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Stells [14]
3 years ago
10

A space station that does not rotate cannot simulate gravity for its occupants.

Physics
1 answer:
ss7ja [257]3 years ago
5 0

<u>Answers:</u>

1. True: If we want to simulate gravity inside a space station, this must rotate about its axis, creating the artificial gravity environment by the centripetal acceleration.  


2. True: The Centrifugal force is a fictitious force that appears when the movement of a body is described in a rotating reference system, or equivalently the apparent force perceived by a non-inertial observer in a rotating reference system.


This means that a non-inertial observer on a rotating platform feels that there is a "force" acting on him, which prevents him from remaining at rest on the platform unless he applies another force directed towards the axis of rotation


3. True: This is possible because the station axis, which is the center of rotation of the system, has the lowest artificial gravity (tending to zero or null gravity). Therefore, the weight of a body in that place will tend to zero, in other words,<u> will be weightless.</u>


4. False: Only having a floor to walk on, is not enough to feel the gravity if the station is not a <u>rotating system</u>.


5.False: The astronaut will experience the centrifugal force, which is a inertial force that will tend to push her back .


This is because the rotation displaces any object or body  from the interior of the station towards its walls, giving the appearance of a gravitational thrust directed towards the exterior. The "push", or centrifugal force is actually a manifestation of the objects inside the ship attempting to move in a straight line due to inertia.

The walls of the station provide the centripetal force required for objects to move in a circle. Therefore, the gravity felt by the objects is a simple reaction of force of the object on the walls reacting with the centripetal force of the wall on the object, according to Newton's third law of motion.



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If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.​
Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

4 0
1 year ago
Which phrase best describes the outer planets?
pav-90 [236]

Answer:

Astronomers have divided the eight planets of our solar system into the inner planets and the outer planets. The 4 inner planets are the closest to the Sun, and the outer planets are the other four – Jupiter, Saturn, Uranus, and Neptune. The outer planets are also called the Jovian planets or gas giants.

Explanation:

6 0
3 years ago
Read 2 more answers
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0
3 years ago
Read 2 more answers
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
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