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zheka24 [161]
3 years ago
5

A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and at

mospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation
Physics
1 answer:
melomori [17]3 years ago
6 0

Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

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One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6
Angelina_Jolie [31]

Solution :

Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.

It is given that :

Successive harmonic frequencies, f = 52.2 Hz

and f' = 60.9 Hz

Therefore, fundamental frequency, F = f' - f

                                                           F = 60.9 - 52.2

                                                          F = 8.7 Hz

Therefore the string which is fixed at both the ends forms all the harmonics.

6 0
2 years ago
Which of these tools is used to measure temperature?
Feliz [49]

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A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla
Yuri [45]

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

5 0
3 years ago
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