If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. find the velocity when t = 1.

The boat takes 7.75 seconds to reach the buoy and its velocity when it reaches the buoy is 0 m/s.

To solve this problem, we can use the equations of motion under constant acceleration. Let's start with part (a).

Given:

Initial velocity, v0 = 31.0 m/s

Final velocity, v = 0 m/s (since the boat reaches the buoy and stops)

Acceleration, a = -4.00 m/s2

We can use the equation:

v = v0 + at

Substituting the given values, we have:

0 = 31.0 - 4.00t

4.00t = 31.0

t = 7.75 s

Therefore, it takes the boat 7.75 seconds to reach the buoy.

Now, let's move on to part (b).

To find the final velocity, we can use the same equation:

v = v0 + at

Substituting the known values:

v = 31.0 - 4.00(7.75)

v = 31.0 - 31.0

v = 0 m/s

Therefore, the velocity of the boat when it reaches the buoy is 0 m/s.

Answer:

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

The minimum thickness of the film required assuming that the wavelength of the light in air is 500 nanometers is; 83.9 nm

We are given;

Index of refraction of polystyrene; η_p = 1.49

Index of refraction of Fabulite; η_f = 2.409

Wavelength of the light in air; λ = 500 nm

In this question, it will be discovered that light will first of all be reflected two times, first from the upper layer of coating and then from the interface between the coating and the polystyrene material.

The additional path travelled by the light in coating is 2t while the path difference is 2tη_p.

Thus, minimum thickness of film required will be gotten from the formula used in destructive interference which is;

2tη_p = λ/2

t = λ/2 * 1/2η_p

t = λ/(4η_p)

t = 500/(4 * 1.49)

t = 83.9 nm

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The force the ground exerts on him is 650 N.

The push or pull on an object with mass that causes it to change its velocity.

According to law of newton

"Every action has an equal and opposite reaction."

This means that the ground is exerting the same amount of force as the man, 650 N.

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To find the initial speed of the car when the brakes were first applied, you can use the formula: v^2 = u^2 + 2as. Plugging in the given values, we find that the car was traveling at approximately 28.3 ft/s (or 19.3 mph) when the brakes were first applied.

To find the initial speed of the car when the brakes were first applied, you can use the formula:

v^2 = u^2 + 2as

Plugging in the values into the formula, we get:

0 = u^2 + 2(8)(100)

Solving for u, we find that the car was traveling at approximately 28.3 ft/s (or 19.3 mph) when the brakes were first applied.

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I believe that L is in feet while t is in seconds. To find for the hang time, all we have to do is to plug in the value of L in feet in the equation.

4 inches is equivalent to 4/12 feet or 0.33 feet, therefore the total L is:

L = 4.33 ft

Using the equation to find for t:

L = 4 t^2

4.33 = 4 t^2

t = 1.04 s

To calculate the hang time for a basketball player with a vertical leap of 4 feet 4 inches, solve the equation l = 4[tex]t^2[/tex] to find that the hang time is approximately 1.041 seconds.

The vertical leap l is related to the hang time t by the equation l = 4[tex]t^2[/tex]. To find the hang time for a basketball player with a vertical leap of 4 feet 4 inches (which is 4.333 feet), we plug this value into the equation and solve for t:

4.333 = 4[tex]t^2[/tex]

[tex]t^2[/tex] = 4.333 / 4

[tex]t^2[/tex] = 1.08325

t = [tex]\sqrt{(1.08325)[/tex]

t ≈ 1.041

Hence, the hang time for this leap is approximately 1.041 seconds.

The acceleration of the bike that speeds up from 0m/s to 9m/s in 3 seconds is 3m/s². Details about acceleration can be found below.

The acceleration of a moving body can be calculated by dividing the change in velocity of the body by the time taken.

According to this question, a bike speeds up from 0m/s to 9m/s in 3 seconds. The acceleration can be calculated as follows:

a = (9m/s - 0m/s)/3 seconds

a = 9m/s ÷ 3s

a = 3m/s²

Therefore, the acceleration of the bike that speeds up from 0m/s to 9m/s in 3 seconds is 3m/s².

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To find the momentum of the electron after collision with a photon, the momentum of the photon must be calculated first using Planck's constant and the wavelength of the photon. The kinetic energy of the electron depends on the specific details of the conservation of energy and momentum in the collision, which are not provided.

The question involves calculating the momentum of the electron and the kinetic energy of the electron after it collides with a photon. This scenario is explored in the context of compton scattering, which is a common topic in high school and college physics classes.

To find the momentum of the electron, we use the relationship for the momentum (p) of a photon, which is given by p = h/λ, where h is Planck's constant (6.63 × 10-34 m² kg/s) and λ is the wavelength of the photon. For an 0.828-nm photon, this yields a momentum that the electron will take after the collision.

However, without additional information on the scattering angle or final energy/wavelength of the photon, we cannot accurately determine the momentum or the kinetic energy of the electron, as these values depend on the specifics of the conservation of energy and momentum in the collision process.

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This problem is about the physics of rotation and equilibrium. To open the trap-door, an upward force that counteracts the door's weight is necessary. The total force exerted on the door equals the applied force minus the net upward force. If the force is applied farther from the door, More force is required due to the increased torque.

The net upward force to open the door is the force that counteracts the gravitational force acting on the door, which can be calculated using the door’s mass and the acceleration due to gravity. The total force exerted on the door by the side is equal to the total force applied on the door minus this net upward force.

In case (a), where the force is applied at the center of the door, the force required to start opening is equal to the weight of the door distributed over the two sides. In case (b), where the force is applied at the center of the edge opposite the door, the force required is larger because the distance from the door to the point where the force is applied is greater, increasing the torque.

In the case of the ladder problem, an analysis of the torques and the net forces, considering the ladder’s weight and the friction with the floor, would be necessary.

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In topics related to power, work or energy, the definition of efficiency is described as:

Efficiency = useful intended output / total input 

Therefore with this definition, we restate that:

Efficiency is energy transferred to light divided by the electrical energy consumed (input).

The potential difference vab between points a and b when the switch s is open can be calculated using the formula VAB = Ed where E is the electric field intensity and d is the distance between the points.

The potential difference vab between points a and b when the switch s is open can be calculated using the formula VAB = Ed, where E is the electric field intensity and d is the distance between the points. The potential difference is measured in volts (V).

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The correct answer is[tex]\(\boxed{\omega \approx 479.0 \text{ rad/s}}\)[/tex].

To determine the angular velocity of the head when[tex]\(\theta = 30^\circ\),[/tex] we need to integrate the angular acceleration with respect to time. However, since angular acceleration is given as a function of [tex]\(\theta\)[/tex], we will use the relationship between angular acceleration [tex](\(\alpha\))[/tex], angular velocity[tex](\(\omega\))[/tex], and angle [tex](\(\theta\))[/tex] to find the angular velocity as a function of [tex]\(\theta\).[/tex]

The relationship between angular acceleration, angular velocity, and angle is given by:

[tex]\[\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \omega \frac{d\omega}{d\theta}\][/tex]

Since[tex]\(\alpha\)[/tex]is a function of [tex]\(\theta\)[/tex], we can write:

[tex]\[\omega \frac{d\omega}{d\theta} = 700\cos\theta + 70\sin\theta\][/tex]

To find [tex]\(\omega\),[/tex] we separate variables and integrate:

[tex]\[\int \omega \, d\omega = \int (700\cos\theta + 70\sin\theta) \, d\theta\][/tex]

Integrating both sides, we get:

[tex]\[\frac{\omega^2}{2} = 700\sin\theta - 70\cos\theta + C\][/tex]

where [tex]\(C\)[/tex] is the constant of integration. To find [tex]\(C\)[/tex], we use the initial condition that the head is initially at rest, which means[tex]\(\omega = 0\)[/tex] when [tex]\(\theta = 0\)[/tex]. Plugging these values into the equation, we find:[tex]\[0 = 700\sin(0) - 70\cos(0) + C\]\[C = 70\][/tex]

Now we have the constant[tex]\(C\)[/tex], and we can find the angular velocity when [tex]\(\theta = \frac{\pi}{6}\) radians (or \(30^\circ\)):[/tex][tex]\[\frac{\omega^2}{2} = 700\sin\left(\frac{\pi}{6}\right) - 70\cos\left(\frac{\pi}{6}\right) + 70\] \[\frac{\omega^2}{2} = 700\left(\frac{1}{2}\right) - 70\left(\frac{\sqrt{3}}{2}\right) + 70\] \[\frac{\omega^2}{2} = 350 - 35\sqrt{3} + 70\] \[\frac{\omega^2}{2} = 420 - 35\sqrt{3}\][/tex]

Multiplying both sides by 2 to solve for[tex]\(\omega^2\)[/tex]:

[tex]\[\omega^2 = 2(420 - 35\sqrt{3})\]\[\omega^2 = 840 - 70\sqrt{3}\][/tex]

Taking the square root of both sides to find [tex]\(\omega\)[/tex]:

[tex]\[\omega = \sqrt{840 - 70\sqrt{3}}\][/tex]

Now we can calculate the numerical value:

[tex]\[\omega \approx \sqrt{840 - 70 \times 1.732}\] \[\omega \approx \sqrt{840 - 121.24}\] \[\omega \approx \sqrt{718.76}\] \[\omega \approx 479.0 \text{ rad/s}\][/tex]

Therefore, the angular velocity of the head[tex]when \(\theta = 30^\circ\) is approximately \(\boxed{479.0 \text{ rad/s}}\)[/tex]

A concave mirror is curved inward in the middle, more like a cave. Because the mirror is curved inward, the angle of the light surface can be focused similar to that of the camera. They can form real images that are projected out in front of the mirror at the place where light focuses. When the object is located at the center of the curvature the image formed will also be at the curvature. The image will be inverted and the magnification value is equal to 1 which will become a real image because the ray of light converges at the location of the formed image.

Answer:

The object must be farther from the mirror than the focal point

Explanation:

gradpoint lesson

The components of Earth's life support system are air, water, soil, and living organisms.  

The life support system is necessary for living organisms. It elements are necessary to promote living organisms and give sustainable life to living organisms.

It is a system that provides all the substances which is essential for the living. It is the combination of equipment that allows the survival in environment.

Life support systems offer food, oxygen, water, and the disposition of carbon and wastes. It includes the atmosphere(air), hydrosphere (water), biosphere (living organisms), and geosphere (soil).

The atmosphere offers air (oxygen) for living organisms to breathe. Hydrosphere offers water for living organisms and the geosphere offers shelter for living organisms.

Hence the components of Earth's life support system are air, water, soil, and oxygen.

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A sound wave is a pressure wave that results from the vibration of the particles o the medium from the source. The motion of the particles in the medium is parallel to the direction of the energy transport. The type of wave formed by a sound wave is the longitudinal wave. A longitudinal wave is characterized by rarefactions. A longitudinal wave is a wave motion wherein the particles in the wave medium are displaced parallel to transport. When motion is detected from the source, the particle next to it vibrates from its rest position and a progressive change in phase vibration is observed at each particle within that wave. The result is that the energy is transported from one region to the other. These combined motions result in the movement of alternating regions of rarefaction in the direction of propagation.      

The electrical potential midway between two point charges is determined by adding the potentials due to each charge, calculated using the formula V = kq/r, where V is potential, k is Coulomb's constant, q is charge, and r is the distance to the point.

The question asks about the electrical potential at the point midway between two point charges of values +3.4 and +6.6 μC (microcoulombs) separated by 0.10 m. To solve this, we can use the formula for electric potential due to a point charge, V = kq/r, where V is the electric potential, k is Coulomb's constant (approximately 8.987 × 10^9 Nm^2/C^2), q is the charge in coulombs, and r is the distance from the charge to the point in question. At the midpoint between the two charges, the potentials due to each charge add up algebraically because they are both positive charges.

The distance from each charge to the midpoint is 0.05 m. Thus, the total potential at the midpoint is V_total = V_1 + V_2 = (kq_1/r_1) + (kq_2/r_2) = (8.987 x 10^9 Nm^2/C^2)(3.4 x 10^-6 C/0.05 m + 6.6 x 10^-6 C/0.05 m). Simplifying, the electric potential at the midpoint is calculated to be the sum of the individual potentials from each charge.

To answer this problem, we will use the formula for gravitational force:

F = G m1 m2 / r^2

Where,

G = gravitational constant (6.67 * 10^-11)

m = mass of a body

r = distance between two bodies

To have a net gravitational force equal to zero acting on body of mass 1 kg, then the forces of 1 kg and 10 kg bodies and 1 kg and 5 kg bodies should be equal.

F1 (1 kg and 10 kg) = F2 (1 kg and 5 kg)

Therefore,

(1 kg) (10 kg) / (2 m)^2 = (1 kg) (5 kg) / r2^2

Calculating for r2 (distance of another body 5 kg from body of mass 1 kg):

r2^2 = 2

r2 = 1.41 m       (ANSWER)

Therefore another body of mass 5 kg should be placed 1.41 m from body of mass 1 kg to have a net gravitational force equal to zero.

Answer:

I = 0.667 A

Explanation:

A capacitor behaves like a short circuit at t = 0 immediately after the switch is closed, therefore we can discard it from the equation, the resistance is given in watts or power units equivalent to current by voltage and the ammeter behave like a short circuit by nature.

Considering all of the above, the circuit looks like a battery in series with a resistor, therefore, we can use the power equation to solve the problem as follow:

P = I*V

4W = I*6V

I = 0.667 A  

The term for the surface or material that underlies a two-dimensional work of art is 'substrate'. It plays a critical role in the final appearance of the artwork, and its understanding is key to appreciating a given piece of art.

The surface or material that underlies a two-dimensional work of art is called the substrate. This term refers to the material on which the artwork is created. For example, in a painting, the substrate could be the canvas. Similarly, for a drawing, the substrate might be a piece of paper. The substrate plays a vital role in the final appearance of the artwork.

One can relate the concept of substrate in art to Aristotle's explanation of substance as a composite of matter and form. In his perspective, the form is the unchanging purpose or idea informing each particular instance, while the substrate (in his terminology, matter or material cause) contributes to the physical properties of the artwork. Therefore, understanding the substrate is an essential aspect of appreciating a given piece of art.

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