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olya-2409 [2.1K]

3 years ago

15

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex

perience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.50 km

1 answer:

sleet_krkn [62]3 years ago

8 0

The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r

g - v²/r = 0

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s

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Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is abo

hoa [83]

To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

$f = 4.11 *10^{12} Hz$

$A = 1.23 * 10^{-11}m$

The angular velocity of a body can be described as a function of frequency as

$\omega = 2\pi f$

$\omega = 2\pi 4.11 *10^{12}$

$\omega=2.582*10^{13} rad/s$

PART A) The expression for the maximum angular velocity is given by the amplitude so that

$V = A\omega$

$V =( 1.23 * 10^{-11})(2.582*10^{13})$

$V = = 317.586m/s$

PART B) The maximum acceleration on your part would be given by the expression

$a = A \omega^2$

$a =( 1.23 * 10^{-11})(2.582*10^{13})^2$

$a= 8.2*10^{15}m/s^2$

4 0

3 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

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2 years ago

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Gre4nikov [31]

Answer:

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(c) Another spring has a spring constant of 250 N/m.

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What is the international standard for measurement it is a modified version of the metric system

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Answer:

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