An astronaut is said to be weightless when he/she travels in a satellite. Does it mean that the earth does not attract him/her?

Physics

1 answer:

Snezhnost [94]2 years ago

6 0

Answer:

If a person is close enough to be pulled by gravity then they will, but if they are weightless and in absolute zero gravity then earth is not close enough to attract them. The earth more likely will be attroacting the satalite in this situation so I think it will probably be some gravitational pull.

Explanation:

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A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m

BaLLatris [955]

Answer: a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass

 m = son's mass = M/3

 V = father's initial speed

 v = son's initial speed  (1/2)MV^2 = (1/2)*(1/2)*m v^2

 M*V^2 = (1/2)(M/3)v^2

 V^2/v^2 = 1/4

 V = v/2  Second equation:

 (1/2)M*(V + 1.4)^2 = (1/2)m*v^2

 = (1/2)*(M/3)*(3V)^2

 cancel out the M's and (1/2)'s

 (V + 1.4)^2 = 3V^2

 V^2 + 2.8V + 1.96 = 3V^2

 V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

3 0

3 years ago

Read 2 more answers

A base is a substance that's the chemical opposite of an acid.<br> True or false

anygoal [31]

True

Explanation:

A base is a substance that is often used as the chemical opposite of an acid.

Both behaves in opposite way to one another.

They can be said to complementary or conjugate chemicals.

According to Bronsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor.
The lewis theory states that an acid is an electron pair acceptor while a base is an electron pair donor.
Acids turn blue litmus paper red and bases turns red litmus paper blue.

learn more :

Acid brainly.com/question/11062486

#learnwithbrainly

6 0

3 years ago

A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and

ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

$f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\$