Ask question

Ask question

All categories

2 years ago

11

A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I_1 is the m

oment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and l_2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that l_1 = l_2. l_l > l_2. l_2 > l_1

1 answer:

Vanyuwa [196]2 years ago

7 0

Answer: $I_2>I_1$

Explanation:

Given

Shape of the object is dumbbell shaped

Moment of Inertia w.r.t an axis passing through center and perpendicular to it

$I_1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

$I_1=\frac{mr^2}{2}$

For the axis which passes through one of the masses

$I_2=mr^2$

so $I_2>I_1$

You might be interested in

I need someone to give me 2 answers,

yanalaym [24]

Uh.. what's the question..?

8 0

3 years ago

Help please help me?

Andrews [41]

Answer:

C. Planet D has the greatest mass and will exert a greater gravitational force

6 0

3 years ago

A plastic bottle partially filled with water floats on water, even though the density of the plastic (1.2 g/cc) is more than tha

Semmy [17]

Answer:

True the plastic will float because of the principle of flotation or buoyancy

Explanation:

Buoyancy explains it all!!

Buoyancy is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.

According to the principle of flotation:

<em>"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"</em>

The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

5 0

3 years ago

9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.

Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength $\lambda$ is the distance between to crests; this means we have $4\lambda$ in $40 m$ :

$40 m=4\lambda$

Clearing $\lambda$ :

$\lambda=\frac{40 m}{4}$

$\lambda=10 m$

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- $T$

Then:

$T=\frac{(1 wave)(16 s)}{10 waves}$

$T=1.6 s$ This is the period of the wave

On the other hand, the frequency $f$ of the wave has an inverse relation with its period $T$ :

$f=\frac{1}{T}$

$f=\frac{1}{1.6 s}$

$f=0.625 Hz$ This is the frequency of the wave

c) The speed $v$ of a wave is given by the following equation:

$v=\frac{\lambda}{T}$

$v=\frac{10 m}{1.6 s}$

Finally:

$v=6.25 m/s$

4 0

3 years ago

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

Harlamova29_29 [7]

We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
$n_1sin(\theta_1)=n_2sin(\theta_2)$
Where $\theta_2$ differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, $\Delta x$ is the distance between both rays.
$\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705$
$\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21$
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
$d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m$
For red we have:
$d_{red}=h.tan(\theta_{2red})\approx 0.0134m$
We finally have:
$\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m$

6 0

2 years ago