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Novosadov [1.4K]
3 years ago
13

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

eights are connected by a thin 0.96-m rod with a mass of 24 kg. Find the moment of inertia of the barbell through the axis perpendicular to the rod at its center, assuming the two weights are small enough to be treated as point masses.
Physics
1 answer:
kondaur [170]3 years ago
6 0

The moment of inertia is 24.8 kg m^2

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

where

m = 50 kg is the mass of the ball

r=\frac{L}{2}=\frac{0.96}{2}=0.48 m is the distance of each ball from the axis of rotation

So we have

I_b = (50)(0.48)^2=11.5 kg m^2

Therefore, the total moment of inertia of the system is

I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

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A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2
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Explanation:

Given that,

Distance 1, r = 100 m

Intensity, I_1=1.24\times 10^{-8}\ W/m^2

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

I=\dfrac{P}{4\pi r^2}.........(1)

Let I' is the intensity at r'. So,

I'=\dfrac{P}{4\pi r'^2}............(2)

From equation (1) and (2) :

I'=\dfrac{Ir}{r'^2}

I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}

I'=1.98\times 10^{-9}\ W/m^2

Intensity level is given by :

dB=10\ log(\dfrac{I'}{I_o}), I_o=10^{-12}\ W/m^2

dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})

dB = 32.96 dB

Hence, this is the required solution.

7 0
3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

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Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.
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Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

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Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

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Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

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Statement B : 2.567km to three significant figures.

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2.567km to 2 S.F = 2.6km

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2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

5 0
3 years ago
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