What is the volume of a cone with a radius of 3 feet and a height of 6 feet use 3.14 for pie round your answer to the nearest hundredth

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for DAC:

⇒

Now for the BAC:

⇒

Now, differentiating w.r.t x:

For maximum angle, = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 =

After solving the above eqn, we get

x =

The observer should stand at a distance equal to x =

Final answer:

For optimum viewing of a painting in a gallery, an observer should position themselves a distance away from the painting calculated using Pythagoras theorem, forming a right-angled triangle with the painting and the floor. This distance can be expressed as c = √[(h/2 + d)² + (h/2)²], where h is the height of the painting and d is the height from the observer's eye to the bottom of the painting.

Explanation:

In the physics of optics, the viewer should position themselves to where they form a right-angled triangle with the ceiling and the painting leading to the best viewing experience. This is widely known as the 'normal viewing distance'.

Given that the painting has a height h and its lower edge is at a distance d above the observer's eye, the observer should stand a distance away from the wall, which can be calculated using Pythagoras' theorem in right triangles, which states that the square of the hypotenuse (c) is equal to the sum of the square of the other two sides (a and b), i.e., c² = a² + b²

Since the painting height and viewer height forms the right-angle in this case, we have: a = (h/2 + d), and b = h/2. Substituting a and b in Pythagoras equation, we can solve for c which is the required distance: c = √[(h/2 + d)² + (h/2)²]

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Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

.....(2)

Dividing equation (1) and (2), we get :

Hence, the correct option is (d) i.e. " 4F/9"

Final answer:

The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.

Explanation:

The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.

Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.

Answer:

1.5 x 10¹⁸hz

Explanation:

Given parameters:

Wavelength  = 2 x 10⁻¹⁰m

Unknown:

Frequency   = ?

Solution:

To find the frequency, use the expression below;

  V  = f x wavelength

V is the speed of light = 3 x 10⁸m/s

f is the frequency

Now;

 Insert the parameters

     3 x 10⁸  = 2 x 10⁻¹⁰ x frequency

   Wavelength =    = 1.5 x 10¹⁸hz

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

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Answer:

1)   I_ pendulum = 2.3159 kg m², 2)  I_pendulum = 24.683 kg m²

Explanation:

In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum

Sphere

They indicate the density of the sphere roh = 37800 kg / m³ and its radius

r = 5 cm = 0.05 m

we use the definition of density

               ρ = M / V

               M = ρ V

the volume of a sphere is

                V = 4/3 π r³

we substitute

              M = ρ 4/3 π r³

we calculate

              M = 37800  4/3 π 0.05³

              M = 19,792 kg

Bar

the density is ρ = 32800 kg / m³ and its dimensions are 1 m,

0.8 cm = 0.0008 m and 4cm = 0.04 m

The volume of the bar is

               V = l w h

              m = ρ l w h

we calculate

              m = 32800 (1   0.008   0.04)

              m = 10.496 kg

Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is

               r_cm = 1 / M (M r_sphere + m r_bar)

               M = 19,792 + 10,496 = 30,288 kg

               r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))

               r_cm = 1 / 30,288 (5,248 + 20,7816)

               r_cm = 0.859 m

This is the center of mass of the pendulum.

1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:

Sphere I = 2/5 M r2

Bar I = 1/12 m L2

parallel axes theorem

                  I = I_cm + m D²

where m is the mass of the body and D is the distance from the body to the axis of rotation

Sphere

      m = 19,792 ka

the distance D is

                D = 1.05 -0.85

                D = 0.2 m

we calculate

               I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168

               I_sphere = 0.811472 kg m²

Bar

m = 10.496 kg

distance D

             D = 0.85 - 0.5

             D = 0.35 m

              I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576

              I_bar = 1.5044 kg m²

The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts

              I_pendulum = I_sphere + I_bar

              I_pendulum = 0.811472 +1.5044

              I_ pendulum = 2.3159 kg m²

this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m

2) The moment is requested with respect to the pivot point at r = 0 m

Sphere

        D = 1.05 m

         I_sphere = 2/5 M r2 + M D2

        I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82

        I_sphere = 21.84 kg m²

Bar

D = 0.5 m

      I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624

      I_bar = 2,84266 kg m 2

The pendulum moment of inertia is

       I_pendulum = 21.84 +2.843

       I_pendulum = 24.683 kg m²

This moment of inertia is about the turning point at r = 0 m

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

Final answer:

The method of braking that involves applying smooth, steady pressure to the brake to bring the vehicle to a smooth stop is called controlled braking. It helps prevent skidding and provides the driver with more control over the vehicle.

Explanation:

The form of braking used to bring a vehicle to a smooth stop by applying smooth, steady pressure to the brake is known as controlled braking. This method of braking involves applying consistent, even pressure to the brake pedal, which allows the car to slow down gently and gradually. It helps prevent uncontrolled skidding and provides the driver with more control over the vehicle's direction and speed during the stop. Unlike other methods like trail braking, threshold braking, or coasting, controlled braking is typically the safest and most effective method for daily driving conditions.

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