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Kryger [21]
3 years ago
11

A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5. What are the dimensions of s

uch a cylinder which has maximum volume
Physics
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 units.

Explanation:

The formula to find the volume of inscribed cylinder is: V  =  (πr^2).y

let 'y' is the height of the cylinder and 'r' be the radius of the cylinder.

Radius of the inscribed cylinder is r = 5-x.

and the value of y is y = (4.5)/5 * x, it is find by the formula of slope because we know the slope of the slanted side of the cone. We go up 4.5 units for every 5 units we move to the right.

Now,

V = π(5-x)^2 × (4.5/5) . x

V = (4.5/5 . π) × (25x + x^3 - 10x^2)

And now to find the maximum volume derivative of V with respect to 'x' will be equal to zero like: dV/dx = 0

dV/dx = (4.5/5 . π) × (25 + 3x^2 - 20x)

(4.5/5 . π) × (25 + 3x^2 - 20x) = 0

25 + 3x^2 - 20x = 0

Write the equation in sequence and then use the factorization to find x

3x^2 - 20x + 25 = 0

3x^2 - 15x - 5x + 25 = 0

3x(x-5) -5(x-5) = 0

(x-5) (3x-5) = 0

x = 5/3 and x = 5

So, we have two values of 'x'. But the value 5/3 gives us the point of maximum by graphing V with 'x'. Because when the value is between 0 to 5 then the left edge of the inscribed cylinder stays properly within the cone.

Hence, at x = 5/3,

y = (4.5)/5 * x =  (4.5)/5 * 5/3

y = 1.5 ≈ 2.

and r = 5 - x = 5 - 5/3 = 10/3.

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 units.

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
NISA [10]

Answer:

Explanation:

We shall consider a Gaussian surface inside the insulation in the form of curved wall of a cylinder having radius equal to 3mm and unit length , length being parallel to the axis of wire .

Charge inside the cylinder = 250 x 10⁻⁹ C .

Let E be electric field at the curved surface , perpendicular to surface .

Total electric flux coming out of curved surface

= 2π r x 1 x E

= 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's theorem , total flux coming out

= charge inside / ε ( 250 x 10⁻⁹C  charge will lie inside cylinder )

= 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²    (  ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹² )

= 11.3 x 10³ weber .

so ,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

= .599 x 10⁶ N /C .

4 0
4 years ago
HELP When the forces are applied in the same direction, how do you determine net force?
aleksklad [387]

Answer:

Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

5 0
3 years ago
You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
3 years ago
A box of books weighing 325N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted
iragen [17]

Answer:The coefficient of kinetic friction is 0.61

Explanation:A state of constant velocity connotes an acceleration equal to zero {this is evident from Acceleration=(final velocity-initial velocity)/time}

Consequently,a balanced forces system is attained

Force of pulling

F(x)=F×cosα=425N×0.817=347N

is equivalent to the resistance force.

F=U(k)×(Fw+Fsinα)=U(k)×(325N+425N×0,576)=U(k)×570N

Note; U(k) is the coefficient of kinetic friction

Equating the two force components,

The coefficient of static friction ×570N= 347N

The coefficient of static friction = 347N/570N

The coefficient of static friction =0.61

Please note that the coefficient of static friction is dimensionless .

This is as a result of the same unit of forces whose division resulted in the determination of the coefficient of kinetic friction.

5 0
3 years ago
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe
gladu [14]

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

4 0
4 years ago
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