20. The hospital is....<br>the police station. (with/on/by/between<br>

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

The electromagnetic wave that delivers a cellular phone callto

spin [16.1K]

Answer:

$E=2.41\cdot 10^{-5} J$

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

$I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}$ (1)

c is the speed of light (3*10⁸ m/s)
μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
B₀ is the magnetic field

$I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}$

$I_{average}=2.68\cdot 10^{-6} W/m^{2}$

Now, let's define the relationship between power (P) and average intensity (I).

$I_{average}=\frac{P}{A}$

P is the power
A is the area crossed

So we can calculate the power.

$P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W$

Finally, energy is the product of P times time, so:

$E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J$

I hope it helps you!

5 0

3 years ago

Why isn't direct current used in transformers

Furkat [3]

Answer:

No, it will not and this has a historical importance. The reason is that transformers work via induction of electrical forces by changes in magnetic fields, so the constat fields produced by dc currents won't work at all

Explanation:

4 0

3 years ago

Read 2 more answers

An astronaut goes out for a space-walk at a distance above the earth equal to the radius of the earth. What is her acceleration

Lera25 [3.4K]

Answer: $\frac{g}{4}$

Explanation:

Let m be the mass of Astronaut

M=mass of earth

G=Gravitational constant

R=radius of Earth

Force Exerted by Earth on Astronaut

$F=\frac{GmM}{R^2}$

acceleration due to gravity is $=\frac{F}{m}=g$

$g=\frac{GM}{R^2}$

When it is at $r=2R$

$g'=\frac{GM}{(2R)^2}$

$g'=\frac{GM}{4R^2}=\frac{g}{4}$

7 0

3 years ago

Gravity is greater when there is

Papessa [141]

-- more mass involved

-- less distance between the two objects

4 0

3 years ago

20. The hospital is....the police station. (with/on/by/between​

20. The hospital is....the police station. (with/on/by/between