Drawing is a form of

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

3 years ago

What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a

Fiesta28 [93]

Answer:

a) $E_photon =0.306 eV$

b) $E_photon =0.166 eV$

Explanation:

The energy of the photon (E) for $n^th$ orbit of the hydrogen atom is given as:

$E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

where,

$E_o$ = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

$E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})$

$E_photon =0.306 eV$

b) Now for the transition from n = 5 to n = 6

$E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})$

$E_photon =0.166 eV$

7 0

3 years ago

A magnet has stronger force of attraction at the poles than in the middle why?

seraphim [82]

Answer:

The molecular magnets are arranged in an open chain so that the north pole or the south pole of the molecular magnets lie in the same direction which gives strong force at the poles whereas two opposite poles are arranged at the middle and the force cancel each other. So, poles have more force than the middle portion.

7 0

3 years ago

A small sphere of

mixas84 [53]

Answer:

θ = 39.7º

Explanation:

In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left

Y Axis

$T_{y}$ -W = 0

$T_{y}$ = W

X axis

- $F_{e1}$ - F_{e2} + Tₓ = 0

let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical

sin θ = Tₓ / T

cos θ = T_{y} / t

Tₓ = T sin θ

T_{y} = T cos θ

let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.

F = ∫ E. dA = $q_{int}$ / ε₀

in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate

F = 2E A = q_{int} / ε₀

let's use the concept of surface charge density

σ = q_{int} / A

we substitute

2E A = σ A /ε₀

E = σ / 2ε₀

this is the field created by each plate. The electric force is

$F_{e}$ = q E

for plate 1 with σ₁ = -30 10⁻⁶ C / m²

F_{e1} = q σ₁ /2ε₀

for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²

F_{e2} = q σ₂ /2ε₀

we substitute and write the system of equations

T cos θ = mg

- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0

we introduce t in the second equations

- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0

mg tan θ = q /2ε₀ (σ₁ + σ₂)

θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)

data indicates the mass of 0.25 g = 0.25 10⁻³ kg

give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²

let's calculate

θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))

θ = tan⁻¹ 8.3 10⁻¹)

θ = 39.7º

5 0

3 years ago

Determine the stopping location of the prize wheel. At this moment it is centered on the number 5. It is spinning with an energy

Maurinko [17]

Answer:

Thanks for the 50 points. Answer eill be always 29.0m

3 0

3 years ago

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Drawing is a form of​

Drawing is a form of