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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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Which option tells the forces that influence the movement of earths plates
Sergio039 [100]

Answer:

Gravity

Explanation:

8 0
3 years ago
Your fingernail grows at a rate of about 0.1 mm per day. How many inches do your fingers grow per year?
forsale [732]

Answer:

1.44

Explanation:

0.1mm x 365 = 3.65mm

3.65mm into inches is 1.44

6 0
3 years ago
Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.
Natalka [10]

Explanation:

Things you need to know:

Accuracy refers to the maximum error encountered when a particular observation is made.

Error in measurement is normally one-half the magnitude of the smallest scale reading.

Because one has to align one end of the rule or device to the starting point of the measurement, the appropriate error is thus twice that of the smallest scale reading.

Error is usually expressed in at most 1 or 2 significant figures.

Tape

Equipment: It is made up of a long flexible tape and can measure objects or places up to 10 – 50 m in length. It has markings similar to that of the rigid rule. The smallest marking could be as small as 0.1 cm or could be as large as 0.5 cm or even 1 cm.

How to use: The zero-mark of the measuring tape is first aligned flat to one end of the object and the tape is stretched taut to the other end, the reading is taken where the other end of the object meets the tape.

Ruler

Equipment: It is made up of a long rigid piece of wood or steel and can measure objects up to 100 cm in length. The smallest marking is usually 0.1 cm.

How to use: The zero-end of the rule is first aligned flat with one end of the object and the reading is taken where the other end of the object meets the rule.

Vernier Caliper

Equipment: It is made up of a main scale and a vernier scale and can usually measure objects up to 15 cm in length. The smallest marking is usually 0.1 cm on the main scale.

It has:

a pair of external jaws to measure external diameters

a pair of internal jaws to measure internal diameters

a long rod to measure depths

How to use: The jaws are first closed to find any zero errors. The jaws are then opened to fit the object firmly and the reading is then taken.

Micrometer Screw Gauge

Equipment: It is made up of a main scale and a thimble scale and can measure objects up to 5 cm in length. The smallest marking is usually 1 mm on the main scale (sleeve) and 0.01 mm on the thimble scale (thimble). The thimble has a total of 50 markings representing 0.50 mm.

It has:

an anvil and a spindle to hold the object

a ratchet on the thimble for accurate tightening (prevent over-tightening)

How to use: The spindle is first closed on the anvil to find any zero errors ( use the ratchet for careful tightening). The spindle is then opened to fit the object firmly (use the ratchet for careful tightening) and the reading is then taken.

5 0
3 years ago
Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours
MariettaO [177]
With that information you can only suppose a uniformly accelerated motion.  This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2)  * t = Vo +  2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration. 
5 0
3 years ago
Neutral hydrogen can be modeled as a positive point charge +1.6×10^−19C surrounded by a distribution of negative charge with vol
Archy [21]

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

4 0
3 years ago
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