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3 years ago

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

Physics

1 answer:

ss7ja [257]3 years ago

5 0

Answers:

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, will be weightless.

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

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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One way to force air into an unconscious person's lungs is to squeeze on a balloon appropriately connected to the subject. What

frozen [14]

Answer:

The force that you must exert on the balloon is 1.96 N

Explanation:

Given;

height of water, h = 4.00 cm = 4 x 10⁻² m

effective area, A = 50.0 cm² = 50 x 10⁻⁴ m²

density of water, ρ = 1 x 10³ kg/m³

Gauge pressure of the balloon is calculated as;

P = ρgh

where;

ρ is density of water

g is acceleration due to gravity

h is height of water

P = 1 x 10³ x 9.8 x 4 x 10⁻²

P = 392 N/m²

The force exerted on the balloon is calculated as;

F = PA

where;

P is pressure of the balloon

A is the effective area

F = 392 x 50 x 10⁻⁴

F = 1.96 N

Therefore, the force that you must exert on the balloon is 1.96 N

3 0

4 years ago

If we increase the amount of work being done, and all other factors remain the same, the amount of power would

BlackZzzverrR [31]

The amount of power would increase
P=W/t
P is directly proportional to W

4 0

4 years ago

Please help me find b) and d):

maks197457 [2]

Answer:

wow it's so hard

8 0

3 years ago

The armature of a motor is accelerated uniformly from rest to a rotational velocity of 1800 rev/min in 10 seconds. the rotationa

Liono4ka [1.6K]

The initial angular velocity of the motor is zero, while the final angular velocity is 1800 rev/min. Let's convert this into rad/s, keeping in mind that
$1 rev = 2 \pi rad$
$1 min = 60 s$
we have
$\omega_f = 1800 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min}=188.4 rad/s$

And so now we can calculate the angular acceleration of the motor:
$\alpha = \frac{\omega_f - \omega_i }{t} = \frac{188.4 rad/s -0}{10 s}=18.8 rad/s^2$

7 0

4 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

3 years ago