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alex41 [277]
2 years ago
5

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

n a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.6 rad/s experiences a centripetal acceleration of 3.3 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of 2.6 rad/s, the trainee experiences a total acceleration equal to 4.8 times the acceleration due to gravity.
(a) How long is the arm of the centrifuge?
(b) What is the angular acceleration in the second training exercise?
Physics
1 answer:
SCORPION-xisa [38]2 years ago
7 0

Answer:

4.7889 m

7.14037 rad/s²

Explanation:

r = Radius

\omega = Angular velocity = 2.6 rad/s

Centripetal acceleration = 3.3g = 3.3\times 9.81=32.373\ m/s^2

\alpha=r\omega^2=3.3g\\\Rightarrow r=\frac{3.3\times 9.81}{2.6^2}\\\Rightarrow r=4.7889\ m

Length is the arm of the centrifuge is 4.7889 m

Acceleration felt by the trainee = 4.8g = 4.8\times 9.81=47.088\ m/s^2

Angular acceleration

\alpha r=\sqrt{47.088^2-32.373^2}\\\Rightarrow \alpha=\frac{34.19456}{4.7889}\\\Rightarrow \alpha=7.14037\ rad/s^2

Angular acceleration acceleration of the second exercise is 7.14037 rad/s²

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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n
Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

f=\frac{v}{\lambda}

where for the dolphin:

v = 1531 m/s is the wave speed

\lambda=1.50 cm=0.015 m is the wavelength

Substituting into the equation,

f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz

C. 13.6 m

Again, the wavelength is given by:

\lambda=\frac{v}{f}

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz

3 0
2 years ago
The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet
Naily [24]

Answer:

Speed of comet before collision is

v_{2_{i}}=-2.5\times10^{3}\quad km/h

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h

Case is considered as partially inelastic collision, by conservation of momentum

m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h

8 0
2 years ago
A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0
2 years ago
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