a ball is kicked upward with an initial velocity of 56 feet per second. the ball's height, h (in feet), from the ground is modeled by h

Nicolaus Copernicus was the first person to propose the heliocentric model of the solar system, in which the Sun, not the Earth, is at the center.

Copernicus lived during the Renaissance and was an astronomer and mathematician. He proposed his heliocentric model in his book "De Revolutionibus Orbium Coelestium" (On the Revolutions of the Celestial Spheres) published in 1543. In this model, Copernicus argued that the Earth and the other planets revolve around the Sun, and not the other way around as was commonly believed at the time. This challenged the geocentric model of the universe, which placed the Earth at the center.

Copernicus' heliocentric model was based on observations and calculations he made using instruments like the astrolabe and the armillary sphere. He also relied on the work of earlier astronomers like Aristarchus of Samos and Ptolemy. Copernicus' model helped to explain phenomena like retrograde motion and the varying brightness of planets.

Copernicus' heliocentric model was a major shift in the way people thought about the universe, and paved the way for later scientific discoveries. It challenged long-held beliefs and led to a greater understanding of our place in the cosmos. Copernicus' work is considered a cornerstone of modern astronomy.

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The common final velocity of the two blocks after the inelastic collision is -2.25 m/s.

In a totally inelastic collision, the two objects stick together and move as a single object with a common velocity after the collision. The final velocity of this combined object can be determined using the law of conservation of momentum.

The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. This means that the total momentum of the two blocks before the collision must equal their total momentum after the collision.

The momentum of each block before the collision can be calculated as follows:

m1 * v1 = 3 kg * 4 m/s = 12 kg m/s

m2 * v2 = 5 kg * -6 m/s = -30 kg m/s

The total momentum before the collision is the sum of the individual momenta:

\(p_initial = m1 * v1 + m2 * v2 = 12 kg m/s + (-30 kg m/s) = -18 kg m/s\)

After the collision, the two blocks have combined into a single object of mass m1 + m2 = 3 kg + 5 kg = 8 kg. The velocity of this combined object is equal to the final velocity of the collision, which we will call v_final.

The total momentum after the collision is equal to the momentum of the combined object:

\(p_final = (m1 + m2) * v_final = 8 kg * v_final\)

Setting p_initial equal to p_final and solving for v_final:

\(-18 kg m/s = 8 kg * v_finalv_final = -18 kg m/s / 8 kg = -2.25 m/s\)

So the common final velocity of the two blocks after the inelastic collision is -2.25 m/s.

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Answer:

0_2 is 0.2 with 0_1 0.1 will 0_3 is 0.3 and with 0_2 0.

Explanation:

2

a) If the density remains constant throughout the fluid field, the derivative of the density with respect to time should be zero.

b) The curl of the velocity vector is not equal to zero. Therefore, the flow is rotational.

c) The acceleration of the fluid particle along the streamline direction is zero.

d) The equation of the streamline is x = tan (2t).

a) Judge if the flow is compressible or incompressible:

The flow is compressible if the density of the fluid changes at any point in space with time. On the other hand, the flow is incompressible if the density of the fluid remains constant throughout the fluid field with time. For an incompressible flow, the continuity equation states that ρ = constant, where ρ is the fluid density. Differentiating ρ = constant with respect to time, we get dρ/dt = 0.

Here, the velocity components are given as u = x² + 1, v = y, and w = x. Comparing with the continuity equation, V = ∂u/∂x + ∂v/∂y + ∂w/∂z. For the given flow, ∂u/∂x = 2x, ∂v/∂y = 1, and ∂w/∂z = 1. Substituting the above values, V = 2x + 1 + 1 = 2x + 2 ≠ 0. Since V ≠ 0, the flow is compressible.

b) Judge if the flow is rotational or irrotational:

A flow is irrotational if the curl of the velocity vector is zero. On the other hand, the flow is rotational if the curl of the velocity vector is not zero. The curl of a vector is given by curl V = (∂w/∂y - ∂v/∂z)i + (∂u/∂z - ∂w/∂x)j + (∂v/∂x - ∂u/∂y)k. For the given flow, u = x² + 1, v = y, and w = x. Comparing with the curl of the velocity vector, ∂w/∂y - ∂v/∂z = 0, ∂u/∂z - ∂w/∂x = 0, and ∂v/∂x - ∂u/∂y = -1.

c) Find the instantaneous acceleration along the streamline direction of the particle located at (x, y, z) = (1, 0, 1) at time t = 2 sec:

We can find the instantaneous acceleration along the streamline direction of the particle using the acceleration formula, a = dv/dt. Here, the velocity components are given as u = x² + 1, v = y, and w = x. At (x, y, z) = (1, 0, 1), the velocity of the fluid particle can be obtained as u = (1)² + 1 = 2, v = 0, and w = 1. Thus, the velocity of the fluid particle is v = 0 m/s, w = 1 m/s, and u = 2 m/s. Substituting the values, we get a = dv/dt = (d/dt)(2i + j + k) = 0.

d) Find the equation of the streamline passing through (x, y, z) = (1, 0, 1) at time t = 2 sec:

The equation of a streamline is given by dx/u = dy/v = dz/w. Thus, ∫ dx/u = ∫ dy/v = ∫ dz/w. Integrating the first term with respect to x, we have ∫ dx/u = ∫ dx/(x² + 1) = tan⁻¹ (x) + C1. Integrating the second term with respect to y, we have ∫ dy/v = ∫ dy/y = ln |y| + C2. Integrating the third term with respect to z, we have ∫ dz/w = ∫ dz/x = ln |x| + C3.

Using the boundary condition, we get C1 = C2 = C3 = 0. Substituting the values, we get ∫ dx/(x² + 1) = 2t + C1. On solving the integral, we get tan⁻¹ (x) = 2t + C1 or x = tan (2t + C1). Applying the boundary condition (x, y, z) = (1, 0, 1), we get x = tan (2t).

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If you run off the pavement, you should turn the steering wheel quickly toward the road and is denoted as option C.

This is a path which is found in almost all roads and is designed to enhance foot traffic in most climes. This is the path where individuals walk so as not to get hit down by a vehicle.

In a situation where a car runs into the pavement, the best choice is to turn the steering wheel quickly toward the road as braking hard could damage the brakes of the car and cause accident in subsequent trips.

This is therefore why option C was chosen as the correct choice.

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Answer: Steer straight and slow down before attempting

Explanation:

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

Electrostatic force is the force exerted between two charged particles. It is also called Coulomb's law. The force between an electron and a proton separated by 0.1 mm can be calculated using the equation of Coulomb's law.

F =\((k × q1 × q2) / d²\) Where, F = Electrostatic force k = Coulomb's constant = 9 × 10⁹ N m² C⁻²q1 and q2 are the magnitudes of charges on two particles separated by distance d.

\(F = (9 × 10⁹ N m² C⁻² × (-1.6 × 10⁻¹⁹ C) × (+1.6 × 10⁻¹⁹ C)) / (0.1 × 10⁻³ m)²F = 2.304 × 10⁻²⁸\) N Therefore, the electrostatic force between an electron and a proton separated by 0.1 mm is\(2.304 × 10⁻²⁸ N.\)

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Answer:

free electrons

Explanation:

Answer:

free electrons

Explanation:

Answer:

350N

Explanation:

Calculate the weight of the box by using this formula;

W = mg

W = 10 × 9.8

W = 98N

Then, total up the weight with the exerted pressure;

98N + 250N

= 348N

≈ 350N (2sf)

I hope this helps! Let me know if I miscalculated anything!

Since the moon's orbit is tipped in relation to how Earth orbits the sun, the moon frequently passes below or above Earth, making it impossible to see a solar eclipse every month.

A solar eclipse happens when the Moon moves in front of the Sun, blocking the Sun's light completely or partially from a small portion of the Earth. A similar arrangement takes place roughly every six months, during the new moon phase of eclipse season, when the Moon's orbital plane is closest to the Earth's plane. Since the moon's orbit is tipped in relation to how Earth orbits the sun, the moon frequently passes below or above Earth, making it impossible to see a solar eclipse every month.

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The force the magnetic field exerts on the particle is 0.072 N with velocity

Charge q = 8 μC is moving with a velocity v = [31 +43] x 106 m/s. If the charge enters a magnetic field B=0.4k T, then the magnitude of the magnetic force (in N) on the charge is given by

F = q x v x B

Given thatq = 8 μC

                   = 8 x 10-6 C

Velocity v = [31 +43] x 106 m/s

                = [31,000,000 + 43,000,000] m/s

                = [74,000,000] m/s

Magnetic field B = 0.4k T

                           = 0.4 x 103 T

TeslaSubstituting these values in the above equation, we get

F = (8 x 10-6) x (74 x 106) x (0.4 x 103)

F = 2.95 N

Therefore, the magnitude of the magnetic force (in N) on the charge is 2.95 N.

Another problem:

Given that a particle with a charge of 5.0 C travels at 2.0 m/s and encounters a magnetic field of 0.0100 T that makes an angle of 46.0 with respect to the x-axis in the xy-plane. We need to find the force the magnetic field exerts on the particle.

The force F on the particle is given by:

F = q x v x B x sin(θ)Where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field and θ is the angle between the velocity of the particle and the magnetic field.

F = 5.0 C x 2.0 m/s x 0.0100 T x sin(46.0)

F = 0.072 N

Therefore, the force the magnetic field exerts on the particle is 0.072 N.

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Answer:

C. The light waves will bend as they hit the water and slow down because water is a denser medium than air.

Explanation:

Refraction, or the bending of light, is caused by a change in the medium through which it passes. Air and water have different optical densities. As a result, light that enters the water is refracted more normally.

The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Given:

Mass of gymnast, m = 65.5 kg

The speed 'v' of gymnast is constant

Solution:

Consider the free-body diagram of the system as shown below.

Balancing forces along the vertical axis we get:

ΣFy = 0

Thus, we get:

F = ma     - (1)

where, m is mass of gymnast

            a is acceleration of gymnast (a = 0m/s², as the speed is constant)              

Also,

F = T - mg          -(2)

where, T is tension in the rope

           g is acceleration due to gravity

Equating (1) & (2), we get:

ma = T - mg

Re-arranging the equation, we get:

T = m(a+g)

Applying values in above equation we get:

T = (65.5 kg)(0 m/s²+9.8 m/s²)

T = 641.9 N

Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

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Let u be a vector and let y be a non-zero vector. Also, let proj y u denote the projection of u onto y. In order to show that u - proj y u is orthogonal to proj y u, the dot product of the two vectors needs to be calculated.

Given that u is a vector and y is a non-zero vector, then the projection of u onto y is given as follows:proj y u = [(u . y) / (y . y)] * yWhere, u . y represents the dot product of u and y, and y . y represents the dot product of y with itself. The vector u - proj y u is given as follows:u - proj y u = u - [(u . y) / (y . y)] * yIn order to show that u - proj y u is orthogonal to proj y u, the dot product of the two vectors needs to be calculated.

This is shown below:(u - proj y u) . proj y u= (u . proj y u) - (proj y u . proj y u)= [(u . y) / (y . y)] * (y . proj y u) - [(u . y) / (y . y)] * (y . proj y u)= 0Hence, u - proj y u is orthogonal to proj y u.

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The force F1 is equal and opposite to the downward force thus, F1 is equal to 60 N. The force F2 is inclined to 30 ° from leftward force and it is equal to 38.97 N in magnitude.

Force is an external agent acting on a body to deform it or to change its state of motion or rest. Force is a vector quantity and it is characterised by its magnitude and direction.

If two forces acting on a body from the same directions, then the net force will be the sum of these two forces. If they are acting from opposite directions, they will cancel each other in magnitude.

The force F1 is equal and opposite to the force acting downward. Thus its magnitude is 60 N. The force F2 is inclined to 30 ° from horizontal direction.

F2 = 45 cos 30 = 38.9 N.

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Answer:

A

Explanation:

Name me brainliest plzzzzz

Answer:

The answers to this question is:

1. inner ear

2. outer ear

3. middle ear

Explanation:

Hope that helps!

Answer:

inner, outer, middle

Explanation:

Answer: b

Explanation: no cule

Answer:

Central City is the county seat and the most populous municipality of Gilpin County. The city is a historic mining settlement founded in 1859 during the Pike's Peak Gold Rush and came to be known as the "Richest Square Mile on Earth".

Answer: A Comet

Explanation: A comet is a small icy body that when passing close to the sun, heats up in releases gases, producing invisible tail.

Answer:

0.248 m

Explanation:

The point where the electric field magnitude is e/4 is located 4 meters away from the isolated point charge.

The electric field due to a point charge decreases inversely with the square of the distance from the charge. Therefore, if the field magnitude is e at a distance of 2 meters, it will be (1/4)e at a distance of 4 meters. This is because the electric field strength is proportional to the inverse square of the distance from the point charge, so the field strength decreases rapidly with increasing distance.

In summary, the point where the electric field magnitude is e/4 is located 4 meters away from the isolated point charge due to the inverse square relationship between the electric field and distance from the charge.

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A 1987 205 cubic inch V6 Mercruiser 4.3 engine with 575 hours on it does not necessarily have too many hours with potential breakdowns in the near future, as long as it has been well-maintained and shows no major signs of wear and tear.

The number of hours on an engine is just one factor to consider when determining the potential for breakdowns. Other factors such as maintenance history, usage conditions, and overall condition of the engine can also play a role.

With that being said, 575 hours on a 1987 4.3 Mercruiser engine is not necessarily an alarming number, as these engines are known for their durability and longevity. However, it is important to have the engine inspected and properly maintained to ensure it continues to run smoothly. Regular maintenance and inspections can help prevent potential breakdowns and extend the life of the engine.

1. Assess the average lifespan of a Mercruiser 4.3 engine. Generally, these engines can last anywhere from 1,500 to 2,000 hours with proper maintenance.

2. Evaluate the maintenance history of the engine. Regular maintenance, such as oil changes, spark plug replacements, and cooling system checks, can significantly prolong the engine's lifespan.

3. Inspect the engine for signs of wear and tear. Check for corrosion, oil leaks, or any other visible issues that may indicate potential breakdowns.

Considering these factors, a 1987 205 cubic inch V6 Mercruiser 4.3 engine with 575 hours on it does not necessarily have too many hours with potential breakdowns in the near future, as long as it has been well-maintained and shows no major signs of wear and tear.

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Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= \(10+1.50\)

= \(11.5 \ m/s\)

Travelling such distance for 300 m will be:

⇒ \(v = \frac{d}{t} \ sot \ t\)

      \(=\frac{d}{v}\)

On putting the values, we get

      \(=\frac{300}{11.5}\)

      \(=26.08 \ s\)

Throughout the opposite direction, when the boat seems to be travelling then,

= \(10-1.50\)

= \(8.5 \ m/s\)

Travelling such distance for 300 m will be:

⇒ \(v=\frac{v}{t} \ sot \ t\)

      \(=\frac{d}{v}\)

On putting the values, we get

      \(=\frac{300}{8.5}\)

      \(=35.29 \ s\)

hence,

The time taken by the boat will be:

= \(26.08+35.29\)

= \(61.37 \ s\)

Answer:

35/9 m/s

Explanation:

21 km = 21000 m

90 min = 90*60 s = 5400 s

21000/5400 = 35/9 m/s

Answer:

Value of Sin(77 degree) = 0.97437006

Degree / Radian Function

(angle in degree) (angle in radian) Sin Cos Tan Cot Sec Cosec SinH CosH TanH CosecH SecH CotH Arc Sin Arc Cos Arc Tan Arc Cosec Arc Sec Arc Cot

Answer:

x=-0.9

Explanation:

Two potential places on Mars where water may be found are the polar ice caps and underground aquifers. This water is important for future human missions because it can be used for drinking, agriculture, and as a resource for producing rocket fuel.

The polar ice caps on Mars, located near the north and south poles, are believed to contain significant amounts of water ice. These ice caps are composed of a mixture of water ice and frozen carbon dioxide, commonly known as dry ice. The presence of water ice at the poles offers a readily accessible source of water for future missions.

It could be extracted and purified for various uses, including drinking water for astronauts, irrigation for plant growth in controlled environments, and as a raw material for producing oxygen and hydrogen for life support systems and rocket propellant.

Another potential source of water on Mars is underground aquifers. Recent scientific evidence suggests that Mars may have large reservoirs of underground water. These aquifers could be accessed by drilling beneath the planet's surface, providing a reliable and potentially abundant supply of water.

Accessing underground water sources would be crucial for sustaining long-duration human presence on Mars. It could support the establishment of habitats, provide water for agricultural activities, and serve as a vital resource for fuel production, enabling the production of rocket propellant for return missions to Earth.

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For the horizontal component we have:

Vx = Vo * cos α

Replacing we have:

Vx = 24 * cos 45°

Resolving:

Vx = 24 * 0.525...

Vx = 12.607

The horizontal velocity is 12,607 m/s.