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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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A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
3 0
3 years ago
What would a force diagram for something WHILE it is being thrown DOWNWARDS look like? <br><br> Ty
Dovator [93]

Answer:

it look the same just to tell you

5 0
3 years ago
An electron in a mercury atom drops
aksik [14]

Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.

<h3>How to calculate the photon energy?</h3>

In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.

Mathematically, the Planck-Einstein equation can be calculated by using this formula:

E = hf

<u>Where:</u>

  • h is Planck constant.
  • f is photon frequency.

In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.

Read more on photons here: brainly.com/question/9655595

#SPJ1

4 0
2 years ago
What is the net charge of a copper atom if it gains 2 electrons?
Alex_Xolod [135]
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
7 0
3 years ago
Ill mark as brainliest.
rewona [7]

Answer: 75V

Explanation:

Given that,

total resistance (Rtotal) = 150Ω

Current (I) = 0.5A

Change in electric potential (V) = ?

Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.

So, apply the formula V = I x Rtotal

V = 0.5A x 150Ω

V = 75V

Thus, the change in electric potential across the circuit is 75 Volts

8 0
3 years ago
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