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Dmitry_Shevchenko [17]
3 years ago
5

Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.40 m . The merry‑go‑round

is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, n 33.0 kg. If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute?
Physics
1 answer:
Irina18 [472]3 years ago
3 0

Answer:

new angular velocity  is 25.20 rpm

Explanation:

Given data

mass = 105 kg

radius =  1.40 m

spinning = 20.0 rpm

masses = 22.0 , 28.0 and 33.0 kg

to find out

new angular velocity

solution

we apply here conservation momentum

L initial  = L final   .........1

we know Ip = 1/2 mr² = 1/2 × 105 ×1.4² = 102.9 kg/m³

and for children initial I(i) = ( 33 + 28 + 22 )=   83 × 1.4² = 162.68 kg/m²

and  I(f) children final = 33+ 22 = 55 × 1.4² = 107.80 kg/m²

so as that from equation 1

Ip  + I(i) × 20 = Ip ×  I(f)  × ω

(102.9  + 162.68 ) × 20 = (102.9 +  107.80 ) × ω

ω =  25.209302

so  new angular velocity  is 25.20 rpm

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xxMikexx [17]

This is another time to look at Newton's 2nd law of motion:

Net Force = (mass) x (acceleration)

If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:

Net Force = (mass) x (zero)

or  Net Force = (zero) .

"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.

So the answer is '<em>yes</em>', and that's why.

6 0
3 years ago
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

7 0
3 years ago
A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u
harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

\displaystyle P=\frac{W}{t}

Since W=F.d

\displaystyle P=\frac{F.d}{t}

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

\displaystyle P=\frac{(75). (4)}{10}=30\ Watt

The second option has a lower power output

7 0
3 years ago
The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc
Ivanshal [37]

1. Frequency: 3.23\cdot 10^{20} Hz

The energy given is the energy per mole of particles:

E=1.29\cdot 10^{11} J/mol

1 mole contains a number of Avogadro of particles, N_A, equal to

N_A=6.022\cdot 10^{23} particles

So, by setting the following proportion, we can calculate the energy of a single photon:

1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J

This is the energy of a single photon; now we can calculate its frequency by using the formula:

E_1 = hf

where

h=6.63\cdot 10^{-34} Js is the Planck's constant

f is the photon frequency

Solving for f, we find

f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz

2. Wavelength: 9.29\cdot 10^{-13} m

The wavelength of the photon is given by the equation:

\lambda=\frac{c}{f}

where

c=3\cdot 10^8 m/s

is the speed of the photon (the speed of light). Substituting,

\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m

6 0
3 years ago
The suns energy reaches the sun by<br> Conduction<br> Convection<br> Radiation <br> Fusion
amid [387]

Answer:

Radiation

Explanation:

The sun energy reaches us by Radiation.

8 0
3 years ago
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