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3 years ago

14

Determine la inercia rotacional de una varilla de 4 m de largo y 2 Kg de mesa si su eje de rotación esta situado a la mitad de l

a varilla.

1 answer:

Maslowich3 years ago

8 0

Answer:

I = 2.667 kg m²

Explanation:

The moment of inertia of a body can be calculated by the expression

I = ∫ L² dm

For high symmetry bodies the expressions of the moment of inertia are tabulated, for a rod with its axis of rotation at its midpoint it is

I = $\frac{1}{12}$ m L²

let's calculate

I = $\frac{1}{12}$ 2 4²

I = 2.667 kg m²

You might be interested in

A wave with a frequency of 32 Hz has a wavelength of 9 meters. At what speed will this

jarptica [38.1K]

Answer:

228 m/s

Explanation:

f = 32 Hz

λ = 9 m

v = ?

v = f λ

v = (32 Hz) (9 m)

v = 288 m/s

6 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

6 0

2 years ago

Cfare eshte energjia

Whitepunk [10]

Answer:

what

Explanation:

5 0

3 years ago

Read 2 more answers

A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,

Tamiku [17]

Answer:

Explanation:

Let 100 m/s be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t + .5 x 9.8 x t²

4.9 t² - 66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t + .5 x 9.8 x t²

4.9 t² - 76.60 t - 2.1 = 0

t = 15.659 s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49 m

So horizontal range is increased , if angle of projection is increased .

8 0

3 years ago