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Oduvanchick [21]
3 years ago
12

If there is gravity where the International Space Station (ISS) is located above Earth, why doesn’t the space station get pulled

back down to Earth?
Physics
1 answer:
nasty-shy [4]3 years ago
7 0

Answer:

Yes. Gravity is 90% of what it is on the earth's surface. ISS moves at such high speeds that the curve of its fall will always match  the curve of the earth and so will never hit the earth.

Explanation:

The location of the International Space Station (ISS) is at an altitude of about 200 or 250 miles above the earth's surface. Since gravity decreases with altitude, it is less at that height. Gravity is still present there. Since the ISS is in orbit around the earth, the condition is of free fall. Result of this free fall is apparent weightlessness. The high speed of the ISS allows it to orbit around in such a way that the gravitational pull of the earth is equal to the centripetal force experienced by the ISS in its orbit.

The conditions which make it remain in space are similar to how the moon stays in the orbit around the earth. Moon is also in free fall but it never falls down to earth.

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A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to
777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0
2 years ago
A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
Nikitich [7]
There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>E=U_{max}= \frac{1}{2}k A^2
<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
</span>v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }=  1.2 m/s
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
</span>E=U+K
<span>we find the kinetic energy when the glider is at this position:
</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>K= \frac{1}{2}mv^2
v=  \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
where \omega= \sqrt{ \frac{k}{m} } is the angular frequency, and A is the amplitude.
The angular frequency is:
\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
and so the maximum acceleration is
a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span>a(t)=\omega^2 x(t)
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find 
</span>a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
8 0
3 years ago
Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put
Volgvan

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

8 0
3 years ago
Read 2 more answers
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Alexxx [7]

Answer:

b

Explanation:

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attashe74 [19]

Answer:

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Explanation:

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2 years ago
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