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Vsevolod [243]
2 years ago
13

A slender rod of length 80.0 cm and 0.600 kg has its center of gravity at its geometrical center. But its density is not uniform

; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia Icm of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate Icm. Therefore, you make the following measurements: You suspend the rod about an axis that is a distance d (measured in meters) above the center of the rod and measure the period T (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of d. When you plot your data as T2−4π2d/g versus 1/d, the data lie close to a straight line that has slope 0.470 m⋅s2. What is the value of Icm for the rod?
Physics
1 answer:
Vlad [161]2 years ago
7 0

Answer:

Icm = 0.0701 kgm^2

Explanation:

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Which of the following is true concerning hurricanes?
densk [106]

Answer:

B.

Explanation:

B. is correct because every time usually when a hurricane hits it causes flooding causing multiple homes to be ruined. It is a known fact that usually hurricanes start close or in along with tornadoes, the water, meaning once they hit land they have some water to flood that land with.

C. is absolutely false hurricanes have VERY strong wind gusts.

D. is absolutely wrong they do alter landscapes by ripping trees and plants and houses out of the ground making the landscapes look different.

A. is wrong they tend to deposit and remove sediment evenly.

<em><u>~ LadyBrain</u></em>

6 0
3 years ago
Read 2 more answers
(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s

The energy in ev/s

\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
3 years ago
A survey in a magazine that asks readers to return the survey would result in:
Angelina_Jolie [31]
D - Most likely. Those who read the magazine can choose whether or not to return the survey.
4 0
3 years ago
At what speed must a 0.6 kg stone be thrown in order that it has a relativistic mass of 0.76 kg? (c = 3.00 x 10^8 m/s) (A) 0.39
exis [7]

Explanation:

It is given that,

Relativistic Mass of the stone, m₀ = 0.6

Mass, m=0.76\ kg

Relativistic mass is given by :

m=\dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}.........(1)

Where

c is the speed of light

On rearranging equation (1) we get :

v^2=c^2(1-(\dfrac{m_o}{m})^2)

v=c\sqrt{ (1-(\dfrac{m_o}{m})^2)}

v=c\sqrt{ (1-(\dfrac{0.6}{0.76})^2)}

v = 0.61378 c

or

v = 0.6138 c

So, the correct option is (c). Hence, this is the required solution.

8 0
3 years ago
Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diame
VashaNatasha [74]

Answer:

a

   V =  5.30 *10^{-2} \ m^3

b

   v_1 = 0.3127 \ m/s

Explanation:

From the question we are told that

   The number of identical drippers is  n =  60

   The diameter of each hole in each dripper is  d =  1.0 \  mm =  0.001 \  m  

   The diameter of the main pipe is  d_m  =  2.5 \  cm  =  0.025 \  m

    The speed at which the water is flowing is  v  =  3.00 \  cm/s =  0.03 \  m/s

Generally the amount of water used in  one hour = 3600 seconds is mathematically represented as

          V =  A *  v  *  3600

Here A is the area of the main pipe with value

         A =  \pi  * \frac{d^2}{4}

=>       A = 3.142   * \frac{0.025^2}{4}

=>        A =  0.0004909 \  m^2

So  

=>   V =  0.0004909  *  0.03  *  3600

=>  V =  5.30 *10^{-2} \ m^3

Generally the area of the drippers is mathematically represented as

       A_1=  n  * \pi \frac{d^2}{4}

=>    A_1 =  60   * 3.142 *  \frac{0.001 ^2}{4}

=>    A_1 =  4.713 *10^{-5} \  m^2

Generally from continuity equation we have that  

         Av =  A_1 v_1

=>      0.0004909 *  0.03 =  4.713 *10^{-5} *  v_1

=>   v_1 = 0.3127 \ m/s

   

     

3 0
3 years ago
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