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Misha Larkins [42]
4 years ago
11

The reason an astronaut in an earth satellite feels weightless is thatThe reason an astronaut in an earth satellite feels weight

less is thatthe astronaut is falling.the astronaut is at a point in space where the effects of the moon's gravity and the earth's gravity cancel.the astronaut's acceleration is zero.this is a psychological effect associated with rapid motion.the astronaut is beyond the range of the earth's gravity.
Physics
2 answers:
ozzi4 years ago
8 0

The reason an astronaut in an earth satellite feels weightless is that the astronaut is falling.

Option a

<u>Explanation: </u>

The other options except Option is not applicable since the gravitational force is a long range force, in which the satellite revolves very close to the surface of the Earth where the gravity is felt.The zero weight experienced by the astronaut in a satellite is due to the earth pulling along with satellite. Due to gravitational force of the Earth,the astronaut falls freely .

But why not the satellite comes down due to gravity when its launched in space. The fact is that the satellite is launched with velocity of tangent direction and it is very high. The centripetal force balances the gravity.

madreJ [45]4 years ago
8 0

Answer:

There is less gravity

Explanation:

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Question 6 of 10
madreJ [45]

Answer:

The answer is "Choice B and Choice C".

Explanation:

In this question, the two features, which make electromagnetic waves effective for interaction, which they don't really need a method of travel of data and move information quite fast. The motion among magnetic and electrical forces is generated by electromagnetic radiation. These waves will spread without the need for a source for their movement, and they have high speed and also the electromagnetic waves it has the same frequency as light.

8 0
3 years ago
To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 ✕ 10−10 m,
Elodia [21]
1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
\Delta V =- \int\limits^{d}_{\infty} {E} \, dr
Where E is the electric field generated by the proton, which is
E=k_e  \frac{q}{r^2} 
where k_e=8.99\cdot10^9~Nm^2C^{-2} is the Coulomb constant and q=1.6\cdot10^{-19}~C is the proton charge.
Replacing the electric field formula inside the integral, we obtain
\Delta V =- \int\limits^{d}_{\infty} {k_e  \frac{q}{r^2} } \, dr = k_e  \frac{q}{d}= 27~V

2) Then, we can calculate the work done by the electric field to move the electron (charge q_e=-1.6\cdot10^{-19}C) through this \Delta V. The work is given by
W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J

5 0
3 years ago
Read 2 more answers
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
Vesnalui [34]

The angle of the planet is mathematically given as

dY= 704 degrees

<h3>What angle has planet Y rotated through during this time?</h3>

With Kepler's third rule, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.

Generally, the equation for the period is  mathematically given as

(periodX / periodY)^2 = (radius X / radius Y)^3

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

\sqrt{(pX / pY )^2}= \sqrt{64}

(pX / pY=8

In conclusion, Because it takes 8 times longer to complete one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time period...

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

Read more about Kepler's third rule

brainly.com/question/1086445

#SPJ1

5 0
2 years ago
I need help with the circled one
Whitepunk [10]
Yes the answer is yes 
7 0
3 years ago
Difference between force and acceleration
galina1969 [7]
Force your doing it purposely and acceleration it’s just happening
7 0
3 years ago
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