex].

Divide the tension on the spring by the displacement to find another expression for the spring constant:

[tex]\begin{aligned}k &= \frac{F}{x} \approx \frac{18.15}{0.725 - L_{0}}\end{aligned}[/tex].

Equate the two expressions for the spring constant [tex]k[/tex] and solve for the unloaded length [tex]L_{0}[/tex]:

[tex]\begin{aligned}\frac{18.15}{0.725 - L_{0}} &= \frac{2.207}{0.15 - L_{0}}\end{aligned}[/tex].

[tex](18.15)\, (0.15 - L_{0}) = (2.207)\, (0.725 - L_{0})[/tex].

[tex]\begin{aligned} L_{0} &\approx \frac{(18.15)\, (0.15) - (0.725) (2.207)}{18.15 - 2.207}\; {\rm m} \\ &\approx 0.070\; {\rm m}\end{aligned}[/tex].

Substitute [tex]L_{0}[/tex] back into one of the two expressions for the spring constant [tex]k[/tex] and evaluate:

[tex]\begin{aligned}k &\approx \frac{18.15}{0.725 - L_{0}} \\ &\approx \frac{18.15}{0.725 -0.070} \approx 28\; {\rm N\cdot m^{-1}} \end{aligned}[/tex].

Apply unit conversion:

[tex]0.070\; {\rm m} = (0.070 \times 10^{2})\; {\rm cm} = 7.0\; {\rm cm}[/tex].

In other words, the unloaded length of the spring would be approximately [tex]7.0\; {\rm cm}[/tex].

The correct answer to part a) is, the force constant of the spring can be found using Hooke's Law and it is 3.83 N/m, and part b) is, the unloaded length of the spring is the same as the length when the 1.85 kg mass is hanging from it, which is 0.725 m.

Part (a) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is equal to the spring constant multiplied by the displacement from equilibrium.

The equation for this is F = kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.

In this case, we can use the two given lengths and masses to find the displacement from equilibrium and the force exerted by the spring.

The displacement for the first mass is 0.15 m - 0.725 m = -0.575 m, and the force is 0.225 kg * 9.8 m/s^2 = 2.205 N. The displacement for the second mass is 0.725 m - 0.725 m = 0 m, and the force is 1.85 kg * 9.8 m/s^2 = 18.13 N.

We can then plug these values into the equation F = kx and solve for k:

2.205 N = k(-0.575 m)
k = -2.205 N / -0.575 m
k = 3.83 N/m

Therefore, the force constant of the spring is 3.83 N/m.

Part (b) To find the unloaded length of the spring, we can use the equation x = F/k, where x is the displacement from equilibrium, F is the force, and k is the spring constant.

Since the unloaded length of the spring is when the force is zero, we can plug in F = 0 and k = 3.83 N/m to solve for x:

x = 0 N / 3.83 N/m
x = 0 m

Since the displacement from equilibrium is zero, the unloaded length of the spring is the same as the length when the 1.85 kg mass is hanging from it, which is 0.725 m.

Therefore, the unloaded length of the spring is 0.725 m, or 72.5 cm.

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More information is needed to be able to answer questions 1 and 3. The knowledge of 'ole" would help to tell that the word arteriole refers to a small  component.

Knowing that the suffix "-ole" means "small or little" would help in defining the term "arteriole". An arteriole is a small or little artery that branches out from an artery and leads to a capillary.

The suffix "-ole" in "arteriole" indicates that it is a small or little version of an artery, which is a larger blood vessel that carries blood away from the heart to other parts of the body.

Therefore, understanding the meaning of the suffix "-ole" helps in breaking down the term "arteriole" into its component parts and understanding its meaning.

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The volumetric strain rate is zero, the vorticity is -1/2μ (∂p/∂x) y, the rate of angular deformation is 1/2μ (∂p/∂x) y, and the shear stress at the plate surface is -1/2 (∂p/∂x) [tex]h^2[/tex].

Given:

x-velocity: u = 1/2μ (∂p/∂x) ([tex]y^2 - h^2[/tex])

y-velocity: υ = 0 (no variation in y-direction)

where,

μ = dynamic viscosity

p = pressure

x, y = coordinates

h = distance between the plates

To determine the volumetric strain rate, we can start by considering the continuity equation for incompressible flow, which states that the product of velocity and cross-sectional area is constant:

u × (h-y) = Q/A

where A is the cross-sectional area and Q is the volumetric flow rate.

Taking the derivative of both sides with respect to time and simplifying, we get:

dQ/dt = -u × dA/dy

Since the y-velocity is zero, we have dA/dy = 0, so:

dQ/dt = 0

This means that the volumetric flow rate is constant and there is no change in volume with time. Therefore, the volumetric strain rate is zero.

The velocity vector's curl is used to define the vorticity.

ω = ∇ x v

where ∇ is the del operator. For two-dimensional flow, the vorticity is a scalar and can be expressed as:

ω = (∂υ/∂x) - (∂u/∂y)

Substituting the given values for u and υ, we get

ω = 0 - (∂/∂y)[1/2μ (∂p/∂x) ([tex]y^2 - h^2[/tex])]

Simplifying and integrating with respect to y, we get:

ω = -1/2μ (∂p/∂x) y

The formula for angular deformation rate is:

D = (∂u/∂y + ∂υ/∂x) = ∂u/∂y

Substituting the given value for u, we get:

D = (∂/∂y)[1/2μ (∂p/∂x) [tex](y^2 - h^2)[/tex]]

Simplifying and integrating with respect to y, we get:

D = 1/2μ (∂p/∂x) y

To find the shear stress at the plate surface, we can use the following relation:

τ = μ (∂u/∂y)|y=h

Substituting the given value for u, we get:

τ = μ (∂/∂y)[1/2μ (∂p/∂x) ([tex]y^2 - h^2[/tex])]|y=h

Simplifying, we get:

τ = -1/2 (∂p/∂x) [tex]h^2[/tex]

Therefore, the volumetric strain rate is zero, the vorticity is -1/2μ (∂p/∂x) y, the rate of angular deformation is 1/2μ (∂p/∂x) y, and the shear stress at the plate surface is -1/2 (∂p/∂x) [tex]h^2[/tex].

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The oceanic crust is not thicker than the continental crust. The thickness of the continental crust is typically 30 km, compared to the oceanic crust's average thickness of 7 km.

There are two different types of crust that cover the Earth: continental and oceanic. The continental crust is typically up to 25 miles thick, whereas the thinner oceanic crust is typically a little over four miles thick.

The thickness of the continental crust is normally 40 km (25 miles), whereas the thickness of the oceanic crust is only 6 km (4 miles). Different densities of lithospheric rock's impact can be seen in

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the magnitude of the average force of air resistance exerted on the object is approximately 1.05 N.

The magnitude of the average force of air resistance exerted on an object depends on various factors such as the shape, size, speed, and density of the object, as well as the density and viscosity of the air.

F = (1/2) * rho * Cd * A * v

we can estimate the density of air at sea level as 1.225 kg/m, and assume a drag coefficient of 0.5 for a spherical object.

F = (1/2) * rho * Cd * A * v

= (1/2) * 1.225 kg/m * 0.5 * pi * (0.1 m)* (8.50 m/s)

= 1.05 N (to two significant figures)

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The Perseid meteors repeat regularly around August 11th or so, despite everything in the solar system moving around, because the debris that creates them follows a consistent orbit.

The orbit of the debris producing the meteor shower is consistent, and it orbits the Sun in the same manner every year. When the Earth crosses through the debris stream, the debris enters the Earth's atmosphere and burns up, producing the Perseid meteor shower.This means that even if the Earth's movement is unpredictable, as it revolves around the Sun, the Perseid meteor shower will occur around the same time every year.

Hence, it's not because of the Earth's movement or other bodies in the solar system that the Perseid meteors repeat regularly around August 11th or so, but because of the consistent orbit of the debris that creates them.

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If the mass of the cylinder is increased, then the temperature of the water will increase because an increase in the mass of the cylinder will result in an increase in the potential energy of the system.

Energy is a fundamental concept in physics that refers to the ability of a system or object to do work. It is a scalar quantity, meaning that it has only magnitude and no direction. Energy can take many forms, including mechanical energy, thermal energy, electrical energy, chemical energy, nuclear energy, and electromagnetic energy. The unit of energy in the International System of Units (SI) is the joule (J), which is defined as the work done by a force of one newton over a distance of one meter (1 J = 1 N × 1 m). Energy can be transferred from one system or object to another, and it can be converted from one form to another. For example, the kinetic energy of a moving object can be converted into thermal energy due to friction, or the chemical energy stored in food can be converted into mechanical energy by the muscles in our bodies. The study of energy and its transformations is a fundamental concept in physics and has important applications in fields such as engineering, environmental science, and renewable energy.

Here,

When the cylinder is heated and placed in the water, its potential energy is converted into thermal energy, which is transferred to the water and causes its temperature to increase. Therefore, an increase in the mass of the cylinder will result in a greater amount of potential energy being converted into thermal energy, leading to a greater increase in the temperature of the water. This hypothesis addresses the lesson question by explaining how an increase in potential energy (due to an increase in the mass of the cylinder) can be converted into thermal energy (due to the heating of the cylinder) in a system.

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The distance between the slits, d = 0.450 mm = 0.00045 m, The distance from the screen, L = 75.0 cm = 0.75 m,The wavelength of the coherent light, λ = 500 nm = 5.00 x 10^-7 m

The distance between the central maximum and the first dark line can be found using the formula: sin θ = mλ/d, where θ is the angle between the line connecting the slits and the screen, m is the order of the dark line (m = 1 for the first dark line), and λ is the wavelength of the light. Rearranging this formula gives: sin θ = mλ/d, θ = sin^-1(mλ/d). For the second dark line, m = 2, and for the third dark line, m = 3. So we can find the angles θ2 and θ3: θ2 = sin^-1(2λ/d) = sin^-1(2 x 5.00 x 10^-7 m / 0.00045 m) ≈ 0.140 radians, θ3 = sin^-1(3λ/d) = sin^-1(3 x 5.00 x 10^-7 m / 0.00045 m) ≈ 0.234 radians. Now we can use trigonometry to find the distances between the second and third dark lines on the screen. The distance between the central maximum and the second dark line is given by: y2 = L tan θ2 ≈ 0.099 m. Similarly, the distance between the central maximum and the third dark line is given by: y3 = L tan θ3 ≈ 0.165 m. Therefore, the distance between the second and third dark lines is: y3 - y2 ≈ 0.066 m. So the distance between the second and third dark lines of the interference pattern on the screen is approximately 0.066 m.

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Answer:

Explanation:

The object does not carry a charge

To move with the same acceleration as the 10kg box, the 50kg box needs higher net force.

The term "net force" refers to the combined effects of all the forces exerted on a moving body, including gravitational, frictional, and normal forces. FNet equals Fa, Fg, Ff, and FN.

Newton's Second Law of Motion states that

A mass-proportional force is needed to accelerate an item.

The following formula determines the net force necessary to move an item at a given acceleration:

net force = mass x acceleration

The 50kg box and the 10kg box both accelerate at the same rate, but the 50kg box's net force need is more than the 10kg box's because of its heavier mass.

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Problems with data management may have a detrimental impact on a variety of issues. Bad risk management choices, data loss, information leakage, unauthorised, data pyramids, compliance with laws, an unsafe environment, a shortage of resources, etc. are instances among these.

What issues surround the handling of data?

These are a few potential issues in managing scientific, financial, or administration data that have been briefly explained.

Technical information not adequately recorded.

The administration of performance specifications is not under the PI's control.

Data not kept on file by the organization.

improper maintenance of financial or administrative data.

What would you say is data management?

Data administration is the act of gathering, arranging, and using data to job in an effective, economy, and judgement call.

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Translation-Adding more gas to a tank, the internal pressure will increase because by increasing the number of particles?

Adding more gas to a tank, the internal pressure will increase because by increasing the number of particles increases collision between particles.

The frequency of collisions between gas particles and the container walls determines the gas pressure. The number of collisions and consequently the pressure will rise if we fill the container with more gas particles.

The Avogadro Principle (V N) states the frequency of collisions with the container walls must increase as the number of gas particles increases. The pressure of the gas then rises as a result of this. Pressure rises as the number of gas molecules rises while the volume of the container stays the same. Gas pressure rises as container volume decreases. The pressure rises as the temperature of a gas inside a rigid container rises.

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Answer:

Pool:
1 lap: 100 m distance, 0 m displacement
1.5 laps: 150 m distance, 50 m displacement
2 laps: 200 m distance, 0 m displacement

Track:
1 lap: 200 m distance, 0 m displacement
1.5 laps: 300 m distance, 50 m displacement
2 laps: 400 m distance, 0 m displacement

Explanation:

Distance and displacement are two different things:

Distance is the full length you have traveled. For example, if you swim one lap around a 50 meter pool, you've traveled a distance of 100 meters (there and back).

However, displacement how far you are from the starting point. If you swim one lap around the same pool, your displacement is 0 meters since you ended up in the same place you had started from.

This lets most of the light through but still blocks some hot air. It will take a long time to pop.

A hypothesis is an educated guess or tentative explanation that serves as a starting point for scientific investigation. It is a statement that suggests a possible relationship between two or more variables that can be tested through research.

A hypothesis should be based on existing knowledge and should be testable through empirical evidence. It should also be falsifiable, meaning that it should be possible to prove it wrong through experimentation or observation. In the scientific method, a hypothesis is usually formulated after conducting a literature review and making observations about the phenomena being studied. Once the hypothesis is established, the researcher can design experiments or collect data to test the hypothesis.

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The labels for vectors A, B, and C ought to be determined by their magnitudes and directions. It is essential to keep track of their angles in relation to the +x axis.

We need to use trigonometry to figure out the y and x-components of each vector in order to find Ay, Ay, B., B, C, and C. The magnitude of the vector is divided by the cosine of its angle with the +x-axis to get the y-component, and the magnitude of the vector is divided by the sine of its angle with the +x-axis to get the x-component.

The Pythagorean theorem can be used to determine the net force. We really want to add the x and y-parts of the vectors to acquire the resultant vector. The Pythagorean theorem can then be used to determine the magnitude of the net force.

In conclusion, we can use the Pythagorean theorem to determine the net force and trigonometry to determine the components of each vector.

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1) The potential energy of the stone when it is at the top of the building is approximately 4313.2 Joules. 2) The height of the hill is approximately 4.09 meter

1.The potential energy of an object at a height h is given by:

PE = mgh

where m is the mass of the object, g is the acceleration due to gravity (approximately 9.81 m/s^2 on Earth), and h is the height.

In this case, the mass of the stone is 2.3 kg, the height is 197 m, and g is approximately 9.81 m/s^2. Therefore, the potential energy of the stone when it is at the top of the building is:

PE = mgh = 2.3 kg * 9.81 m/s^2 * 197 m = 4313.2 J

Therefore, the potential energy of the stone when it is at the top of the building is approximately 4313.2 Joules.

2.The total mechanical energy of a ball rolling down a hill is conserved, meaning that the sum of its potential energy and kinetic energy at any point is constant. Therefore, we can use the following formula to solve for the height of the hill:

PE_initial + KE_initial = PE_final + KE_final

where PE is potential energy, KE is kinetic energy, and "initial" and "final" refer to the starting and ending points of the ball's motion.

At the top of the hill, the ball is not moving, so its initial kinetic energy is zero. Therefore, we can simplify the formula to:

PE_initial = PE_final + KE_final

At the bottom of the hill, the ball has a velocity of 9.2 m/s. The final potential energy is zero, since the ball is at the bottom of the hill. Therefore, we can substitute in the known values and solve for the initial potential energy:

PE_initial = PE_final + KE_final

PE_initial = 0 + (1/2) * m * v^2

PE_initial = (1/2) * m * v^2

PE_initial = (1/2) * 0.5 kg * (9.2 m/s)^2

PE_initial = 20.08 J

Now, we can use the formula for potential energy to solve for the height of the hill:

PE_initial = mgh

h = PE_initial / (mg)

h = 20.08 J / (0.5 kg * 9.81 m/s^2)

h = 4.09

Therefore, the height of the hill is approximately 4.09 meter

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Answer:

The other force acting on a meteor as it enters the Earth's atmosphere is air resistance or drag force.

Explanation:

Air resistance or drag force is the force that opposes the motion of an object as it moves through a fluid (such as air or water). When a meteor enters the Earth's atmosphere, it encounters a large amount of air resistance due to the high speed at which it is moving relative to the atmosphere.

As the meteor moves through the atmosphere, air molecules collide with the meteor, creating a resistance force that acts in the opposite direction to the meteor's motion. This force increases as the meteor's speed increases, and can cause the meteor to slow down and heat up due to friction with the surrounding air molecules.

The force of air resistance is particularly important for meteors because they are typically travelling at very high speeds relative to the atmosphere, and can experience significant heating and deceleration as they enter the denser lower atmosphere. Without air resistance, meteors would continue to travel at their original speeds and would not experience the bright trails or fireballs that are commonly associated with meteors.

Use a rope or twine to fasten a spring scale to the box. Make sure the rope or string is parallel to the surface and the scale is perpendicular to the surface.

A push force from the outside is applied to a box when it slides, and a frictional force is also applied to the box. The contact sliding motion between the object and the ground causes the frictional force.

Kinetic friction is a kind of sliding friction that occurs often. Kinetic friction is a force that opposes the motion of an object sliding along a surface whenever two objects' surfaces are in contact with one another.

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The dog home would measure 5 feet long, 4 feet wide, and 3 feet high (height) or 4 feet (length) x 5 feet (width) x 3 feet (height).

To find the best dimensions for a dog house with a volume of approximately 60 cubic feet, we need to consider several factors such as the size of the dog, the available space for the dog house, and the materials available for construction.

Assuming that the dog is of average size, we can start by considering the shape of the dog house. A rectangular prism is a common shape for a dog house, so we can begin with that shape.

Let's define the dimensions of the dog house as length (L), width (W), and height (H). We can express the volume of the dog house as:

Volume = L x W x H

Given that the volume of the dog house is approximately 60 cubic feet, we can try different combinations of the dimensions to find the best one. One way to do this is to keep one dimension constant and vary the others.

For example, if we keep the length and width constant at 4 feet and 5 feet, respectively, we can find the height by solving the equation:

60 = 4 x 5 x H

H = 3 feet

Therefore, the dimensions of the dog house would be 4 feet (length) x 5 feet (width) x 3 feet (height).

Alternatively, we could keep the length and height constant at 5 feet and 3 feet, respectively, and vary the width:

60 = 5 x W x 3

W = 4 feet

Therefore, the dimensions of the dog house would be 5 feet (length) x 4 feet (width) x 3 feet (height).

Both options result in a volume of approximately 60 cubic feet and provide a comfortable space for the dog. The choice between them would depend on the available space for the dog house and the materials that are available for construction.

Finding the best dimensions for a dog house requires consideration of various factors, such as the size of the dog and the available materials. By experimenting with different combinations of dimensions, we can find the optimal size and shape that provide enough space and comfort for the pet.

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The equation for the acceleration of the two connected blocks in terms of m1, m2, and the acceleration due to gravity is a = (m1 + m2)g/m2.

The equation shows that the acceleration of the two connected blocks is directly proportional to the total mass of the two blocks and the acceleration due to gravity. In other words, the more mass in the two blocks, the higher the acceleration.

Similarly, as the acceleration due to gravity increases, the acceleration of the two connected blocks increases. To understand this further, consider the example of two blocks, one with mass m1 and the other with mass m2, being accelerated by the same force due to gravity.

The equation shows that the acceleration of the two blocks is a = (m1 + m2)g/m2, where m1 and m2 are the masses of the two blocks and g is the acceleration due to gravity. This means that if the mass of the first block is doubled, the acceleration will double. Similarly, if the acceleration due to gravity is doubled, the acceleration of the two blocks will also double.

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In this collision, momentum is conserved as there is no external force acting on the system. We can use the conservation of momentum to find the final velocity of puck 2.

pinitial = m1v1 + m2v2

where m1 and m2 are the masses of puck 1 and puck 2 respectively, and v1 and v2 are their initial velocities. Since puck 2 is initially at rest, we have:

pinitial = m1×v1

Substituting the given values, we get:

pinitial = (0.35 kg)×(2.5 m/s) = 0.875 kg m/s

After the collision, puck 1 is deflected at an angle of 30 degrees with a speed of 1.7 m/s. We can use trigonometry to find the x and y components of its final velocity:

v1x = v1cos(30) = 1.7cos(30) = 1.47 m/s v1y = v1sin(30) = 1.7sin(30) = 0.85 m/s

The final momentum of the system is:

pfinal = m1v1 + m2v2'

where v2' is the final velocity of puck 2. Since momentum is conserved, we have:

pinitial = pfinal

Substituting the given values, we get:

0.875 kg m/s = (0.35 kg)×(1.47 m/s) + (0.51 kg)×v2'

Solving for v2', we get:

v_2' = (0.875 kg m/s - 0.51 kg*(1.47 m/s)) / 0.51 kg = -0.94 m/s

Therefore, the final velocity of puck 2 is -0.94 m/s in the negative x direction (i.e., towards the left).

Consider a collision between two pucks on a frictionless air hockey table. Puck 1 has a mass of 350 g and an initial velocity of 2.5 m/s in the +x direction. Puck 2 has a mass of 510 g and is initially at rest.

After the collision puck 1 has been deflected by 30 degrees and has a speed of 1.7 m/s. What is the speed and direction of puck 2 after the collision?

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The advantage of having a blowhole on top of a dolphin's head is that it allows them to breathe while swimming without having to break the surface of the water, which is crucial for their survival.

Dolphins are aquatic mammals and need to breathe air to survive. The location of their blowhole on the top of their head allows them to take in air without having to break the surface of the water, which is critical for their survival in the wild. When a dolphin needs to breathe, it can simply surface and quickly exhale and inhale through its blowhole.

This enables them to maintain a constant speed while swimming, and also helps them avoid detection from predators or prey that may be lurking at the surface. In addition, having a blowhole on top of their head allows them to quickly clear their airways of water and mucus, which can be useful for diving and hunting.

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Electric potential in y direction (a) will be constant and in x direction (b) will decrease as you move in direction of the electric field.

The electric field E in a region is uniform if it is the same at all points. Specifically, if E has a magnitude of 5 V/m and points in the x direction, then we can say the following about the behavior of the electric potential:

a) In the y direction, the electric potential will be constant. This is because the electric field is perpendicular to the y direction, and therefore does not affect the electric potential in that direction.

b) In the x direction, the electric potential will decrease as you move in the direction of the electric field.

This is because the electric field is parallel to the x direction, and therefore affects the electric potential in that direction. Specifically, the electric potential will decrease by 5 V for every meter you move in the x direction.

To summarize, the electric potential will be constant in the y direction, and will decrease in the x direction as you move in the direction of the electric field. This is due to the fact that the electric field is uniform and points in the x direction.

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The free-body diagram for the crate  shows all the forces acting on the crate like the force applied by the person pushing the crate, the force of friction between the crate and the floor, and the force of gravity acting on the crate.

The free-body diagram for the crate should show all the forces acting on the crate, including the force applied by the person pushing the crate, the force of friction between the crate and the floor, and the force of gravity acting on the crate.

Since the crate is not moving, the force applied by the person pushing the crate must be equal in magnitude and opposite in direction to the force of friction acting on the crate.

This means that the net force on the crate is zero, and the free-body diagram should reflect this.

Here is a description of the forces acting on the crate and a corresponding free-body diagram:

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Answer:

c

Explanation:

step by step hope this helps you got this!!

F = ma, or force equal to mass times acceleration, is Newton's second law of motion.

Second Law of Movement by Newton Since it shows how powers and movement are connected, F=ma is essential. You can use it to determine an object's velocity and position, as well as its acceleration with known forces. For inventors, scientists, and engineers, This is extremely helpful.

As per Newton's Second Law of Movement, when a power works on a mass, the mass speeds up (gains speed) (object). When you ride a bicycle, you can see this law of motion in action in a great way. Your bike makes up the mass. Your leg muscles press against the bicycle's pedals to produce the force.

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The direction of their (joint) velocity after the collision is to the northeast.

To solve this problem, we need to use the conservation of momentum principle.

In this case, the truck and car form a closed system, and we can use the following equation to solve for their joint velocity:

(m1 * v1) + (m2 * v2) = (m1 + m2) * v

where:

First, we need to break down the initial velocities of the truck and car into their x and y components.

For the truck:

vx1 = 0 m/s (moving north)

vy1 = 26.0 m/s (moving north)

For the car:

vx2 = 13.0 m/s (moving east)

vy2 = 0 m/s (not moving in the y-direction)

Next, we need to find the total x and y momentum of the system before the collision.

Px = m1 * vx1 + m2 * vx2

= (3520 kg) * 0 m/s + (1480 kg) * 13.0 m/s

= 19240 kg*m/s (to the right)

Py = m1 * vy1 + m2 * vy2

= (3520 kg) * 26.0 m/s + (1480 kg) * 0 m/s

= 91520 kg*m/s (moving north)

The total momentum of the system before the collision is therefore:

P = √(Px^2 + Py^2)

= √((19240 kgm/s)^2 + (91520 kgm/s)^2)

= 97397.59 kg*m/s

Now we can use the conservation of momentum principle to solve for the joint velocity of the truck and car after the collision:

(m1 * v1) + (m2 * v2) = (m1 + m2) * v

(3520 kg * 0 m/s) + (1480 kg * 13.0 m/s) + (3520 kg + 1480 kg) * v = 97397.59 kg*m/s

v = (3520 kg * 0 m/s + 1480 kg * 13.0 m/s) / (3520 kg + 1480 kg)

= 8.60 m/s (to the northeast)

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The 2m/s is the cat's speed v² when she reaches the tοp οf the incline.

The definitiοn οf velοcity is the rate at which a bοdy mοves in a particular directiοn. Velοcity is the rate at which a distance changes in relatiοn tο time. A vectοr quantity with bοth magnitude and directiοn is velοcity.

The rate οf a directiοnally changing οbject's lοcatiοn. The SI unit οf speed is created by cοmbining the fundamental units οf length and time. Meters per secοnd (m/s) is the unit οf speed in the metric system.

Therefοre, 2m/s is the cat's speed v2 when she reaches the tοp οf the incline.

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The force exerted by the pitcher on the 0.15 Kg ball, given that the ball accelerated from rest is 357.35 N

First, we shall determine the acceleration of the ball. Details below:

v² = u² + 2as

90² = 0² + (2 × a × 1.7)

8100 = 0 + 3.4a

8100 = 3.4a

Divide both side by 3.4

a = 8100 / 3.4

a = 2382.35 m/s²

Finally, we shall determine the force exerted. Details below:

Force = mass × acceleration

Force exerted = 0.15 × 2382.35

Force exerted = 357.35 N

Thus, the force exerted is 357.35 N

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Complete question:

A pitcher throws a 0.15-kg baseball accelerating it from rest to a speed of about 90 m/s in a distance 1.7 m . Estimate the force exerted by the pitcher on the ball